Understanding the quotient ring nomenclature
Tim BandTech.com
Keywords: ideal, ring, quotient ring.
Question One: Is the quotient ring a construction of a space or is it
a model of a previously constructed space? If this question is poorly
worded please feel free to correct the context. Preferably you would
not mimic the definitional statements but provide a more tangible
motivation.
Question Two: If we discuss the ring R[x] is this a reference to the
real numbers or is this an entirely abstract R[x]?
Question Three: Is the usage of the word quotient actually a division
operation?
- Tim
Andrew Tomazos
> I am seeking help understanding the quotient ring nomenclature.
Toni...@yahoo.com
> I am seeking help understanding the quotient ring nomenclature.
> Keywords: ideal, ring, quotient ring.
>
> Question One: Is the quotient ring a construction of a space or is it
> a model of a previously constructed space? If this question is poorly
> worded please feel free to correct the context. Preferably you would
> not mimic the definitional statements but provide a more tangible
> motivation.
No, a quotient ring is not "a space" in general, but rather an
abstract algebraic construction from a given ring and a given ideal in
it
>
> Question Two: If we discuss the ring R[x] is this a reference to the
> real numbers or is this an entirely abstract R[x]?
>
The second option, unless you're specifically working only with real
numbers.
> Question Three: Is the usage of the word quotient actually a division
> operation?
>
Not division as understood in school: 3/4 or 25:5, but it can be seen
as an abstract kind of division of abstract algebraic structures. In
fact, quotient group, or quotient ring, or quotient vector space, etc.
is an algebraic structure constructed on the set of equivalence
classes in some group, ring, vector space, etc. under a very specific
equivalence relation there.
This already seems too long and involved to be explained here, so you
better grab some algebra book and read it there.
Regards
Tonio
> - Tim
Tim BandTech.com
> On Jun 3, 5:42 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > I am seeking help understanding the quotient ring nomenclature.
> > Keywords: ideal, ring, quotient ring.
>
> > Question One: Is the quotient ring a construction of a space or is it
> > a model of a previously constructed space? If this question is poorly
> > worded please feel free to correct the context. Preferably you would
> > not mimic the definitional statements but provide a more tangible
> > motivation.
>
> No, a quotient ring is not "a space" in general, but rather an
> abstract algebraic construction from a given ring and a given ideal in
> it
question:
Is the quotient ring a construction on an algebra or is it a model of
a previously constructed algebra?
Arturo Magidin
> On Jun 3, 12:05 pm, Tonic...@yahoo.com wrote:> On Jun 3, 5:42 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > I am seeking help understanding the quotient ring nomenclature.
> > > Keywords: ideal, ring, quotient ring.
>
> > > Question One: Is the quotient ring a construction of a space or is it
> > > a model of a previously constructed space? If this question is poorly
> > > worded please feel free to correct the context. Preferably you would
> > > not mimic the definitional statements but provide a more tangible
> > > motivation.
>
> > No, a quotient ring is not "a space" in general, but rather an
> > abstract algebraic construction from a given ring and a given ideal in
> > it
>
> You've corrected the context but ignored the question. To rephrase the
> question:
>
> Is the quotient ring a construction on an algebra or is it a model of
> a previously constructed algebra?
A ring is an ordered tuple, (R,+,-,0,*), where R is a set, + is a
binary operation on the set, - is a unary operation on the set, 0 is a
zeroary operation on the set, and * is a binary operation, which
satisfy certain identities (the "ring axioms).
Given a congruence ~ on (R,+,-,0,*) (this is an equivalence relation
on the set R which satisfies certain properties related to the four
operations on R), the "quotient ring of R modulo ~" is a ->new<- ring.
This new ring is defined in terms of both (R,+,-,0,*) and ~, but it is
something different: it is a tuple, (S,&,!,O,x), where the set S
consists of the equivalence classes of R modulo ~, & is a binary
operation on S defined by [a]&[b] = [a+b], where [a] denotes the
equivalence class of the element a of R modulo ~; ~ is a unary
operation on S defined by ![a] = [-a]; O is a zeroary operation on S
whose value is [0]; and x is a binary operation on S defined by [a]x[b]
=[a*b]. One needs to show these operations are well defined and that
they give you a ring, which is a consequence of those "certain
properties related to the four operations on R" that the equivalence
relation ~ must satisfy in order to be a congruence.
As you can see, the quotient ring is *constructed* from a given ring
and a given congruence on that ring. Whether or not it is a "model of
a previously constructed algebra" will depend on which algebras you
may have constructed before, but it is not, in general, a "model" for
the one you started with (namely, (R,+,-,0,*) ).
--
Arturo Magidin
Tim BandTech.com
Thanks Arturo. There is an example of R[X]/(X^2 + 1) representing the
complex numbers through the usage of polynomials. This example is very
challenging to me. There would be a dimensional increase occuring in
the quotient space, whereas a modulo effect cuts a ring down to a
smaller algebra.
In http://en.wikipedia.org/wiki/Quotient_ring#Examples
"Now consider the ring R[X] of polynomials in the variable X with
real coefficients, and the ideal I = (X^2 + 1) consisting of all
multiples of the polynomial X^2 + 1."
Here they have neither specified what R[X] is nor what X is. They've
merely described polynomials in X having real valued coefficients. Yet
they claim that this is a description of a quotient ring which is
isomorphic to the complex numbers. It does not seem logical that we
are free to choose any ring R[X] since choosing the integers will not
yield the continuum of complex numbers. Is it true that they mean to
be speaking of the real numbers as R[X]? Would this then mean that X
is real?
Will this reasoning simply break down to the standard construction of
the complex numbers at some point with a study of the square roots of
negative numbers and supposing them to exist?
- Tim
KMF
Hope someone can correct/comment on my post. I
am trying to see if this finally filtered down
correctly to my brain:
R stands for real numbers. So the coefficients are
real ( but their names have been changed to protect
the innocent :) ). X is just a variable. R[X] is the
ring of polynomials in one variable, with sum defined
coefficient-wise and product defined in the "standard
way" , i.e., term-by-term multiplication :
:(a_0+a_1x+..+a_nx^n)(b_0+b_1x+...+b_mx^m)=
a_0b_0+ (a_0b_1+a_1b_0)x+.....+a_nb_mx^(n+m)
They've
> merely described polynomials in X having real valued
> coefficients. Yet
> they claim that this is a description of a quotient
> ring which is
> isomorphic to the complex numbers.
The quotient identifies/collapses all polynomials
with the same remainder when dividing by (x^2+1);
i.e., p(x)~q(x) iff. (def.) (p(x)-q(x))=t(x)(x^2+1)
where p(x),q(x),t(x) are polys. in one variable
with real coefficients.
So the elements in R[X]/(x^2+1) are the classes
of possible remainders when dividing a real-valued
poly. by x^2+1. By the Euclidean algorithm,(R is a
field, so R[X] is a Euclidean ring) the
possible remainders when dividing by x^2+1 are of
the form ax+b (a,b in R), i.e., all polys. in R[X]
with degree less than X^2+1. This is how you get the
equivalence with the complexes:
ax+b <-->ai+b
And notice that when a=0 , you get the real numbers
back.
It does not seem
> logical that we
> are free to choose any ring R[X] since choosing the
> integers will not
> yield the continuum of complex numbers. Is it true
> that they mean to
> be speaking of the real numbers as R[X]? Would this
> then mean that X
> is real?
>
> Will this reasoning simply break down to the standard
> construction of the complex numbers at some point with a study of the square roots of
> negative numbers and supposing them to exist?
>
You sort of construct the root of negative numbers:
When you mod out by a polynomial p(x), the class of the
polynomial p(x) is the 0 class, or the 0 element of the
quotient ring ( which is a field in this case).
This implies specifically that:
(x^2+1)=0
So x is a square root of -1 in this sense.
> - Tim
Maybe some
Arturo Magidin
> Thanks Arturo. There is an example of R[X]/(X^2 + 1) representing the
> complex numbers through the usage of polynomials.
If you start with the ring R[X] of polynomials over the real numbers,
then the quotient ring R[X]/(X^2+1) is *isomorphic* to what you
usually think of as the complex numbers (numbers of the form a+bi,
with a and b real numbers, whose addition is defined by (a+bi) + (x
+yi) = (a+x) + (b+y)i, and with multiplication given by (a+bi)*(x+yi)
= (ax-by) + (ay+bx)i ).
The isomorphism is given by the realization that the congruence class
of X in the quotient satisfies [X]^2 = [-1]. Given any element p(x) in
R[X], then by performing long division we can always write p(x)
uniquely as p(x)=q(x)(x^2+1) + r(x), where r(x) is either 0, or a
polynomial of degree at most 1 in R[X]. It then follows that [p(x)] =
[r(x)] in R[X]/(X^2+1).
Then, the isomorphism between R[X]/(X^2+1) and the complex numbers is
realized by the function that takes any equivalence class of
polynomials in R[X], [p(x)], then finds the polynomial r(x) = a+bx
defined above (a and b real numbers), and maps [p(x)] = [r(x)] = [a
+bx] to the complex number a+bi. One has to show this is a well-
defined ring isomorphism, of course, but that is really what it's all
about.
> This example is very
> challenging to me. There would be a dimensional increase occuring in
> the quotient space,
No, there is no dimension at play here *at all*. You are confusing
many notions, (dimension in the geometric/topological sense is not at
play here, and neither is the linear algebra notion). And even then,
while the complex numbers are "2 dimensional" when identified with the
plane, that is, 2 dimensional over the reals, ther are also 1
dimensional when considered over the complex numbers).
> whereas a modulo effect cuts a ring down to a
> smaller algebra.
Not when you are dealing with infinite sets. The number of elements in
R[X] and in R[X]/(X^2+1) is exactly the same, just as the number of
real numbers and of complex numbers is exactly the same. Cardinality
is what is at play here, and for infinite underlying sets, the
quotients may have the same cardinality as the original.
>
> Inhttp://en.wikipedia.org/wiki/Quotient_ring#Examples
> "Now consider the ring R[X] of polynomials in the variable X with
> real coefficients, and the ideal I = (X^2 + 1) consisting of all
> multiples of the polynomial X^2 + 1."
>
> Here they have neither specified what R[X] is nor what X is.
Yes, they have: R[X] is the ring of polynomials in the variable X with
real coefficients. Those words have *precise*, specific meanings. That
means that they have in fact *completely specified* what R[X] and what
X are.
> They've
> merely described polynomials in X having real valued coefficients.
That *is* what they are.
> Yet
> they claim that this is a description of a quotient ring which is
> isomorphic to the complex numbers. It does not seem logical that we
> are free to choose any ring R[X]
You are not choosing "any" ring R[X]. You are taking exatly the ring
of polynomials with real coefficients. There is one and only one (up
to isomorphism) such ring. There is absolutely no liberty of choice
here. What makes you think there is?
Ah; the letter "R". Here, R does not stand for "arbitrary ring", it
stands for "The real numbers". And they say so explicitly.
> since choosing the integers will not
> yield the continuum of complex numbers. Is it true that they mean to
> be speaking of the real numbers as R[X]? Would this then mean that X
> is real?
No, R is the ring of real numbers, X is an indeterminate variable, and
R[X] is the ring of polynomials with coefficients in R. Given any ring
S, and any set T of objects not in S, one defines the ring S[T] of
polynomials in T with coefficients in S in a standard way. That is
what they are doing here.
--
Arturo Magidin
leland....@gmail.com
> "Now consider the ring R[X] of polynomials in the variable X with
> real coefficients, and the ideal I = (X^2 + 1) consisting of all
> multiples of the polynomial X^2 + 1."
>
> Here they have neither specified what R[X] is nor what X is. They've
> merely described polynomials in X having real valued coefficients. Yet
> they claim that this is a description of a quotient ring which is
> isomorphic to the complex numbers. It does not seem logical that we
> are free to choose any ring R[X] since choosing the integers will not
> yield the continuum of complex numbers. Is it true that they mean to
> be speaking of the real numbers as R[X]? Would this then mean that X
> is real?
> Will this reasoning simply break down to the standard construction of
> the complex numbers at some point with a study of the square roots of
> negative numbers and supposing them to exist?
KMF and Arturo have provided some nice answers, but given the nature
of your questions and apparent misunderstandings I fear their answers
may be somewhat opaque to you. What I'm about to give as an answer is
not actually true, but it will help put your intuitions closer to what
is actually going on, and will hopefully at least be more
comprehensible to you than I fear the other excellent and actually
technically accurate replies will be.
First you need to get some sense of what R[X] means here. The 'R' is
specifically referring to the real numbers. The '[X]' building a far
far larger object out of the real numbers. You seem to want to think
in terms of "space" and "dimension" so, if you are picturing the real
numbers as a one dimensional line, and the complex numbers as a 2
dimensional plane, then you can think of R[X] as an *infinite*
dimensional space.
Now the (X^2 + 1) refers not just to the polynomial, but to an entire
subspace of R[X] that is generated by the polynomial X^2 + 1. Now the
space (X^2 + 1) is also infinite dimensional. It just so happens that
the ratio of dimension of R[X] to (X^2 + 1) is two to one. This means
that when you form R[X]/(X^2 + 1) the result is two dimensional. Now
this resulting two dimensional space is not actually the complex
numbers, but there is very obvious (if you know enough about rings
etc.) way to got back and forth between the two dimensional space and
the complex numbers in such a way that all the algebraic operations
all match up perfectly.
Now, none of that (aside from the final part about going back and
forth with the complex numbers) is actually true, but it bears some
resemblance to what is actually going on. To actually understand how
it really works I'm afraid there is no choice but to sit down and
learn the basics about rings and ideals. Hopefully, however, this has
been helpful in at least pointing your vague intuitions in a direction
that is closer to the truth that they were.
Bill Dubuque
>"Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>>
>> In http://en.wikipedia.org/wiki/Quotient_ring#Examples
>> "Now consider the ring R[X] of polynomials in the variable X with
>> real coefficients, and the ideal I = (X^2 + 1) consisting of all
>> multiples of the polynomial X^2 + 1." Here they have neither
>> specified what R[X] is nor what X is.
>
> Yes, they have: R[X] is the ring of polynomials in the variable X with
> real coefficients. Those words have *precise*, specific meanings. That
Part of the problem here is that "variable" is an overloaded term.
We're stuck with it for historical reasons. In this context X can
denote any indeterminate, i.e. any element of some ring containing
R that is transcendental (not algebraic) over R. It is better to
avoid this confusion by defining polynomials by their coefficient
sequences, i.e. they are functions N -> R with finite support that
are added pointwise and multiplied by Cauchy product (convolution).
Then X = (0,1,0,0,0...), and X^n is the sequence having 1 in the
n'th place and 0 elsewhere; r = (r,0,0,0...) for constants r in R.
Now the question "what is X?" has a clear and rigorous answer.
When learning about these ring constructions it is essential to
understand that it is not the particular construction that matters
but rather the essential properties, i.e. one should view the objects
as solutions to a universal mapping problem. This yields tremendous
power, e.g. see some of my prior posts [1] on universal techniques.
--Bill Dubuque
[1] http://google.com/groups/search?q=author%3Adubuque+universal+polynomial
Tim BandTech.com
Thanks. It certainly does feel vague. For instance if I were to build
a polynomial in this X with real coefficients such as
- 5.2 XXX + 2.3 XX - 1.1 X - 0.23
I'd have rather a lot of trouble evaluating this for a given X if X is
undefined. On the other hand, if X is real then regardless of what I
put in this expression the result is likewise real. Somehow
mathematicians are comfortable working with objects which are not
clearly instantiated. I'm afraid that at this level of discussion I'm
dealing in several of these stacked uninstantiated objects.
Above here Arturo states:
"R is the ring of real numbers, X is an indeterminate variable, and
R[X] is the ring of polynomials with coefficients in R."
yet this does not wash since there are three devices described in a
notation of just two devices. The polynomial has a clear description
as a sum
sum over n ( a(n) X ^ n )
yet is never actually invoked within the description. The meaning of X
has been scuttled. Your interpretation seems more believable but you
say it is not the truth. Opaque does certainly describe my own
misunderstanding. But there is one other problem here. I am not
convinced that my misunderstanding is invalid. I'm not going to just
take this information and accept it because it has been written in a
book and transcribed into other books. In that mathematics is to be
fundamental and error free I'd like to get to the bottom of this with
a real feeling of understanding rather than a double check
verification of mimicry.
I guess that one simple question to get answered is:
Is X real valued?
KMF above here gives alot of results but says that
" ax+b <--> ai+b "
and if x were real this would not be true. I'm pretty sure that my
trouble is going back to the usage of polynomial in this context and
then topping that misunderstanding with more of the same in a
different symbolism.
- Tim
cbr...@cbrownsystems.com
> >> "Now consider the ring R[X] of polynomials in the variable X with
> >> real coefficients, and the ideal I = (X^2 + 1) consisting of all
> >> multiples of the polynomial X^2 + 1." Here they have neither
> >> specified what R[X] is nor what X is.
>
> > Yes, they have: R[X] is the ring of polynomials in the variable X with
> > real coefficients. Those words have *precise*, specific meanings. That
> > means that they have in fact *completely specified* what R[X] and X are.
>
> Part of the problem here is that "variable" is an overloaded term.
> We're stuck with it for historical reasons. In this context X can
> denote any indeterminate, i.e. any element of some ring containing
> R that is transcendental (not algebraic) over R. It is better to
> avoid this confusion by defining polynomials by their coefficient
> sequences, i.e. they are functions N -> R with finite support that
> are added pointwise and multiplied by Cauchy product (convolution).
> Then X = (0,1,0,0,0...), and X^n is the sequence having 1 in the
> n'th place and 0 elsewhere; r = (r,0,0,0...) for constants r in R.
> Now the question "what is X?" has a clear and rigorous answer.
>
I first encountered this interpretation in Stanley's Enumerative
Combinatorics where he describes the ring of formal power series
(where /every/ natural power of x^n has a coefficient a_n) in this
way, and I then found it most enlightening to apply to polynomial
rings as well.
Cheers - Chas
Mariano Suárez-Alvarez
Mathematicians have spent quite a lot of time, each of
them, studying the subject. If you would do so, you'd also
see that there is nothing "not clearly instantiated"---whatever
that verb may mean.
-- m
Axel Vogt
Certainly my answer only rephrases others replies:
1) Try to understand the notion of quotient *group* first.
Then (and if you already know what are rings and ideals)
verify that that this construction gives a ring, if your
'subgroup' is an ideal.
As an example try the integers Z, google for examples and
work them out by paper and pencil.
2) Try to view at R[X] as finite sequences (= the tails
are all 0), entries are from R. Addition is element wise.
This is a group. As others already you can also make it
a ring (by introducing a certain 'multiplication').
As you 'see': that R can be any ring, for example R=Reals
but also R = Integers.
That approach is always puzzling for students (I hated
it first ...)
3) The word quotient has historical reason, division with
remainder. Say for integers n: n/3 = m + a/3 where m is
an integer and a is some 0, 1 or 2 ( 7/3 = 2 + 1/3, as
you learn it in school to give a reduced form, n=7 and
m=2, a=1).
Now 'Quotient' means 'inverse ratio' and having in mind
our '3' we get a = 0, 1 or 2.
This way one says 'quotient group' when we talk about the
remainders.
leland....@gmail.com
> Thanks. It certainly does feel vague. For instance if I were to build
> a polynomial in this X with real coefficients such as
> - 5.2 XXX + 2.3 XX - 1.1 X - 0.23
> I'd have rather a lot of trouble evaluating this for a given X if X is
> undefined.
Okay, well here is your first problem. You don't need to "evaluate"
the polynomial for any X ever. X is a formal symbol. It isn't
anything, it just "is". I'll leave this for now, but we'll keep coming
back to this point again and again as I answer your other questions.
> On the other hand, if X is real then regardless of what I
> put in this expression the result is likewise real. Somehow
> mathematicians are comfortable working with objects which are not
> clearly instantiated. I'm afraid that at this level of discussion I'm
> dealing in several of these stacked uninstantiated objects.
In the polynomial ring R[X] we have X clearly instantiated as a formal
symbol. The objects are polynomials where the X doesn't stand for
anything other than itself.
> Above here Arturo states:
>
> "R is the ring of real numbers, X is an indeterminate variable, and
> R[X] is the ring of polynomials with coefficients in R."
>
> yet this does not wash since there are three devices described in a
> notation of just two devices. The polynomial has a clear description
> as a sum
> sum over n ( a(n) X ^ n )
> yet is never actually invoked within the description. The meaning of X
> has been scuttled.
I'm not really sure at all what you mean here. A polynomial indeed has
a clear description. Each and every such polynomial that you can
possibly imagine is a unique object in R[X]. Thus R[X] is very very
large indeed.
Perhaps it might be better for you to picture this using the sort of
ideas Bill Dubuque proposed. We can think of R[X] not as polynomials,
but instead as a vector space. We can describe elements of the plane
by 2 dimensional vectors or the form (x,y). A 3 dimensional space can
be thought of as having elements that are vectors of length 3:
(x,y,z). R[X] can be thought of as having elements that are vectors of
*infinite* length: (a_1,a_2,a_3,...). For reasons that I won't go
into, we will require the added constraint that we only consider
vectors (a_1,a_2,a_3,...) such that *all* but finitely many of the a_i
are 0. Now for R[X] to be a ring we need to be able to "add" and
"multiply" such vectors, and we can. We define (a_1,a_2,a_3,...)+
(b_1,b_2,b_3,...) = (a_1+b_1,a_2+b_2,a_3+b_3,...). Multiplication is a
little trickier; the product (a_1,a_2,a_3,...)*(b_1,b_2,b_3,...)=
(c_1,c_2,c_3,...) where c_n = sum_{i=0 to n} a_i * b_{n-i}. The key
here is that we have, in some sense, an infinite vector space and a
way to add and multiply vectors of infinite length such that we have a
ring structure (i.e. the ring axioms are satisfied). This is what I
was hand waving toward when I talked about R[X] being infinite
dimensional. The technical hurdles here are that we have that "all but
finitely many components of a vector must be zero" and a weird looking
multiplication defined, so it isn't really quite an infinite
dimensional vector space. It may help of you to think of it this way
though, and ignore the X's and polynomials.
So why do mathematicians talk about X's and polynomials? Partly there
is some historical holdover, but also it is a very handy way to look
at things. You see if we have an infinite length vector
(a_1,a_2,a_3,...) then we can associate it with the polynomial a_1 +
a_2 X + a_3 X^2 + .... And certainly given a polynomial we could
associate it to a particular vector (and the vectors would need to be
infinitely long because the polynomial may have an arbitrarily large
power of X in it. Note that if we think of the polynomial a_1 + a_2 X
+ a_3 X^2 + ... simply as a another way of writing the vector
(a_1,a_2,a_3,...) then the X's aren't anything, they're just a way of
separating out the components of the vector. Now, a catch: by
definition a polynomial can only have finitely many terms. That means
that, in fact, we can't translate any vector into a polynomial. We can
only deal with vectors that only have finitely many non-zero
components (resulting in only finitely many terms to the polynomial)
-- but that's what we wanted anyway! So one technical hurdle is dealt
with by default if we work in terms of polynomials instead of vectors.
What of the other hurdle, having to definie multiplication of vectors
(and having it be so weird looking)? If we translate the vectors to
polynomials and then simply multiply the polynomials in the standard
way, then translate back to the vector we find we actually get the
vector product we had defined! Thus if we write the vectors as
polynomials we can simply use standard algebra to deal with products
-- the second technical hurdle is then conveniently covered. So
polynomials actually provide a nicer fit than vectors, but you have to
realise that the X's in the polynomials are really just a book-keeping
matter and don't actually stand for any number in any way shape or
form, they are simply there to make multiplication clear an easy.
> Your interpretation seems more believable but you
> say it is not the truth. Opaque does certainly describe my own
> misunderstanding. But there is one other problem here. I am not
> convinced that my misunderstanding is invalid. I'm not going to just
> take this information and accept it because it has been written in a
> book and transcribed into other books. In that mathematics is to be
> fundamental and error free I'd like to get to the bottom of this with
> a real feeling of understanding rather than a double check
> verification of mimicry.
Then I suggest you either get a very good introductory book on modern
abstract algebra (I quite like "A First Course in Abstract Algebra" by
John Fraleigh), or take a course. Otherwise you are jumping into the
deep end of the pool without having learned to swim, and you'll only
confuse yourself more. Head back to the beginning and understand
everything there solidly first. All of this mathematics is on very
solid ground, but if you don't understand how the foundations are
built and can't be bothered to go back and look then yes, you'll not
see how solid the structure is. Okay, I think I mixed a few to many
metaphors there, but the point remains.
> I guess that one simple question to get answered is:
> Is X real valued?
No, X is a formal symbol, a book-keeping device if you like. It means
nothing other than itself.
> KMF above here gives alot of results but says that
> " ax+b <--> ai+b "
> and if x were real this would not be true. I'm pretty sure that my
> trouble is going back to the usage of polynomial in this context and
> then topping that misunderstanding with more of the same in a
> different symbolism.
Yes, I think your issues begin with a desire to think of high school
algebra polynomials where the x stands for some number that you have
to solve for, or which you evaluate at some particualr number. These
polynomials are not of that kind at all. The X stands not for number,
but is instead simply a symbol.
Consider a + ib. We can think of that as a polynomial in "i". What
does the "i" stand for? Is it real valued? Or is it just a symbol to
which we ascribe certain properties (such as i*i = -1)? Our X is much
like that -- just a symbol to which we ascribe certain properties.
Hopefully some of this will help to ease your confusion. There really
is no other way than to start at the beginning and learn it properly
however.
Arturo Magidin
> Thanks. It certainly does feel vague. For instance if I were to build
> a polynomial in this X with real coefficients such as
> - 5.2 XXX + 2.3 XX - 1.1 X - 0.23
> I'd have rather a lot of trouble evaluating this for a given X if X is
> undefined.
You are confusing polynomial *functions* with polynomials. Polynomials
are not "evaluated". You don't need to "evaluate for a given X". (And,
on the other hand, if you have a polynomial *function*, then a "given
X" would, *of course*, not be "undefined": it would be *given*).
> On the other hand, if X is real
X is not a real. These things are not *functions*, they are
*polynomials*.
> then regardless of what I
> put in this expression the result is likewise real.
These are not expressions to be evaluated, they are polynomials.
> Somehow
> mathematicians are comfortable working with objects which are not
> clearly instantiated.
Polynomials *are* "clearly instantiated". The problem here is not
about what mathematicians are or are not comfortable working with, but
rather that you are confusing two things (not your fault: in high
school most people are not told of the distinction). Polynomials and
polynomial functions are *different things*. See Bill Dubuque's post
if you want a particular way of instantiating polynomials as almost
null sequences.
> I'm afraid that at this level of discussion I'm
> dealing in several of these stacked uninstantiated objects.
I'm afraid that the real problem is that you are a bit confused about
the objects in question, and that there are some rather large lacunae
in your knowledge of the things you need to understand these objects.
Again, not your fault, but you are trying to run a marathon without
knowing how to move your feet.
>
> Above here Arturo states:
>
> "R is the ring of real numbers, X is an indeterminate variable, and
> R[X] is the ring of polynomials with coefficients in R."
>
> yet this does not wash since there are three devices described in a
> notation of just two devices.
If by "this does not wash" you mean "I do not understand", then you
are correct. The problem here is that you are not understading, not
that anyone is hiding anything from you.
The ring of polynomials S[T], with S a ring and T a set is
*completely* determined by specifying S and T. The "third object" is
*constructed* from the first two.
> The polynomial has a clear description
> as a sum
> sum over n ( a(n) X ^ n )
> yet is never actually invoked within the description.
And now you are confusing a *particular* polynomial with the ring of
*all* polynomials. R[X] is the ring of *all* polynomials with real
coefficients, not a particular one. See? It's not that what I said
"does not wash": it's that you are not familiar enough with these
things to understand what is being said.
> The meaning of X has been scuttled.
Nonsense.
> Your interpretation seems more believable but you
> say it is not the truth.
Look: nobody is lying to you. The problem here is that you are simply
not aware of the context and the background necessary to understand
the answers you are given, but you are now blaming those answering you
and even accusing them of lying to you, because you don't understand
the answers.
Nobody is lying to you. Nobody is hiding the truth. What you need to
do is learn a bit more about rings and polynomial rings in order to
understand the answers you are being given.
> I guess that one simple question to get answered is:
> Is X real valued?
Sigh. You are not dealing with a specific polynomial nor with
polynomial functions. Your question may be simple, but it is in fact
quite meaningless in this context. Just as if you were asking "what
color are ideas that dream furiously?".
--
Arturo Magidin
Mariano Suárez-Alvarez
> 3) The word quotient has historical reason, division with
> remainder. Say for integers n: n/3 = m + a/3 where m is
> an integer and a is some 0, 1 or 2 ( 7/3 = 2 + 1/3, as
> you learn it in school to give a reduced form, n=7 and
> m=2, a=1).
>
> Now 'Quotient' means 'inverse ratio' and having in mind
> our '3' we get a = 0, 1 or 2.
>
> This way one says 'quotient group' when we talk about the
> remainders.
Interestingly, old sources spoke of "difference groups"
and "difference modules", and even wrote M - N for what
we know write M/N. IIRC Zariski's book on commutative
algebra still follows this tradition in 1958; for rings,
though, he uses R/I.
-- m
Denis Feldmann
I know that one, I do. Can I answer, teacher? Can I ?
> --
> Arturo Magidin
Virgil
<5d90e8be-ba08-4cde...@l28g2000vba.googlegroups.com>,
"Tim BandTech.com" <tttp...@yahoo.com> wrote:
Let (R, 'add', 'mult') be a ring with addition operator 'add' and
multipliction operator 'mult'.
Consider the set of functions, P, from the set of zero-origin
naturals, N = {0,1,2,3,...}, to set R which have finite carrier (only
finitely many non-zero values in R).
Define addition. "add" in P that h = f "add" g is the function for
which h(n) = f(n) 'add' g(m) for all n in N
Define multiplication in P so that h = f "mult" g is the function for
which h(n) = sum_(i,j in N and i + j = n) f(i) 'mult' g(j) for all n in N
It will transpire that (P,"add", "mult") is isomprphic to the ring of
polynomials in one variable over the ring (R,'add','mult')
But (P,"add", "mult") does not need an X. So there is no need of an X in
a polynomial ring, though it may be convenient to allow one.
--
Virgil
Virgil
<5093c3fb-4709-4fb0...@l28g2000vba.googlegroups.com>,
Wasn't M - N used only for commutative groups in Zaritski?
--
Virgil
Mariano Suárez-Alvarez
> In article
> <5093c3fb-4709-4fb0-b937-fb17e870c...@l28g2000vba.googlegroups.com>,
> Mariano Suárez-Alvarez <mariano.suarezalva...@gmail.com> wrote:
>
>
>
> > On Jun 6, 4:19 pm, Axel Vogt <&nore...@axelvogt.de> wrote:
> > > 3) The word quotient has historical reason, division with
> > > remainder. Say for integers n: n/3 = m + a/3 where m is
> > > an integer and a is some 0, 1 or 2 ( 7/3 = 2 + 1/3, as
> > > you learn it in school to give a reduced form, n=7 and
> > > m=2, a=1).
>
> > > Now 'Quotient' means 'inverse ratio' and having in mind
> > > our '3' we get a = 0, 1 or 2.
>
> > > This way one says 'quotient group' when we talk about the
> > > remainders.
>
> > Interestingly, old sources spoke of "difference groups"
> > and "difference modules", and even wrote M - N for what
> > we know write M/N. IIRC Zariski's book on commutative
> > algebra still follows this tradition in 1958; for rings,
> > though, he uses R/I.
>
> > -- m
>
> Wasn't M - N used only for commutative groups in Zaritski?
I've just checked: he uses it for modules too.
-- m
Arturo Magidin
> I am seeking help understanding the quotient ring nomenclature.
> Keywords: ideal, ring, quotient ring.
It has become clear that you are actually missing a lot of background
and information, and that you are not in a position to understand
these notions at present as they apply to the construction of the
complex numbers as a quotient of the ring of polynomials with real
coefficients, which seems to have been your original interest.
So, let's forget everything we have tried to tell you, because you are
not in a position to process it just yet, and let's built some of
those foundations instead.
I will assume that you understand the real numbers, with their usual
sum + and their usual multiplication *. That is where we will begin.
We will use R to denote the set of real numbers (in a book or
blackboard you would usually see it in bold or in "blackboard bold",
but that is impossible to do in ASCII), and by (R,+,*) the set of real
numbers togethere with its two operations we already mentioned.
We will now construct a new object. The elements of this object are
"infinite tuples of real numbers", that is, sequences:
(a_0, a_1, a_2, ..., a_n, ... )
where each a_i is a real number. You can think of these as the
"graphs" of functions a:N->R (where N is the set of natural numbers, N=
{0,1,2,...,n,...}, and a_i represents the value of a at i), or as
ordered tuples.
We define a binary operation on the set of these tuples. I will denote
it by & so we don't confuse it with the original + of real numbers. We
are going to add these tuples "coordinate-wise", that is, by
definition we will have:
(a_0,a_1,...,a_n,..) & (b_0, b_1, ..., b_n, ... ) = (c_0, c_1,
c_2, ..., c_n, ...)
where c_i = a_i + b_i for all i.
It is not hard to check that:
(i) & is commutative (basically, because + is commutative in the real
numbers);
(ii) & is associative (again, essentially because + is associative);
(iii) The tuple (0,0,0,..., 0,...) has the property that for all
tuples (a_0,a_1,..., a_n,...),
(a_0, a_1, ..., a_n, ...) & (0, 0, ..., 0, ...) = (0, 0, ...,
0, ...) & (a_0, a_1, ..., a_n, ...)
= (a_0,
a_1, ..., a_n, ...)
(essentially, because for every real number a, 0+a = a+0 = a);
(iv) For every tuple (a_0,a_1,..., a_n,...) there exists a tuple
(b_0, b_1, ..., b_n, ...) such that
(a_0, a_1, ..., a_n, ... ) & (b_0, b_1, ..., b_n, ...)
= (b_0, b_1, ..., b_n, ...) & (a_0, a_1, ..., a_n, ...)
= (0,0,..., 0,...)
namely, the tuple with b_i = -a_i for all i. (Again, this is
inherited from the fact that for real
numbers, for every real number a there is a real number b
[namely b=-a], such that a+b=b+a = 0).
This means that the tuples, with the operation &, form an "abelian
group".
Next, we define a "multiplication" on this set of *all* tuples. I will
use x to denote it, again to prevent confusion with the operation *
for reals. We define x as follows: given tuples (a_0, a_1,...,
a_n, ...) and (b_0, b_1,..., b_n, ...), we let
(a_0,a_1,..., a_n, ...) x (b_0, b_1, ..., b_n, ...) = (c_0,
c_1, ..., c_n, ...)
where for each m, c_m = a_0*b_m + a_1*b_{m-1} + ... + a_m*b_0
= Sum(i=0 to m)[ a_i*b_{m-i} ].
Again, it is not hard to show that x is commutative and associative;
that the tuple (1,0,0,...,0,...) has the property that:
(a_0, a_1, ..., a_n, ...) x (1, 0, ..., 0,...)
= (1, 0, ..., 0, ...) x (a_0, a_1, ..., a_n, ...)
= (a_0, a_1, ..., a_n, ....)
for all tuples (a_0, a_1, ..., a_n, ...). And that x distributes over
&.
These things mean that the set of all tuples, together with the
operations & and x, form a "commutative ring with 1".
Now, we will consider a subset of this bigger ring. We will consider
exactly all the tuples (a_0,a_1,...,a_n,...) such that a_n is zero for
all, except perhaps a finite number of n. These are called the
"almost null sequences". Note that a tuple is almost null if and only
if there exists a positive integer N, which may depend on the tuple,
such that a_m = 0 for all m>N.
Now, it is not hard to show that if (a_0,a_1,...,a_n,...) and
(b_0,b_1,...,b_n,...) are both almost null, then (a_0,a_1,...,a_n,...)&
(b_0,b_1,...,b_n,...) and (a_0,a_1,...,a_n,...)x(b_0,b_1,...,b_n,...)
are also almost null; so that if we restrict & and x to the almost
null sequences, they are also operations on them. So the collection of
almost null sequences, with & and x, also forms a commutative ring
with 1.
We will denote the set of all almost null sequences by the notation R
[X] (for historical reasons; don't worry about it now, it's just the
name we give it).
Now, if we consider the collection of all tuples (a_0,a_1,...,a_n,...)
such that a_i=0 for all i>0, then this set is also closed under & and
x; it forms a subring. In fact, this subcollection behaves essetially
the same way that our original real numbers R did, because under the
definitions we gave, it turns out that
(a_0, 0,...,0,...) & (b_0, 0, ..., 0,...) = (a_0 + b_0, 0, ...,
0, ...)
(a_0, 0,...,0,...) x (b_0, 0, ..., 0,...) = (a_0*b_0, 0, 0, ...,
0, ...).
So, by abuse of notation, if a is a real number, we are going to write
a instead of (a,0,0,...,0,...); but "a" still means the almost null
sequence (a,0,...,0,...), it's just hat we are lazy and we don't want
to write all those zeros (that don't really matter all that much
anyway).
Note some nice things here: if (b_0, b_1, b_2,...,b_n,...) is any
tuples (almost null or not), we have that
a x (b_0, b_1, ..., b_n, ...) = (a,0,...,0,...)x(b_0,b_1,...,b_n,...)
= (a*b_0, a*b_1, ..., a*b_n, ...).
Another nice thing: let's consider the tuple (0,1,0,...,0,...) (a 1 in
the entry indexed by 1, and 0's elsewhere). Then
(0,1,0,...,0,...) x (0,1,0,...,0,...) = (0,0,1,0,...,0,...)
(0,1,0,...,0,...) x (0,0,1,0,...,0,...) = (0,0,0,1,0,...)
etc; that is, if you multiple (0,1,0,...,0,...) by itself n times, you
get the almost null tuple that has a 1 in the n-th entry and zeros
elsewhere. Just because we are lazy, and we don't want to write all of
that all the time, we are also going to write a short-hand: we are
going to use X to denote the tuple that has a 1 in the 1-entry, and
0's elsewhere; X^2 the tuple that has a 1 in the 2-entry and 0's
elsewhere (that is, (0,0,1,0,...,0,...)). X^3 will be the tuple with 1
in the 3-entry and 0's elsewhere; etc. In general, for any positive
integer n, X^n is the result of multiplying X by itself n times, and
is the tuple that has a 1 in the n-entry and 0's elsewhere.
Now, suppose that (a_0,a_1,...,a_n,...) is an element of R[X]. that
means that it is an almost null tuple, so that there exists an N such
that a_m = 0 for all m>N. So we have
(a_0, a_1, ..., a_n, ..., a_N, 0, 0, ...)
Now, we can write this as:
(a_0, a_1, ..., a_N, 0, ...) = (a_0, 0, 0, ..., 0, ...)
& (a_1,0,...,0,...) x
(0,1,0,...,0,...)
& (a_2,0,...,0,...) x
(0,0,1,...,0,...)
& ... & (a_N,0,0,...,0,...) x
(0,0,...,0,1,0,...)
(where the last one has a 1 in the N-th entry, 0's elsewhere). Using
our "shorthand" from above, this is the same as
(a_0, a_1, ..., a_N, 0,...) = a_0 & (a_1 x X) & (a_2 x X^2) & ... &
(a_N x X^N).
So, that's what R[X] *is*. The collection of all almost null tuples of
real numbers, with operations & and x.
With me so far?
--
Arturo Magidin
KMF
Just a small point to make for Timothy: the assignment
ax+b<->ai+b is not purely trivial, in that
x^2+1 =0 ( (x^2+1)=(x^2)+(1)=(0)), same as
i^2+1 =0. This, as part of an overall argument
showing that the quotient ring we constructed is
structurally identical (meaning isomorphic as a
ring) to the complex numbers.
Tim BandTech.com
Hi Arturo. Yes, I am with you so far. But for all of the 'nifty' stuff
that you've demonstated(and I do see it) there is the oddity that you
are stuck in infinite dimension. For instance, if we were to call a
value such as
( a1, a2, a3, 0, 0, ... )
a three dimensional value since as you say the upper zeroes are quite
meaningless then when we take a product of two such values AB = C then
C will be six dimensional for three dimensional A and B. In effect
you've been forced to remain in the infinite dimensional form, whereby
the high products go off to what some will worry about as a super
infinite dimensional space. This behavior is markedly different from
other structures such as calculus where we are free to specify n and
see a semblance of the infinite limit. There is no simple form beneath
this construction. I guess that helps explain the usage of quotient
ring a bit. Yet you will have contructed a lower dimensional space
from an infinite dimensional space which was constructed from the one
dimensional space in the case of real coefficients.
Still, I can accept this polynomial construction as a generalized
form. It is very easy to overlook the lack of specification of X in
reading this material and particularly that there is no answer to the
question
What is X?
is troubling. When you suggest that a real coefficient can be
multiplied by such an X and yet have no way of doing it then I do
think a misnomer was created by using the word polynomial. Especially
as a polygon can have many sides but it need not have infinite sides
so should a polynomial. This is an infinite dimensional structure that
you are describing, not a polynomial. So the word was coopted- fine.
There are many instances of this in math.
I am seeing the modulo form a bit now in the quotient ring such that
the six dimensional value that I showed above could become three
dimensional through a mod 3, though it's not so simple as the integers
taken as modulo math. In the integer case there is no need ot sum
components. They just dissapear. Here they accumulate. I still don't
understand the ideal concept, but I do get your polynomial
presentation. Thanks. Now when they talk about the ring R[X] I'll be
sure to keep in mind that even when it is specified X is not specific
when discussing polynomials.
- Tim
Axel Vogt
My Zariski (& Samuel) uses the concurrent notation, but it is
the newer Springer edition.
Actually the notions are not unique (I learned 'quotient ring'
in the sense of 'quotient field', i.e. localization)
In German it is/was more common to call G/N a factor group and
looking in a quite old van der Waerden "Algebra" he calls it
"Restklassengruppe" ~ group of (class of) remainders in §8.
I also vaguely remember the explanation that G/N is called the
quotiemt group only because of the notation (looking like some
qutient).
Bill Dubuque
>>> Mariano Suarez-Alvarez <mariano.suarezalva...@gmail.com> wrote:
>>>> On Jun 6, 4:19 pm, Axel Vogt <&nore...@axelvogt.de> wrote:
>>>>>
>>>>> 3) The word quotient has historical reason, division with
>>>>> remainder. Say for integers n: n/3 = m + a/3 where m is
>>>>> an integer and a is some 0, 1 or 2 ( 7/3 = 2 + 1/3, as
>>>>> you learn it in school to give a reduced form, n=7 and
>>>>> m=2, a=1).
>>>>> Now 'Quotient' means 'inverse ratio' and having in mind
>>>>> our '3' we get a = 0, 1 or 2.
>>>>> This way one says 'quotient group' when we talk about the
>>>>> remainders.
>>>> Interestingly, old sources spoke of "difference groups"
>>>> and "difference modules", and even wrote M - N for what
>>>> we know write M/N. IIRC Zariski's book on commutative
>>>> algebra still follows this tradition in 1958; for rings,
>>>> though, he uses R/I.
>>>> -- m
>>> Wasn't M - N used only for commutative groups in Zaritski?
>> I've just checked: he uses it for modules too.
>> -- m
>
> My Zariski (& Samuel) uses the concurrent notation, but it is
> the newer Springer edition.
>
> Actually the notions are not unique (I learned 'quotient ring'
> in the sense of 'quotient field', i.e. localization)
"Quotient ring" isn't ideal terminology for localizations either
because not every subring of a quotient field is a localization.
Indeed, this holds true iff the domain is a Prufer domain with
torsion class group - see the reviews in my prior post [1].
> In German it is/was more common to call G/N a factor group and
> looking in a quite old van der Waerden "Algebra" he calls it
> "Restklassengruppe" ~ group of (class of) remainders in �8.
Or residue(-class) groups/rings in English.
> I also vaguely remember the explanation that G/N is called the
> quotiemt group only because of the notation (looking like some
> qutient).
I suspect rather that the choice of notation was motivated by the
"quotient-like" behavior given by the Third Isomorphism Theorem,
namely A > B > C => (A/C)/(B/C) = A/B
--Bill Dubuque
[1] http://groups.google.com/group/sci.math/msg/b1924d26d95b6ae0
http://google.com/groups?selm=y8z4qqmya7b.fsf%40nestle.csail.mit.edu
Denis Feldmann
And isomorphisms of the shape A = (A/B) x B ? Products are actually much
more general than quotients, after all ...
Arturo Magidin
> > So, let's forget everything we have tried to tell you, because you are
> > not in a position to process it just yet, and let's built some of
> > those foundations instead.
>
> > I will assume that you understand the real numbers, with their usual
> > sum + and their usual multiplication *. That is where we will begin.
Note the above.
[.snip ring structure for function N-->R, and almost null subring.]
In the entire development I did, not once did I mention
"coefficients", "polynomials", or "dimension". These concepts do not
appear as of yet in the development. Yet, your very first comment, is:
> Hi Arturo. Yes, I am with you so far. But for all of the 'nifty' stuff
> that you've demonstated(and I do see it) there is the oddity that you
> are stuck in infinite dimension. For instance, if we were to call a
> value such as
> ( a1, a2, a3, 0, 0, ... )
> a three dimensional value since as you say the upper zeroes are quite
> meaningless then when we take a product of two such values AB = C then
> C will be six dimensional for three dimensional A and B.
Which means that you are *not* in fact listening to what I said, but
instead are going on with your original misconceptions and ideas. I
never said "the upper zeroes are quite meaningless", nor words to that
effect at all. I never talked about dimensions.
So... I have to ask: are you actually interested in learning this
stuff that you claimed you wanted to see developed form the ground up
(and not take it for granted because it was in a textbook somewhere,
insinuating that the rest of us are merely parroting things we never
checked)? Or are you going to continue to bring in extraneous notions
and misconceptions to the party?
I am willing to spend what is a considerable amount of time showing
you the basics, but there is no point if you have no interest in
actually listening.
Everything you said in your reply yells at high volume that you are
not actually interested in listening, that you are not paying
attention, and that you would rather rely on your vague intuition
(that mistakenly believes, apparently, that "degree" and "dimension"
mean the same thing, among many other errors). Nothing you said in
your reply is relevant or makes sense.
Now, you don't have to learn it from me. You can learn it form a book.
Contrary to your insinuation in a previous post, any mathematics
textbook above the calculus level will not merely *tell* you stuff
copied down from other textbooks, it will *define*, *develop*,
*construct*, and *prove* everything it uses. You can see how
polynomial rings are defined in any reasonable abstract algebra book:
van der Waerden, Fraleigh, Humphreys, Isaacs, Herstein, and countelss
others, if you would rather do it in the privacy of your own home.
But, whatever it is you decide to do, please bear the following in
mind: you have little idea of just how silly the things you are saying
are. You actually know little or nothing about polynomial rings,
please don't think you do. And definitely, do not invoke my name if
you tell others what you think you know about them, because you have
listened to nothing that I've said, and I don't want to be blamed for
your misconceptions.
--
Arturo Magidin
Bill Dubuque
I can't make any sense of what you wrote. What do you mean?
Perhaps concrete examples would serve to clarify your remarks.
The reason I mention the fraction-like nature of the identity
in the Third Isomorphism Theorem is because this theorem is
essentially the universal property of quotient rings and thus
characterizes them. So it's right at the heart of the matter.
Denis Feldmann
> Denis Feldmann <denis.feldm...@neuf.fr> wrote:
>>>> I also vaguely remember the explanation that G/N is called the
>>>> quotiemt group only because of the notation (looking like some
>>>> qutient).
>>> I suspect rather that the choice of notation was motivated by the
>>> "quotient-like" behavior given by the Third Isomorphism Theorem,
>>>
>>> namely A > B > C => (A/C)/(B/C) = A/B
>> And isomorphisms of the shape A = (A/B) x B ?
>> Products are actually much more general than quotients, after all...
>
> I can't make any sense of what you wrote. What do you mean?
That, for instance, there is an isomorphism between Z (the integers) and
the ring product of 2Z (the even integers) and Z/2Z (the field with two
elements) (the isomorphism being (2p, a) -> 2p +a, where a is 0 or 1)
Of course, this works well only for *algebraic* quotient structures, but...
> Perhaps concrete examples would serve to clarify your remarks.
>
> The reason I mention the fraction-like nature of the identity
> in the Third Isomorphism Theorem is because this theorem is
> essentially the universal property of quotient rings and thus
> characterizes them. So it's right at the heart of the matter.
Yes, I agree, but I still think that products come before quotients :-)
Tim BandTech.com
> On Jun 7, 8:38 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > So, let's forget everything we have tried to tell you, because you are
> > > not in a position to process it just yet, and let's built some of
> > > those foundations instead.
>
> > > I will assume that you understand the real numbers, with their usual
> > > sum + and their usual multiplication *. That is where we will begin.
>
> Note the above.
>
> [.snip ring structure for function N-->R, and almost null subring.]
>
> In the entire development I did, not once did I mention
> "coefficients", "polynomials", or "dimension". These concepts do not
> appear as of yet in the development. Yet, your very first comment, is:
>
> > Hi Arturo. Yes, I am with you so far. But for all of the 'nifty' stuff
> > that you've demonstated(and I do see it) there is the oddity that you
> > are stuck in infinite dimension. For instance, if we were to call a
> > value such as
> > ( a1, a2, a3, 0, 0, ... )
> > a three dimensional value since as you say the upper zeroes are quite
> > meaningless then when we take a product of two such values AB = C then
> > C will be six dimensional for three dimensional A and B.
>
> Which means that you are *not* in fact listening to what I said, but
> instead are going on with your original misconceptions and ideas. I
> never said "the upper zeroes are quite meaningless", nor words to that
> effect at all.
Actually you did:
"So, by abuse of notation, if a is a real number, we are going to
write
a instead of (a,0,0,...,0,...); but "a" still means the almost null
sequence (a,0,...,0,...), it's just hat we are lazy and we don't want
to write all those zeros (that don't really matter all that much
anyway)."
I guess the word 'meaningless' is a paraphrase, but an appropriate
one.
I had no idea that my interpretation is so far off as your language
would have us believe.
Still, as I read your initial comment on usage of 'coefficient' and so
forth I am willing to carry on, but I had no idea that this was going
to be so strict. Still You're clearly putting forth a lot of energy
and I am definitely interested in gleaning everything I can. Still I
do think that this criticism that I've given does hold, though I was
off by saying that 3D product would yield a 6D result. Its more like a
5D result. Still, I understand that this language is offensive to you
though I don't honestly understand why.
> ...I never talked about dimensions.
>
> So... I have to ask: are you actually interested in learning this
> stuff that you claimed you wanted to see developed form the ground up
> (and not take it for granted because it was in a textbook somewhere,
> insinuating that the rest of us are merely parroting things we never
> checked)? Or are you going to continue to bring in extraneous notions
> and misconceptions to the party?
I don't believe that what I am discussing is extraneous and if
anything your strong response may be validation that I've hit a sore
point with a master of this math. The dimensional criticism that I
have presented is accurate. Still, this dimensional argument seems to
be the misinterpretation that you are unhappy with so I do apologize
though again I do not understand what misuse I commit. On the parrot
thing you're mostly supporting my argument so far by trying to nail a
ruler accross my knuckles here rather than identify the mistruth you
believe I reflect. Perhaps my reflection is accurate and the mistruth
is buried in the construction itself, albeit subtly. I must remain
open to this possibility.
>
> I am willing to spend what is a considerable amount of time showing
> you the basics, but there is no point if you have no interest in
> actually listening.
I know, and I am listening. You have a problem with the 5D from 3D
claim that I have made.
Yet so far I see no content to dismiss it. By removing the word
dimension then I would simply restate that the three term truncated
series (with trailing zeroes) as general products will yield five term
truncated series.
>
> Everything you said in your reply yells at high volume that you are
> not actually interested in listening, that you are not paying
> attention, and that you would rather rely on your vague intuition
> (that mistakenly believes, apparently, that "degree" and "dimension"
> mean the same thing, among many other errors). Nothing you said in
> your reply is relevant or makes sense.
Yikes. I had no idea I was this far off. Seriously I did study every
line of your carefully typed out work. I'll admit that my reflection
of that cannot be a perfect mirror image and that you are driving on a
more fundamental point than the quotient ring, what I thought was the
polynomial representation so I thought using the term 'coefficient'
was OK. Also the dimension interpretation has been sprinkled in other
posts (perhaps by others). Sorry.
>
> Now, you don't have to learn it from me. You can learn it form a book.
> Contrary to your insinuation in a previous post, any mathematics
> textbook above the calculus level will not merely *tell* you stuff
> copied down from other textbooks, it will *define*, *develop*,
> *construct*, and *prove* everything it uses. You can see how
> polynomial rings are defined in any reasonable abstract algebra book:
> van der Waerden, Fraleigh, Humphreys, Isaacs, Herstein, and countelss
> others, if you would rather do it in the privacy of your own home.
>
> But, whatever it is you decide to do, please bear the following in
> mind: you have little idea of just how silly the things you are saying
> are. You actually know little or nothing about polynomial rings,
> please don't think you do. And definitely, do not invoke my name if
> you tell others what you think you know about them, because you have
> listened to nothing that I've said, and I don't want to be blamed for
> your misconceptions.
>
> --
> Arturo Magidin
Cripes Arturo, I've got to go reread my post...
I stand by what I have written there. Yes, your notation thus far
lacks any usage of X, coefficient, dimension, or polynomial. But I had
no idea that the translation to these things would require another
step. I thought this form that you've set up is directly translatable.
If I stand to be corrected then you ought to correct me. So far all of
your indignance is lacking a direct error on my part. Please select
some text that you wish to identify as conflicted on my part. That is
exactly what I will do for all of your text and that is exactly as it
should be for this subject. I have no idea why you do not simply
concede the point about the higher dimensional product. It is a true
position. Again, while I know that the chances are very slim that this
body of math is errant since it has been checked over millions of
times by others that does not mean that I should not do so myself and
so I must remain open to the possibility of a mistake if I am to
detect a mistake. Otherwise I do become a mimic. While I deeply
appreciate your detailed presentation Arturo I cannot remove my doubt.
Is this not the position of an ideal mathematician? Computers do not
do this yet, but humans can and must. Otherwise the mistruths that
will be perpetrated on future generations will only accumulate. Your
energy is not wasted upon me, but also on the upside you could save
your careful presentation for reuse since it is worthy.
I apologize for offending you.
- Tim
Tim BandTech.com
wrote:
> Bill Dubuque a écrit :
Nice point Dennis. I don't really understand the motivations of
abstract algebra so I may not understand fully what you are
discussing. I do see the cartesian product as a fairly primitive
usage, though I will agree that it is not necessary. Clearly this
quotient is not the inverse of the cartesian product since then
dimensions could be subtracted out. I've just caught on to the idea
that the polynomial is always infinite dimensional. Then moding down
with accumulation of the coefficients builds a lower dimension space.
Still, it can be argued that the infinite dimensional space was merely
built up from whatever dimension coefficient is used in the polynomial
construct. Mostly I just don't get the quotient ring; neither its
motivation nor its usage.
The polynomial throws me for a loop as well since its X is completely
irrelevant and unspecified. Multiplying a real or complex coefficient
by this X was valid for real or complex X, but it cannot be the same
operation. The X as I am witnessing it on this thread is used as a
rotational unit vector, rotating into each nth orthogonal dimension as
X^n just as Beresford's terplex algebra does, but infinite
dimensional.
I know I stand to be corrected so I welcome your context and
identification of my errors.
This statement ought perhaps to be stated universally just as when I
write something there is no need to preface "I think" before
everything I think.
- Tim
KMF
Maybe I can suggest that after having given your best
effort to understand quotients, you try to learn how
to work with them, and with time, the motivation and
understanding that you are looking for may come through.
I have also always tried to build myself bottom-up,
but when I haven't been able to, just doing some work
with the material , and time, helps the brain make
sense, and figure out the motivations and the big picture.
Denis Feldmann
1)You guess wrong
2) You keep doing what Arthuro tell you not to do
3) At this stage, you are probably a crank, or perhaps a troll ; the 5
vs 6 dimensions comment below *provs* that
4) Anyway, you are hopeless. *plonk*
Arturo Magidin
No, it's not. "Meaningless" means that they lack meaning. But here
they do *not* lack meaning. Ommitting them because they are understood
(because we are lazy) is not the same thing as "they are meaningless".
If you truly believe that saying that they are "meaningless" is an
"appropriate paraphrase", then let me put it plainly: you are not just
wrong, you are *damn* wrong.
> I had no idea that my interpretation is so far off as your language
> would have us believe.
Would that be the royal "us", there? For sure, *you* are completely
lost. I would not hazard a guess as to how many people are in there
with you, though you seem to think that it is the rest of the posters
in this thread who are lost and alone.
> Still, as I read your initial comment on usage of 'coefficient' and so
> forth I am willing to carry on, but I had no idea that this was going
> to be so strict.
You *explicitly* requested the details so you could follow them from
first principles. That *requires* that you be not only "strict", but
attentive and careful. For example, did you notice that I started all
my tuples with the index 0? Apparently not, since *you* started your
one and only example with an index 1, showing carelessness.
Correct mathematics *demands* care. If you cannot give it the care it
needs, then you have no business trying to do it.
> Still You're clearly putting forth a lot of energy
> and I am definitely interested in gleaning everything I can.
Then go back to my long post, reread it, and stop trying to jump ahead
with your ill-conceived and ill-digested misconceptions, confusions,
and misunderstandings.
> Still I
> do think that this criticism that I've given does hold, though I was
> off by saying that 3D product would yield a 6D result.
Your "criticisms" do not hold, because they are based on your utter
confusion and willful ignorance. I know you think they hold, but
that's just because you have no idea what you are talking about, and
you decided to not bother trying to learn it when you were presented
with the beginning of it.
> Its more like a
> 5D result.
No, it's not. It has nothing to do with dimension.
> Still, I understand that this language is offensive to you
> though I don't honestly understand why.
>
Because you understand nothing. It's not the "language" that is
offensive, is your attitude of superiority based on ignorance. You
asked to be shown the details, and then you ignored them and prefered
to continue plowing on in your ignorance, dmeonstrating utter contempt
for the help that was offered to you.
That is why you don't understand why: because you haven't bothered to
put any effort into listening, you prefer to continue being ignorant
and blame the rest of the world for your ignorance.
> > ...I never talked about dimensions.
>
> > So... I have to ask: are you actually interested in learning this
> > stuff that you claimed you wanted to see developed form the ground up
> > (and not take it for granted because it was in a textbook somewhere,
> > insinuating that the rest of us are merely parroting things we never
> > checked)? Or are you going to continue to bring in extraneous notions
> > and misconceptions to the party?
>
> I don't believe that what I am discussing is extraneous
That's because you don't know enough about the subject ot even begin
to realize just how far off the mark you are, and instead of listening
to the exposition of the subject being given to you from first
principles, you prefered to go ahead and repeat your misconceptions
and ignorant comments.
> and if
> anything your strong response may be validation that I've hit a sore
> point with a master of this math.
Ah, yes, the old "You get annoyed, so I must have hit a sore spot with
my criticisms." No, what you've hit is a sore spot with your attitude;
I get very annoyed when someone shows so much deep contempt for the
effort made to try to help him. You are not just a fool, not just a
damn fool, but a *willful* damn fool. And that's what annoys me.
> The dimensional criticism that I
> have presented is accurate.
The "dimensional criticism" you have presented is complete and utter
nonsense, based on the fact that you don't even know what "dimension"
means, dear.
> Perhaps my reflection is accurate and the mistruth
> is buried in the construction itself, albeit subtly. I must remain
> open to this possibility.
Perhaps what you should try is to be open to learning *something*
instead of persisting in your ignorance. I won't hold my breath,
though, given that I *presented* you with the "construction itself",
and it clearly flew several miles above your head.
> > I am willing to spend what is a considerable amount of time showing
> > you the basics, but there is no point if you have no interest in
> > actually listening.
>
> I know, and I am listening.
No, you aren't listening. You are not paying attention, and you are
not listening.
Stop fooling yourself into believing you are. *Everything* you have
said is complete and utter nonsense.
> Yikes. I had no idea I was this far off. Seriously I did study every
> line of your carefully typed out work.
Then your efforts are lacking and insufficient. Your continued
insistence that you understand and that your nonsensical comments
about dimensions 'stand' either put the lie to your claim of studying
"every line" of what I wrote, or demonstrate that you simply have not
put the effort necessary into understanding them (or that you lack the
ability to do so, but that seems a bit hard to believe).
> So far all of
> your indignance is lacking a direct error on my part. Please select
> some text that you wish to identify as conflicted on my part.
*EVERYTHING* you have said is nonsense. I set everything up, but have
not even begun to deal with quotients or polynomials. But you think I
did. Demonstrating that you do not know what you are talking about.
*EVERYTHING* you have said about "dimensions", "polynomials", X,
coefficients, etc., lacks foundation and is "conflicted" (when not
wrong, when not incoherent, when not utterly nonsensical).
> I have no idea why you do not simply
> concede the point about the higher dimensional product.
Because, dear, it's NONSENSE. That's why I don't "concede it". Because
it's complete and utter *NONSENSE*. And the fact that you insist on it
is precisely why you aren't worth the effort.
> While I deeply
> appreciate your detailed presentation Arturo I cannot remove my doubt.
Because you ain't listening, dear. That's why you cannot remove your
doubt. Because the willfully ignorant always remain ignorant, that's
why.
> I apologize for offending you.
You did not offend me personally; what you did was show contempt for
my time and waste my time. If you are to apologize for anything,
apologize for that. Good day, and have fun being an ignoramus.
--
Arturo Magidin
Bill Dubuque
>Bill Dubuque a ?crit :
>>Denis Feldmann <denis.feldm...@neuf.fr> wrote:
>>>Bill Dubuque a ?crit :
>>>>Axel Vogt <&nor...@axelvogt.de> wrote:
>>>>>
>>>>> I also vaguely remember the explanation that G/N is called the
>>>>> quotiemt group only because of the notation (looking like some
>>>>> qutient).
>>>>
>>>> I suspect rather that the choice of notation was motivated by the
>>>> "quotient-like" behavior given by the Third Isomorphism Theorem,
>>>>
>>>> namely A > B > C => (A/C)/(B/C) = A/B
>>>
>>> And isomorphisms of the shape A = (A/B) x B ? Products are actually
>>> much more general than quotients, after all...
>>
>> I can't make any sense of what you wrote. What do you mean?
>
> and the ring product of 2Z (the even integers) and Z/2Z (the field with
> two elements) (the isomorphism being (2p, a) -> 2p +a, where a is 0 or 1)
No, you seem to be quite confused about the notion of "ring product".
What you claim above cannot possibly be true simply because nontrivial
ring products have zero divisors so they are never domains (such as Z).
E.g. in your 2Z x Z/2Z we have (2,0) * (0,1) = (0,0).
>> The reason I mention the fraction-like nature of the identity
>> in the Third Isomorphism Theorem is because this theorem is
>> essentially the universal property of quotient rings and thus
>> characterizes them. So it's right at the heart of the matter.
>
> Yes, I agree, but I still think that products come before quotients
Product and quotient structures are distinct concepts. There is
nothing at all conceptually prior about one vs. the other. Indeed,
as we saw above, domains are never closed under products but they
are closed under quotients by prime ideals.
--Bill Dubuque
Marc Olschok
> On Jun 6, 9:04 pm, Arturo Magidin <magi...@member.ams.org> wrote:
|_______________________________________________________________ we are
|> > going to use X to denote the tuple that has a 1 in the 1-entry, and
|> > 0's elsewhere;
|Here it was.
|
|>[...]
|> form. It is very easy to overlook the lack of specification of X in
|> reading this material and particularly that there is no answer to the
|> question
|> What is X?
P.S.: if you are uncomfortable with the tuple notation, you can also
write such sequences as maps f: N --> R from the natural numbers (with 0)
to the reals (or some other ring). In this notation, sum and product
of such maps are given by
(f+g)(n) = f(n) + g(n)
(fg)(n) = sum( f(i)g(n-i) | i=0..n )
Write R^N for the resulting ring of all sequences.
A map f: N --> R is a polynomial iff f(n) =/= 0 only for finitely many n.
In particular the map X defined via
X(1) = 1
X(n) = 0 for all n =/= 1
is a polynomial.
You can then show from this definition, that the k-th power X^k of X
indeed satisfies
X^k(k) = 1
X^k(n) = 0 for all n =/= k
Also there is an injective ring homomorphism const: R --> R^N
defined by
const(r)(0) = r
const(r)(n) = 0 for all n =/= 0
which allows us to identify R with the image of const(R).
Then R[X] can also be described as the smallest subring of R^N that
contains R and X, which probably also motivated the notation.
Anyway, the main point is that the "X" in R[X] is just an ordinary
polynomial and not something "unspecified" thing from outer (math) space.
--
Marc
Mariano Suárez-Alvarez
But the same works if you use the "difference" notation :-)
(A - C) - (B - C) = A - C
just as well!
-- m
Arturo Magidin
<mariano.suarezalva...@gmail.com> wrote:
> > I suspect rather that the choice of notation was motivated by the
> > "quotient-like" behavior given by the Third Isomorphism Theorem,
>
> > namely A > B > C => (A/C)/(B/C) = A/B
>
> But the same works if you use the "difference" notation :-)
>
> (A - C) - (B - C) = A - C
>
> just as well!
You mean "A-B", surely... (-:
--
Arturo Magidin
Denis Feldmann
> Denis Feldmann <denis.feldm...@neuf.fr> wrote:
>> Bill Dubuque a ?crit :
>>> Denis Feldmann <denis.feldm...@neuf.fr> wrote:
>>>> Bill Dubuque a ?crit :
>>>>> Axel Vogt <&nor...@axelvogt.de> wrote:
>>>>>> I also vaguely remember the explanation that G/N is called the
>>>>>> quotiemt group only because of the notation (looking like some
>>>>>> qutient).
>>>>> I suspect rather that the choice of notation was motivated by the
>>>>> "quotient-like" behavior given by the Third Isomorphism Theorem,
>>>>>
>>>>> namely A > B > C => (A/C)/(B/C) = A/B
>>>> And isomorphisms of the shape A = (A/B) x B ? Products are actually
>>>> much more general than quotients, after all...
>>> I can't make any sense of what you wrote. What do you mean?
>>> Perhaps concrete examples would serve to clarify your remarks.
>> That, for instance, there is an isomorphism between Z (the integers)
>> and the ring product of 2Z (the even integers) and Z/2Z (the field with
>> two elements) (the isomorphism being (2p, a) -> 2p +a, where a is 0 or 1)
>> Of course, this works well only for algebraic quotient structures, but...
>
> No, you seem to be quite confused about the notion of "ring product".
Yes , sorry about that. I was not thinking. For *abelian groups*,
though, it still works (but perhaps only because sums and products are
quite confused in this case)
> What you claim above cannot possibly be true simply because nontrivial
> ring products have zero divisors so they are never domains (such as Z).
> E.g. in your 2Z x Z/2Z we have (2,0) * (0,1) = (0,0).
>
>>> The reason I mention the fraction-like nature of the identity
>>> in the Third Isomorphism Theorem is because this theorem is
>>> essentially the universal property of quotient rings and thus
>>> characterizes them. So it's right at the heart of the matter.
>> Yes, I agree, but I still think that products come before quotients
>
> Product and quotient structures are distinct concepts. There is
> nothing at all conceptually prior about one vs. the other.
Again, I was refering to terminology (ie reference to ordinary
arithmetic). For concepts, of course, I agree. But I still feel that,
somehow, simple quotient structure are "inverses" of product structures,
therefore the name...
Virgil
<14695621-ebef-45ae...@g20g2000vba.googlegroups.com>,
Mariano Su�rez-Alvarez <mariano.su...@gmail.com> wrote:
> > I suspect rather that the choice of notation was motivated by the
> > "quotient-like" behavior given by the �Third Isomorphism Theorem,
> >
> > �namely � A > B > C �=> �(A/C)/(B/C) = A/B
>
> But the same works if you use the "difference" notation :-)
>
> (A - C) - (B - C) = A - C
>
> just as well!
Don't you mean "(A - C) - (B - C) = A - B" ?
--
Virgil
Bill Dubuque
>>> the newer Springer edition.
>>
>>> Actually the notions are not unique (I learned 'quotient ring'
>>> in the sense of 'quotient field', i.e. localization)
>>
>> "Quotient ring" isn't ideal terminology for localizations either
>> because not every subring of a quotient field is a localization.
>> Indeed, this holds true iff the domain is a Prufer domain with
>> torsion class group - see the reviews in my prior post [1].
>>
>>> In German it is/was more common to call G/N a factor group and
>>> looking in a quite old van der Waerden "Algebra" he calls it
>>> "Restklassengruppe" ~ group of (class of) remainders in �8.
>>
>> Or residue(-class) groups/rings in English.
>>
>>> I also vaguely remember the explanation that G/N is called the
>>> quotiemt group only because of the notation (looking like some
>>> qutient).
>>
>> I suspect rather that the choice of notation was motivated by the
>> "quotient-like" behavior given by the Third Isomorphism Theorem,
>>
>> namely A > B > C => (A/C)/(B/C) = A/B
>
> But the same works if you use the "difference" notation :-)
>
Of course, but I was talking about _notation_, and as such,
the two-dimensional notation for fractions is better suited
to making obvious such cancellations and related algebra. The
right choice of notation can go a long way. Gian-Carlo Rota
often told many colorful stories about notation, some of which
I think survive in some of his published work.
--Bill Dubuque
Tim BandTech.com
So lets consider the product
( a0, a1, a2, 0, 0, ...) x ( b0, b1, b2, 0, 0, ... )
which has terms
( c0, c1, c2, c3, c4, 0, 0, ... )
where
c4 = a2 b2
right? You are not willing to consider this? Rather than go back to
your own definition of the product let's use your simple rules
(0,1,0,...,0,...) x (0,1,0,...,0,...) = (0,0,1,0,...,0,...)
(0,1,0,...,0,...) x (0,0,1,0,...,0,...) = (0,0,0,1,0,...)
which expose that
(0,0,1,0,...) x (0,0,1,0,...) = (0,0,0,0,1,0,...).
Here we are five terms into the series where c4 as a2b2 sits right? I
have no idea how I'm wrong on this but rather than calling me an
ignoramus it might be wiser to try and communicate on this particular
misconception. I understand that I need to respect your tuple notation
but I can't help going over to the interpretation as infinite
dimensional space which others have alluded to. I see no distinction
between this and the polynomial form
c0 + c1 X + c2 XX + c3 XXX + c4 XXXX
where we can see that the product
( a0 + a1 X + a2 XX )( b0 + b1 X + b2 XX )
will FOIL out into the five term form. Especially given that your
tuple notation is identical to a coordinate representation I have no
idea why this is so wrong. If X will never actually be evaluated then
the dimensional interpretation holds doesn't it? You refuse this
concept without ever providing any detail of why this interpretation
is wrong. Maybe I'm just jumping the gun. Beyond that all that I see
is you asking me to bow down to you here. That's fine. I bow down to
you. From my perspective I still see no error on my part and all that
I see from both you and Dennis is flat out denial without any
substance. I understand that you say I am speaking nonsense but it
would be more instructive to pick some words and really point out the
conflict directly rather than just shouting conflict!
- Tim
leland....@gmail.com
Sure, but where is the problem? The characterisation of dimension
isn't technically accurate, but even if we play a little fast and
loose, we have: an "infinite dimensional" vector multiplied by an
"infinite dimensional" vector giving and "infinite dimensional"
vector. No problems there. If you insist on thinking in terms of
dimension (which I only introduced because you were already asking
about dimension, and I attempted to provide caveats as to it's
technical inaccuracy) then please think of every element of R[X] as
being "infinite dimensional". There's no "increasing dimension" with
products because we're consistently in the "infinite dimensional"
case. Alternatively: sure, the product of two polynomials gives a
polynomial of higher degree, but why does degree have to remain the
same -- all we are asking for is that the product of two polynomials
is again a polynomial, and guess what: it is! Asking for anythign else
is asking for things that were neither given, nor expected.
Your problem, and what I believe is frustrating others, is your desire
to leap well ahead of yourself rather than dealing explicitly with
what has been presented. Arturo simply gave infinite tuples, and ways
to add and multiply them. Wanting to think of different infinite
tuples as having different dimension is leaping ahead of what was
presented, and getting it wrong in doing so. To then expect your
(incorrect) notion of dimension to have any behaviour with respect to
products, let alone some particular arbitrary one you've decided to
require, is to go so well beyond anything Arturo presented as to be
nonsense. Effectively you've taken what Arturo said, made up a bunch
of your own stuff on top of that, and then complained when your made
up ideas don't behave as you arbitrarily expected. How else is one to
respond but to say: well of course it doesn't behave as you expect,
its just nonsense that you made up.
You seem to have come to this with your own deep set ideas about what
should happen, and what things should mean. You need to wipe all of
that away, start with a blank slate, and read and work with only and
exactly with what is given.
Tim BandTech.com
I have not actually said that I have a problem with this feature.
I understand that polynomials behave this way.
I don't think it would be wise to wipe this feature under the carpet.
In fact rather than own this detail Arturo refuses to budge.
It seems that there will be no discussion.
Now, how did I get here? There is an example of a quotient ring which
'builds' the complex numbers.
R[X] / ( XX + 1 )
is the notation that is used. I did not understand this but have made
some progress. Still I have more work to do. Now Arturo comes along
with a long description of polynomials which has been helpful. I've
merely pointed out this feature. Particularly troubling is the usage
of notation by Arturo
(a_0,a_1,...,a_n,...) x (b_0,b_1,...,b_n,...)
= (c_0,c_1,...,c_n,...)
which obfuscates the detail that I've pointed out. The symmetry of
this form is misleading.
Rather than meaningless zeros (the first accusation of my
misunderstanding) I quote Arturo:
"it's just that we are lazy and we don't want to write
all those zeros (that don't really matter all that much
anyway). "
while we know because of the feature that I've pointed out that this
does not apply generally, though it can be used on the very first
element which is how Arturo uses it. Still, this 'nifty' feature is
offset by the feature that I've presented.
Rather than concede this point Arturo has declared all of my
discussion meaningless whereas I see a legitimate concern. This makes
me wonder about the validity of this branch of math. If a master is in
denial of this behavior then what else is he in denial of? A little
handwaiving here added to a little handwaiving somewhere else nearby
can become problematic. Arturo has dodged the X form altogether. I do
still see the usage of X as mysterious, until X is evaluated as a real
value or some such specific domain, at which point the dimensional
issue that I am worried about collapses, but until then X is very much
a dimensional construct, just as Arturo's tuple notation is. These are
shared features. That's why others on this thread have used them. So
Arturo wants me to follow himself and follow him perfectly I guess.
That is a mimicry stance. I say that I will be respectful but that the
dismissal of my criticism is a poor method, whereas instead Arturo
would have us believe that what he has written is perfect and thus
cannot be questioned, which I definitely dismiss as the mimicry
problem.
- Tim
leland....@gmail.com
> I have not actually said that I have a problem with this feature.
No, but you do talk about it as "a legitimate concern" and so on. Yes,
what you describe happens, in the sense that the degree of the product
of polynomials is the sum of the degrees of the polynomials, but so
what? Why is that a concern? Why worry? Why even bring it up? It has
no bearing at all on the soundness of what Arturo presented. It's a
little like explaining natural numbers and arithmetic via Peano axioms
to someone, and having them respond with "The sum of two even number
is even but the sum of two odd numbers is even as well! This would
seem to be an important issue that you didn't address. Perhaps all of
arithmetic is wrong!". You either get Arturo's sort of response: "Why
are you even talking about odd and even numbers, I didn't even mention
them! Look at what I actually said and you'll see it's quite sound.",
or mine: "Yes, that's true, so what?".
Drop your expectations, see that there is actually no problem in what
Arturo actually presented, only in your expectations of how you think
it ought to behave. Arturo presented nearly null infinite tuples and
shows how to add and multiply them, and showed that the set of such
nearly null infinite tuples is closed under those operations (you will
always get a nearly null infinite tuple out of any sum or product of
nearly null infinite tuples). That's all there is to it, and for now
that's all you need to worry about -- expecting any sort of behaviour
out of them is to presume too much. They behave as they behave and we
would have seen what that behaviour was had Arturo been willing to
continue.
> I understand that polynomials behave this way.
Yes, yes they do, and yet polynomials don't seem to be hobbled and
broken: they work just fine for adding and multiplying, so why is
there any problem with what Arturo presented? There isn't.
> I don't think it would be wise to wipe this feature under the carpet.
He wasn't, it's just irrelevant and taking a pointless tangent based
on your apparent expectations of how things should behave that had
little or no bearing on what was actually presented.
> In fact rather than own this detail Arturo refuses to budge.
No, he just feels you aren't willing to actually let him work through
the details of the whole construction. There was a lot more work to go
to eventually get to why R[X]/(X^2 + 1), but you are already raising
"concerns" about why you feel the whole thing is broken and doesn't
work based on your own introduced irrelevancies rather than just
letting Arturo develop the ideas. I suspect he came to the view that
you would simply spend your time derailing the whole discussion into
what you see as "deep problems" that have little or nothing to do with
what Arturo actually presented, but rather your perceptions of how
things ought to behave (which are more than likely wrong since, as you
admit, you don't actually understand any of this). I also suspect your
arrogance in the matter was the final straw: you, who admit you don't
understand any of this, presume that there are problems and the whole
thing may be fundamentally broken and we are all just parrotting
texts. If you actually understood it and made some relevant arguments
not based on faulty understandings and assumptions that would be one
thing, but you aren't doign that. You are waving your lack of
understanding out as a failure of the mathematics. Have at least a
little faith is the thousands and thousands of people who *have*
studied and understood this subject (and I assure you, didn't just
parrot it from books). If something seems wrong to you, it is most
likely because you are wrong, or misunderstanding, or brining your own
expectations into things.
> It seems that there will be no discussion.
Well I suspect he simply doesn't want to get into protracted arguments
about material that is not even what he presented.
> Now, how did I get here? There is an example of a quotient ring which
> 'builds' the complex numbers.
> R[X] / ( XX + 1 )
> is the notation that is used. I did not understand this but have made
> some progress. Still I have more work to do.
And yet you presume to know more than those who are trying to explain
it to you. This is a very off putting attitude on your part.
> Now Arturo comes along
> with a long description of polynomials which has been helpful. I've
> merely pointed out this feature.
Sure, and it has nothing to do with the soundness of what Arturo
presented. You are expecting it to behave in some way and it doesn't.
The problem is in your expectations, not in the behaviour of the
construction. I'm sorry that it doesn't behave as you wish it would --
well, not really, because then it wouldn't work the way it actually
does which is very useful and powerful.
> If a master is in
> denial of this behavior then what else is he in denial of? A little
> handwaiving here added to a little handwaiving somewhere else nearby
> can become problematic. Arturo has dodged the X form altogether. I do
> still see the usage of X as mysterious, until X is evaluated as a real
> value or some such specific domain, at which point the dimensional
> issue that I am worried about collapses, but until then X is very much
> a dimensional construct, just as Arturo's tuple notation is. These are
> shared features. That's why others on this thread have used them. So
> Arturo wants me to follow himself and follow him perfectly I guess.
> That is a mimicry stance. I say that I will be respectful but that the
> dismissal of my criticism is a poor method, whereas instead Arturo
> would have us believe that what he has written is perfect and thus
> cannot be questioned, which I definitely dismiss as the mimicry
> problem.
This is the arrogance I spoke of. It isn't behaving as you thought,
because you don't understand enough about the whole yet to have any
understanding of what you should be expecting. Based on this failure
of your expectations, you presue the entire subject is riddled with
contradiction and we are simply mimicing the errors that are cast in
stone. Do you not think that those of us who do understand the subject
actually bother to check that everything works as we learn it? That we
don't revisit it as we recall it, work with it, or try and explain it
to others? I know this may come as a shock, but we are not, in fact,
mental dwarfs in comparison to you, and can actually do mathematics on
our own without blindly parrotting books. Arutro is just frustrated
with your apparent impatience with his explanations and your desire to
take what he laid out and fit it into whatever misconcieved framework
you have in mind. He could have been more polite or more helpful about
it, but these things happen.
I am willing to keep trying to explain things, but you have to accept
that things may not be even remotely close to your expectation of how
things ought to work.
Tim BandTech.com
OK Leland. I accept the attitude accusation.
Somehow the quotient ring acts as a modulo operator through an ideal
right?
So that getting out say a 3D algebra out of the polynomial ring R[X]
would require a mod 3 type of operation put into this X^n nomenclature
as x ^ ( n mod 3 ). Geeze, already I don't like what I've just said
because the notation R[X] means Ring X right? Therefore R[X] should
mean merely the singular value X, whereas really we're discussing
R[ sum over n from 0 to inf ( a(n) X^n )]
right? This line reads ring of polynomials in X. I can get over that
if what I am saying is accurate, but if this is inaccurate then
obviously I am misunderstanding something. Nobody has bothered to
respond to this point and it has added to my confusion.
- Tim
Arturo Magidin
> On Jun 9, 8:57 am, leland.mcin...@gmail.com wrote:
>
> > On Jun 9, 8:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> > I am willing to keep trying to explain things, but you have to accept
> > that things may not be even remotely close to your expectation of how
> > things ought to work.
>
> OK Leland. I accept the attitude accusation.
> Somehow the quotient ring acts as a modulo operator through an ideal
> right?
No; there is no "modulo operator"; the quotient ring is a particular
structure defined in terms of a given ring and a given ideal. You are
not yet at a stage where you can understand that, since you are having
trouble understanding what the ring R[X] is, so you should certainly
not begin there. You can't discuss the finer points of the the
specific words originally used by Tolstoi in "War and Peace" if you
don't even know the cyrillic alphabet, but that's what you are trying
to do.
> So that getting out say a 3D algebra out of the polynomial ring R[X]
> would require a mod 3 type of operation put into this X^n nomenclature
> as x ^ ( n mod 3 ). Geeze, already I don't like what I've just said
> because the notation R[X] means Ring X right?
No.
> Therefore R[X] should
> mean merely the singular value X, whereas really we're discussing
> R[ sum over n from 0 to inf ( a(n) X^n )]
> right?
No.
R[X] means "the ring whose elements are *all* the almost null
sequences with entries in R, with coordinate-wise addition and
multiplication as defined earlier". It has nothing to do with
"singular value X", nor with "sum over n from 0 to inf ( a(n) X^n )". R
[X] is the *name* of the set of almost null sequences with addition
and multiplication as defined, just like "R" is the name of the ring
of all real numbers, and not a particular real number, and not a
synonym for "dimension 1", and not anything else.
> This line reads ring of polynomials in X. I can get over that
> if what I am saying is accurate, but if this is inaccurate then
> obviously I am misunderstanding something.
What you say is wildly inaccurate. You are misunderstanding more than
one thing. Starting with the notion of "dimension". Dimension can mean
one thing in geometry, something else in linear algebra, something
else in topology; and *none* of those meanings applies to the ring
structure you are trying to understand; the linear algebra meaning
*can*, by a suitable point of view, have *some* bearing, but you need
to know the thing you are looking at before you can take the "suitable
point of view". You aren't there yet, by a long shot.
> Nobody has bothered to
> respond to this point and it has added to my confusion.
Your first confusion lies in confusing polynomials with polynomials
functions. Your second confusion lies in confusing a particular
polynomial with the ring of *all* polynomials (a bag of apples is not
an apple itself).
Your third confusion is misunderstanding the notion of
"dimension" (from linear algebra), and trying to apply it where it
does not apply (rings and ideals); to give you an idea of what you are
arguing, you are arguing that there is something fundamentally weird
and wrong about vectors on the plane, because (1,0) is "1-
dimensional", (0,1) is also "1-dimensional", and when you add them you
get (1,1) which is "2-dimensional". It's all nonsense from beginning
to end, because "dimension" is a property of vector spaces, not
individual vectors, and if they are to be applied to vectors at all
then one would have to say that all of (1,0), (0,1), (0,0), (1,1), and
in general (a,b) for arbitrary real numbers a and b are "2-dimensional
objects" (though much better to call them by their proper name: they
are *ordered pairs*, they have two *entries*). Likewise, you are
confusing the *degree* of a polynomial (which has not yet been
defined) with the notion of 'dimension'. Polynomials, as almost null
sequences, "live" in the set of all almost null sequences. As a vector
space, this space is *infinite dimensional* over the real numbers, and
"dimension" applies to the vector space as a whole, not to its
elements. Individual vectors do not have "dimension", vector spaces
do. And individual polhynomials do not have "dimension", they have
*degree*, which is something else entirely.
Your next confusion is believing that nobody has bothered to respond
to your points, when many people have: you just haven't understood the
responses because you are misusing the words and misunderstanding the
concepts. For example, at least three times I pointed out that you are
misusing "dimension" and confusing "degree of a polynomial" with
dimension.
You keep wanting to fly, when you are having trouble crawling, and
blaming those who would try to help you stand up.
--
Arturo Magidin
Tim Little
> Geeze, already I don't like what I've just said because the notation
> R[X] means Ring X right?
No, the R in R[X] means Real. The X is just a formal parameter, with
primary purpose just being to distinguish which terms are which.
By comparison Z[X] would be the ring of polynomials over X having
integer coefficients (i.e. coefficients in Z).
- Tim Little
(the other Tim)
leland....@gmail.com
> On Jun 9, 8:57 am, leland.mcin...@gmail.com wrote:
>
> > On Jun 9, 8:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> > I am willing to keep trying to explain things, but you have to accept
> > that things may not be even remotely close to your expectation of how
> > things ought to work.
>
> OK Leland. I accept the attitude accusation.
> Somehow the quotient ring acts as a modulo operator through an ideal
> right?
On some level yes, presuming you have the right understanding of
modulo, and are clear on what that means with respect to an ideal. I
am under the strong impression that you do not have the right
understanding, and don't know enough about ideals to know where to
begin, so let's go with: No, not at all. However you are thinking it
works, it doesn't work like that -- you need to properly understand
rings and ideals and equivalence classes to think about it in terms of
modulo operators, and it will only cause confusion and
misinterpretation on your part to try and think of it that way. It is
best if you completely ignore such ideas.
But you are getting well ahead of yourself anyway. It isn't clear at
all that you've actually even properly digested what the ring R[X] is,
so trying to move on the quotient ring is only going to lead to even
deeper confusion.
> So that getting out say a 3D algebra out of the polynomial ring R[X]
> would require a mod 3 type of operation put into this X^n nomenclature
> as x ^ ( n mod 3 ).
Not in the least. It is, in fact, nothing like that at all. You've
started with a poor guess at how you think things should work, and
have quickly progressed to building on top of those preconceptions
despite my warnings about jumping ahead not working with what is
actually explained to you.
> Geeze, already I don't like what I've just said
> because the notation R[X] means Ring X right?
No, it doesn't. Which was my point earlier -- you haven't bothered to
actually understand the first construction and are already worrying
about far more complex constructions that the people who are trying to
explain this to you haven't even touched yet.
> Therefore R[X] should
> mean merely the singular value X,
I'm not sure where you got that impression. Certainly not from
anything anyone here has told you.
> whereas really we're discussing
> R[ sum over n from 0 to inf ( a(n) X^n )]
> right?
No, that's not right either. You're obviously deeply confused about
all of this. I suggest you step back and stop making guesses about how
it all works and instead actually read what people write.
> This line reads ring of polynomials in X.
Well no, it doesn't, but this is just poor and confused use of
notation on your part. Perhaps this part I can clear up relatively
easily. In the notation 'S[Y]' we are referring to a particular ring,
which is the ring of polynomials in Y with coefficients in S. The
symbol 'S' should refer to a ring, it could be the real numbers, if
could be the integers, it could the finite field of 4 elements. The
square brackets '[' and ']' mean that we are now referring to the ring
of polynomials with coefficients in S. That is, 'S[.]' means the ring
formed by taking *all* nearly null infinite tuples where entries in
the tuple are elements of the "base ring" S. The 'Y' denotes the
formal symbol we will be using for book-keeping in the polynomials*.
The 'Y' is arbitrary: we are simply picking out which symbol we will
be using in our polynomials and is nothing ore than a symbol*.
Thus, in our case, R[X] refers specifically to the ring formed by
taking all nearly null infinite tuples with real number entries (since
we are assuming R here refers to the real numbers). The X in the
notation is because mathematicians often prefer to work in polynomials
rather than nearly null infinite tuples (and Arturo showed they were
equivalent) and we need to state which symbol we'll be using in our
polynomials. We could just as well write R[Y] or R[A] and be referring
to exactly the same ring -- the only change would be that we would
write the polynomials using Y or A instead of X. If we wrote, say, Z
[X] however, we would be referring to something completely different,
since the 'S' symbol denotes an actual object, and here Z would mean
the integers. Thus, you see, the R matters and refers to something
concrete, but the X is simply a formal symbol, for which we could
substitute anything. Most importantly the '[]' part means we are
referring to the 'ring of polynomials with coefficients in ...' or,
equivalently, the 'ring of all nearly null infinite tuples with
entries in ...'. Does this make more sense now? I don't think we
should move on to anything more complicated until you fully understand
at least what the notation even means.
> I can get over that
> if what I am saying is accurate,
The key being that none of what you said was even remotely accurate,
and was simply your own speculation that has little or nothing to do
with what anyone has actually told you.
> but if this is inaccurate then
> obviously I am misunderstanding something.
You are very obviously misunderstanding a lot of things, the problem
being that you are not even understanding the most basic points, but
are busy trying to understand the things built from those basic points
anyway, and simply confusing yourself even more. Look, this is not
something simple that you can expect to understand overnight unless
you have some appropriate background. It is becoming clear to me that
you don't have any of the appropriate background, and are trying to
leap ahead anyway. Can I ask if you have a clear understanding of what
a ring is? Could you give me a few examples of different rings to show
what you understand to be the full range of diversity in what a ring
could be like? Please take this seriously and spend the time to come
up with good examples as it will help me establish in what terms I
should actually be explaining things so as to not add to your
confusion.
> Nobody has bothered to
> respond to this point and it has added to my confusion.
It isn't even clear to me what your "point" is, since you rattled off
a whole list of misunderstandings right across the range from basic
notation through to what a quotient by an ideal should mean. I've
decided to pick on notation for now, since if we can't get that
straight then clearly the rest is beyond hope.
* (For the pedants, not Tim: Yes, in some situations the R[a] notation
has 'a' not being a formal symbol but rather an element from some ring
S (containing R) to adjoin to the ring, but I figured it's best not to
confuse that particular point right now)
Tim Little
> right?
Yes, though you may need to be careful exactly what you mean by
"modulo operator". Each element of a quotient ring A/B is a *set*,
not an element of A. In particular, the elements of the quotient ring
are not remainders like those produced the "modulo operator" in
computing.
So, for example, the set {..., -12, -5, 2, 9, ...} is a single element
of the quotient ring Z_7 (given by the quotient Z / (7)). For
notational convenience these sets are often labelled using one
particular element from the set. E.g. that element may be labelled
variously as "[2]_7", "2" with a bar over the top, "2+7Z", or even
simply "2" if it is felt that context is sufficient to distinguish it
from the integer 2.
Each element of R[x]/(x^2+1) is an uncountable set of polynomials.
For example the polynomial x is in one of those sets - and the
polynomials x^2+x+1 and 2x^3+3x are also in the same set.
- Tim
Tim BandTech.com
> On Jun 9, 4:39 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> Well no, it doesn't, but this is just poor and confused use of
> notation on your part. Perhaps this part I can clear up relatively
> easily. In the notation 'S[Y]' we are referring to a particular ring,
> which is the ring of polynomials in Y with coefficients in S. The
> symbol 'S' should refer to a ring, it could be the real numbers, if
> could be the integers, it could the finite field of 4 elements. The
> square brackets '[' and ']' mean that we are now referring to the ring
> of polynomials with coefficients in S. That is, 'S[.]' means the ring
> formed by taking *all* nearly null infinite tuples where entries in
> the tuple are elements of the "base ring" S. The 'Y' denotes the
> formal symbol we will be using for book-keeping in the polynomials*.
> The 'Y' is arbitrary: we are simply picking out which symbol we will
> be using in our polynomials and is nothing ore than a symbol*.
Ohhhh... I'm sorry. So when discussing
" the ring R "
that means the real numbers right? I got a neg. on this back a while
so that everything I read with an R in it I was confused on whether it
was R the ring or R the reals.
Thanks Leland for at least getting me past this [] usage. I've been
reviewing
http://www.math.uiuc.edu/~r-ash/Algebra.html
which is a link you gave me awhile back. He defines the quotient ring
in terms of the group ring, which is OK, but there is a list of
vocabulary that feels tentative to me so that working through it all
with confidence is not happening.
>
> Thus, in our case, R[X] refers specifically to the ring formed by
> taking all nearly null infinite tuples with real number entries (since
> we are assuming R here refers to the real numbers). The X in the
> notation is because mathematicians often prefer to work in polynomials
> rather than nearly null infinite tuples (and Arturo showed they were
> equivalent)
Sure, I see this.
> and we need to state which symbol we'll be using in our
> polynomials. We could just as well write R[Y] or R[A] and be referring
> to exactly the same ring -- the only change would be that we would
> write the polynomials using Y or A instead of X. If we wrote, say, Z
> [X] however, we would be referring to something completely different,
> since the 'S' symbol denotes an actual object, and here Z would mean
> the integers. Thus, you see, the R matters and refers to something
> concrete, but the X is simply a formal symbol, for which we could
> substitute anything. Most importantly the '[]' part means we are
> referring to the 'ring of polynomials with coefficients in ...' or,
> equivalently, the 'ring of all nearly null infinite tuples with
> entries in ...'. Does this make more sense now? I don't think we
> should move on to anything more complicated until you fully understand
> at least what the notation even means.
Yes, I understand. We could construct R[R] and nearly have grade
school polynomials.
>
> > I can get over that
> > if what I am saying is accurate,
>
> The key being that none of what you said was even remotely accurate,
> and was simply your own speculation that has little or nothing to do
> with what anyone has actually told you.
>
> > but if this is inaccurate then
> > obviously I am misunderstanding something.
>
> You are very obviously misunderstanding a lot of things, the problem
> being that you are not even understanding the most basic points, but
> are busy trying to understand the things built from those basic points
> anyway, and simply confusing yourself even more. Look, this is not
> something simple that you can expect to understand overnight unless
> you have some appropriate background. It is becoming clear to me that
> you don't have any of the appropriate background, and are trying to
> leap ahead anyway. Can I ask if you have a clear understanding of what
> a ring is? Could you give me a few examples of different rings to show
> what you understand to be the full range of diversity in what a ring
> could be like? Please take this seriously and spend the time to come
> up with good examples as it will help me establish in what terms I
> should actually be explaining things so as to not add to your
> confusion.
Yeah, a ring obeys some fundamental algebraic properties, which are
not necessarily tied to the actual shape of a ring, as in an 'O'
graphical representation. So getting
a ( b + c ) = a b + a c
is consistent with a ring behavior regardless of what domains a,b,and
c are in. From
http://www.math.uiuc.edu/~r-ash/Algebra/Chapter2.pdf
"A ring R is an abelian group with a multiplication operation
(a, b) → ab
that is associative and satisfies the distributive laws:
a(b + c) = ab + ac
and
(a + b)c = ab + ac
for all a, b, c ∈ R."
Darn it, there is that usage of R not as the real numbers but as a
generic right?
I'll tell you what, for generations of evolution to clean up
formalities, this nomenclature is poor. This does worry me. Anyway, at
some level we are free to define as we go and for this reason I will
defer to you and your interpretation on R[X] as you did with the fine
instance S[Y], which seems very sensible since there are then three
items in the description 'ring of polynomials in Y with coefficients
in S', rather than this book quote which puts R as meaning Ring which
lessens the descriptive content by one item.
Anyway, here we see an advancement past the group stage, where just
one binary operation necessarily exists. Now we have two operators,
traditionally sum and product, though as I understand it these can be
'overloaded' to be any consistently defined behavior with the
requirements as defined within the term 'ring'.
The reals are a ring but not all rings are reals.
The integers are a ring.
Modulo forms from the integers are rings, whether you eliminate
redundancies or not.
The link gives a set of rules of an R F form with inverses that forms
a ring.
The polysign numbers are rings.
Particularly, these last have been described in terms of the quotient
ring most formally by Hagen von Eitzen
http://www.von-eitzen.de/math/PolysignNumbers.pdf
but I have yet to understand that form. Close by is the
R[X] / ( XX + 1 )
form of the complex numbers, though the polysign build the complex
numbers as a mod 3 form rather than a mod 2 form. I do not mean to
promote polysign here but this is my true motive for understanding the
quotient ring form and helps explain my own resistance to the quotient
form. Especially the interpretation of whether the quotient form truly
constructs number systems or merely in hindsight is consistent is of
interest to me, but breaking through to understanding the ring
nomenclature cleanly is difficult. If I can get through it then I
might attempt a magnitudinal form of lets say M[X] where M are
unsigned magnitudes which will then satisfy the ring requirements and
commutative behavior as well, though the polynomial form will have a
new behavior after the modular quotient effect where for say, the mod
3 system
( 1, 1, 1 ) = 0
in the tuple form which extends generally to a mod n system.
>
> > Nobody has bothered to
> > respond to this point and it has added to my confusion.
>
> It isn't even clear to me what your "point" is, since you rattled off
> a whole list of misunderstandings right across the range from basic
> notation through to what a quotient by an ideal should mean. I've
> decided to pick on notation for now, since if we can't get that
> straight then clearly the rest is beyond hope.
>
> * (For the pedants, not Tim: Yes, in some situations the R[a] notation
> has 'a' not being a formal symbol but rather an element from some ring
> S (containing R) to adjoin to the ring, but I figured it's best not to
> confuse that particular point right now)
Thanks Leland for putting up with my nonsense and for putting forth
this energy.
I do not see some of the finer distinctions you make about the
dimension discrimination but you concede that it is consistent
somewhat so that is good enough. I tend to think informationally where
format is of the essence so that
( 1.23, 2.34, 3.45 )
as a tuple or as a coordinate system still confers the same quantity
of information. Still, I work in a format where this is tweaked to be
one less dimension since
( 1, 1, 1 ) = 0
so I can concede a difference yet I am also able to substantiate how
the difference exists whereas in the indeterminate polynomial form I
cannot see any discrepancy informationally between a dimensional
interpretation and the tuple or polynomial representation. Yes, the
product is uniquely defined for the polynomial form, but that form is
still consistent with vector principles, up to the discrepancy which I
pointed out which requires the general form to remain in (super)
infinite dimension.
- Tim
Tim BandTech.com
OK Tim. Jeeze, it's like I'm talking to myself...
Umm... I guess someone would quibble with you about leaving the zeroes
off the end of your instances of polynomials. Still, this the funny
thing about the construction. For instance with the polynomial ring
quotient construction it is possible to build n dimensional spaces
consistent with the ring requirements, yet the underlying components
will not play out this way with a limit like n since there will be
terms up to 2n-1 in the product domain so that to build a finite form
is not possible until the quotient is taken. Thus the instances you
describe must already be quotient forms within the formality and so
their products would then wrap at some dimension as well. Somewhere in
this hairy nuance different people are taking differing
interpretations and I've got to try and stay in one context so I'm
going to try to follow Leland for awhile. He does not like the modulo
interpretation at all but I am caught in it for now. Still, I've got
to compartmentalize more. Somewhere mixed up in this is the
indeterminate form and so long as there is no definition of X within
your polynomials there is a dual haze for me. Upon collapsing to a
real x everything collapses and I am told you cannot do that with this
form.
- Tim
Jeeze, now it's like you're talking to myself...
Arturo Magidin
> From
> http://www.math.uiuc.edu/~r-ash/Algebra/Chapter2.pdf
> "A ring R is an abelian group with a multiplication operation
> (a, b) → ab
> that is associative and satisfies the distributive laws:
> a(b + c) = ab + ac
> and
> (a + b)c = ab + ac
> for all a, b, c ∈ R."
>
> Darn it, there is that usage of R not as the real numbers but as a
> generic right?
> I'll tell you what, for generations of evolution to clean up
> formalities, this nomenclature is poor. This does worry me.
As many people have told you (myself at least twice), the common
nomenclature for the real numbers is either a boldface "R" or a
"double struck" or "blackboard bold" "R". You can see the double
struck typeface at http://www.w3.org/TR/MathML2/double-struck.html
The symbol for the real numbers is unicode 0211D.
However, you are posting in USENET, which uses ASCII. ASCII has
neither boldface nor doublestruck, nor other such alternative
typefaces, so it is usual to use a simple capital "R", and specify
explicitly (as we did) that we are using that letter to denote the
real numbers.
It is also common, when refering to generic rings, to use either a
capital (roman or italic) "R" or "A" (for 'ring', or for 'annel', the
french word for ring, respectively). There is normally no possible
cause for confusion since the typeface of the letter is different. In
the context of Usenet there is the possibility of confusion, which is
why one would say things like "Let R be a ring" or "Let R be the real
numbers", to be very clear what it is that "R" denotes.
There is very little chance for problems in any case, when one knows
the context at work. There are only finitely many letters, but
infinitely many concepts; we cannot have a one-to-one correspondence
between notation and ideas, except by compartmentalizing them. So the
word "normal" has precise technical meanings, but different precise
technical meanings in different contexts. Just like "negative" means
one thing in medicine and something completely different in
arithmetic. Is that also an example of "poor nomenclature" that
"troubles you"?
But we are so pleased that you could come along and point out all
those problems that you are perceiving; clearly, the entire field of
mathematics is full of idiots who can't see past their noses, it took
a genius of your caliber, one who by self-admissions knows *nothing*
about the subject to point out the problems. So glad you could come
along.
--
Arturo Magidin
Mariano Suárez-Alvarez
You have managed to write 218 words which mean
absolutey nothing. Good job.
-- m
Mariano Suárez-Alvarez
No you do *not* understand. That you write "R[R]" is
a great proof of that!
-- m
Tim BandTech.com
> On Jun 9, 11:34 pm, leland.mcin...@gmail.com wrote:
>
> > You are very obviously misunderstanding a lot of things, the problem
> > being that you are not even understanding the most basic points, but
> > are busy trying to understand the things built from those basic points
> > anyway, and simply confusing yourself even more. Look, this is not
> > something simple that you can expect to understand overnight unless
> > you have some appropriate background. It is becoming clear to me that
> > you don't have any of the appropriate background, and are trying to
> > leap ahead anyway. Can I ask if you have a clear understanding of what
> > a ring is? Could you give me a few examples of different rings to show
> > what you understand to be the full range of diversity in what a ring
> > could be like? Please take this seriously and spend the time to come
> > up with good examples as it will help me establish in what terms I
> > should actually be explaining things so as to not add to your
> > confusion.
> The integers are a ring.
> Modulo forms from the integers are rings, whether you eliminate
> redundancies or not.
> The link gives a set of rules of an R F form with inverses that forms
> a ring.
above back.
Tim Little
> Ohhhh... I'm sorry. So when discussing
> " the ring R "
> that means the real numbers right?
Sometimes it just means any ring. Context helps in determining which
is intended, and many typeset texts use a doubled-line symbol for R
when denoting the reals (looking a bit like "|R").
In the context of going from reals to complex numbers though, it is
much more obvious that the R in R[x] denotes the reals and not some
arbitrary ring.
> I'll tell you what, for generations of evolution to clean up
> formalities, this nomenclature is poor. This does worry me.
There is only a fairly small set of letters and other symbols easily
available to publishers (especially historically), and enormously many
mathematical concepts to represent using those symbols. Multiplicity
of assigned meanings to symbols is inevitable.
Also many distinct fields that originally developed independent
symbolism later came into collision as links were developed between
them. Mathematics is like that. There has not just been generations
to clean up nomenclature, there has been generations of developing new
concepts requiring new nomenclature and cross-connecting previous
ones.
Any good textbook or paper will define meanings of any potentially
ambiguous symbols before they are used.
> Especially the interpretation of whether the quotient form truly
> constructs number systems or merely in hindsight is consistent is of
> interest to me,
Constructing rings via quotients of polynomials to produce novel
algebraic structures has been of great practical use in developing
many of the error-correcting codes in use today - as one small example
of a great many applications.
- Tim (Little)
Tim Golden
Yeah, and that was a short post for me.
- Tim
Tim Little
I must say I did a double-take once some time ago when reading one of
your posts and coming to the last line where you signed your name in
exactly the same way I normally do. A moment of "did I write that!?"
> Umm... I guess someone would quibble with you about leaving the zeroes
> off the end of your instances of polynomials.
Quite possibly; I'm sure that on Usenet there is always someone who
can quibble about nearly anything.
> Still, this the funny thing about the construction. For instance
> with the polynomial ring quotient construction it is possible to
> build n dimensional spaces consistent with the ring requirements,
I really don't know where you're getting this "dimension" thing. You
appear to be using it in the sense of dimension of a real vector
space. In this particular case the space of real polynomials does
satisfy the axioms of a real vector space, but in that case it has
infinite dimension, not any natural number n. It isn't a very useful
concept here.
> yet the underlying components will not play out this way with a
> limit like n since there will be terms up to 2n-1 in the product
> domain so that to build a finite form is not possible until the
> quotient is taken.
Yes, the particular quotient R[x]/(x^2+1) can also be interpreted as a
real vector space of dimension 2, but again dimensionality is not a
very useful concept in the context of quotient rings.
> Thus the instances you describe must already be quotient forms
> within the formality and so their products would then wrap at some
> dimension as well.
The polynomial instances I mentioned are two example members of a
single element of the quotient ring. The quotient does not have
polynomials as elements of the ring, it has infinite sets (equivalence
classes) of polynomials.
In general, if A is a ring and I is an ideal of that ring, then you
can divide up A into equivalence classes so that elements x and y from
A are in the same class exactly when x-y is in I. Each class is an
element of the quotient ring A/I.
- Tim (Little)
Dik T. Winter
...
> If I can get through it then I
> might attempt a magnitudinal form of lets say M[X] where M are
> unsigned magnitudes which will then satisfy the ring requirements and
> commutative behavior as well,
M does not have all the required ring properties. It is not a group
under addition.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Tim Golden BandTech.com
> On 2009-06-10, Tim BandTech.com <tttp...@yahoo.com> wrote:
>> OK Tim. Jeeze, it's like I'm talking to myself...
>
> I must say I did a double-take once some time ago when reading one of
> your posts and coming to the last line where you signed your name in
> exactly the same way I normally do. A moment of "did I write that!?"
>
>
>> Umm... I guess someone would quibble with you about leaving the zeroes
>> off the end of your instances of polynomials.
>
> Quite possibly; I'm sure that on Usenet there is always someone who
> can quibble about nearly anything.
>
>
>> Still, this the funny thing about the construction. For instance
>> with the polynomial ring quotient construction it is possible to
>> build n dimensional spaces consistent with the ring requirements,
>
> I really don't know where you're getting this "dimension" thing. You
> appear to be using it in the sense of dimension of a real vector
> space. In this particular case the space of real polynomials does
> satisfy the axioms of a real vector space, but in that case it has
> infinite dimension, not any natural number n. It isn't a very useful
> concept here.
There is an insistence on remaining in the 'indeterminate form' whereby
X is never evaluated. In effect X is undefined. But now that I've read
on and it's fairly obvious anyway the polynomial factors
a X + b
are evaluatable aren't they? Is it true that in polynomial of real
coefficients that X is real? This is the only way to evaluate a factor
isn't it? Yet this will then collapse any polynomial into a real value
for any X. We could then procede to simply study
y / ( x^2 + 1 )
with x and y real couldn't we? We'd eventually get to a square route of
a negative value and be forced to concede i in the traditional manner.
This seems to be in the interpretations on the indeterminate form
anyway. At some point they introduce i.
>
>
>> yet the underlying components will not play out this way with a
>> limit like n since there will be terms up to 2n-1 in the product
>> domain so that to build a finite form is not possible until the
>> quotient is taken.
>
> Yes, the particular quotient R[x]/(x^2+1) can also be interpreted as a
> real vector space of dimension 2, but again dimensionality is not a
> very useful concept in the context of quotient rings.
Well, this is troubling that you say this. I have worked out some math
which develops dimensional behaviors without the need of any cartesian
product. This math has been expressed in terms of ring theory by Hagen
von Eitzen through a ring quotient and nobody has found any fault with
that math. This is somewhat how I arrive at puzzling this all out. I'd
like to understand how dimension is developed through this math. So back
at my o.p. I ask specifically is this ring quotient considered an actual
construction or is it more like a hindsight type of thing where a given
construction fits into this form and behaves consistently with it? I
understand that the polysign numbers are rings. I have no problem with
that. But can the ring quotient form build the polysign numbers from a
more primitive ring? The example of the complex numbers as
R[X] / ( X X + 1 )
is nearly an identical but more familiar instance. Whether the ring form
builds a number system or whether it merely reflects a number system's
behaviors is crucial to understanding this nomenclature.
This is all about dimension to me. Still, if you pick a domain for X
then I understand the dimension of the classes and elements being
discussed should collapse to that dimension assuming that the arithmetic
product XX is in the same domain as X.
>
>
>> Thus the instances you describe must already be quotient forms
>> within the formality and so their products would then wrap at some
>> dimension as well.
>
> The polynomial instances I mentioned are two example members of a
> single element of the quotient ring. The quotient does not have
> polynomials as elements of the ring, it has infinite sets (equivalence
> classes) of polynomials.
This double abstraction is not doing much for me. What seems to matter
is determining the domain X. If the quotient ring is capable of defining
higher complexity X from simpler complexity X then understanding this
feature would be useful to me. If it is merely describing behaviors
consistent with these domains regardless of their complexity as common
features then that is an important distinction.
Going back to the primitive operators of superposition and product it is
fairly clear that any math that follows these operators as their well
known behaviors will be similarly behaved. The construction of those
maths could come into question here. Who constructed who as the most
primitive is then a problem and either may construct the other on its
own terms. For instance the polysign numbers construct the reals, yet
many insist upon constructing the polysign numbers from the reals. The
ring quotient seems to be an ultimately abstracted X form but the
nuances I am trying to understand arise. Polysign shares the polynomial
product behavior but is inherently in modulo n form so they do not
suffer the 2n-1 behavior. These then become a rotational geometry which
is dimensional through simplex geometry where a balanced coordinate
system defines the quality of sign as much as it does dimension
(1,1,1,...) = 0 .
- Tim Golden
Tim BandTech.com
> In article <e08c1671-1323-4698-a3ac-d35d613d3...@o20g2000vbh.googlegroups.com> "Tim BandTech.com" <tttppp...@yahoo.com> writes:
> ...
> > If I can get through it then I
> > might attempt a magnitudinal form of lets say M[X] where M are
> > unsigned magnitudes which will then satisfy the ring requirements and
> > commutative behavior as well,
>
> M does not have all the required ring properties. It is not a group
> under addition.
> --
> dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/
Oh,yes, there it is. The additive inverse does not exist.
Thanks Dik.
Still I have to try and understand this quotient ring construction to
understand Hagen's paper.
- Tim
Arturo Magidin
wrote:
> There is an insistence on remaining in the 'indeterminate form' whereby
> X is never evaluated. In effect X is undefined. But now that I've read
> on and it's fairly obvious anyway the polynomial factors
> a X + b
> are evaluatable aren't they? Is it true that in polynomial of real
> coefficients that X is real?
Sigh.
X is the name of a particualr element of the polynomial ring, that is,
of a particular almost null sequence.
There is no "evaluation", because polynomials are not, repeat, *not*
functions.
There is a notion of "polynomial function". This arises from a certain
"universal property" that the ring of polynomials has. Namely, if S is
any ring, S[X] is the ring of polynomials with coefficients in S
(constructed in the same manner as the ring of polynomials with real
coefficients was, by taking the set of almost null sequences of
elements of S), if T is any ring, if f:S-->T is a ring homomorphism (a
function from S to T such that for all a,b in S, f(a+b) = f(a)+f(b)
and f(ab) = f(a)f(b) ), and t is *any* element of T, then there exists
a unique ring homomorphism F:S[X]-->T such that F(a) = f(a) for every
a in S (identified as the almost null sequence (a,0,0,0,...), and F(X)
= t.
If you now consider the ring of polynomials with real coefficients R
[X], and take the identity map R-->R, then each choice of element t in
R defines a unique ring homomorphism R[X] --> R by mapping r to itself
for every real number r (that is, mapping the almost null sequence (r,
0,0,...,0,...) to r), and mapping X (the almost null sequence
(0,1,0,0,...,0,...) ) to t. This is called the "evaluation map at t".
Denote it by e_t.
A function g:R-->R is called a "polynomial function" if and only if
there exists an element p in R[X] such that for each t in R, g(t) is
equal to e_t(p).
In the real numbers, polynomial functions are in one to one
correspondence with elements of R[X]. In other situations, there may
be more than one polynomial corresponding to the same function. For
example, if you work in the integers modulo 3, then X^3-X and 0 are
two distinct polynomials that define the same polynomial function.
The key is: POLYNOMIALS ARE NOT FUNCTIONS. POLYNOMIAL FUNCTIONS ARE
NOT THE SAME THING AS POLYNOMIALS.
That's why there is no talk of "evaluating X" when discussing
polynomials. Because polynomials ARE NOT FUNCTIONS. Polynomials can be
used to define functions, but those functions are not polynomials,
any more than the fact that we can use letters to define functions
makes letters into functions.
--
Arturo Magidin
Mariano Suárez-Alvarez
wrote:
> Tim Little wrote:
Only in your imagination has anything been called
into question... and your imagination has but your
ignorance of elementary algebra to build upon--you'll
agree that that is not a very sound basis to build
anything.
> Who constructed who as the most
> primitive is then a problem and either may construct the other on its
> own terms. For instance the polysign numbers construct the reals, yet
> many insist upon constructing the polysign numbers from the reals.
I would say that what many keep doing is showing you,
by exhibiting a construction of your beloved
polysigned things using the most elementary algebra
available, that they do not provide anything new.
This quest for "primitiveness" of yours is of no real
interest: it may have had a certain charm a few
centuries ago, but nowadays mathematicians have
understood that the only important thing is that
objects actually exist, and not how they are shown to
exist: that the specific construction of an object
is irrelevant.
-- m
Leland McInnes
There are an awful lot of misconceptions and misunderstandings in all
of that. Unfortunately I am very busy with other things at the moment
and don't have time to address all of them. I think, however, that we
will have to go back and start with just rings (not polynomial rings,
and certainly not quotient rings thereof) and weed out all your
misconceptions. You are starting with far to many expectations of what
is supposed to happen which is biased heavily by your single minded
focus on your polysigned numbers. You need to let go of all of that
and come at the subject fresh and just learn about abstract rings for
a while if you ever hope to get a handle on this. As it is you are
just making a meal of it.
cbr...@cbrownsystems.com
<snip>
> The key is: POLYNOMIALS ARE NOT FUNCTIONS. POLYNOMIAL FUNCTIONS ARE
> NOT THE SAME THING AS POLYNOMIALS.
What this interchange makes me reflect on is the pedantic value of
teaching group theory before approaching ring theory; it eases the
task of UN-learning certain things about the meaning of "using a
notation".
It seems natural to start with talk about how we can have rules about
strings (or if you prefer, actions), their concatenations, and their
legal substitutions like:
xx = e
xo = ox
ooooo = e
and then it's at most an innocent convenience that we would like to
write "x x o^3 o = x^2 o^4 = xxoooo" and thus we begin to loosen up
the mental preconception of "o^3" (or "x^3") as necessarily being a /
numeric/ function.
Cheers - chas
Tim BandTech.com
I like whay you've said here Mariano. I do accept that I am back a few
centuries.
The polysign representation is only slightly different than the
polynomial representation, but it does already work in low n with no
need of requiring an infinite series. The modulo two behavior of the
real number is obvious but ignored as merely an inverse.
But I am not trying to push polysign here. I would like to make it
through this ring quotient construction but it seems it's a number of
courses and as Leland says I am trying to make one meal out of it. I
accept that criticism. It is awfully difficult to get through this. I
can accept that it is my own failing too. But your stance here on
modern math not caring about fundamentals is really fascinating. I do
sometimes think of many of these math topics as cryptologies. People
are learning a language, yes. How arbitrary a language is though ought
to be a concern of a mathemtatician. Fundamentals are important and so
the primitives in use and how they are used do weigh heavily in
mathematics.
- Tim
Musatov
>
> - Show quoted text -
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Re: 2 Pi^2 - 0.75 = INVERSE/EXVERSE CONSTANTJun 11, 2009 ... Inverse
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inverse Fourier transform is defined by replacing $\mF$ with $\mF^\ast
$ in \eqref formula_Fourier. ...... Theorem~ tells us that we need
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combinatorial ...... On a theorem of Ivasev-Musatov III, ...... \cite
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[PDF] AUTHOR INDEX, VOLUMES 51–75File Format: PDF/Adobe Acrobat - View
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The quotient of a free product of groups by a single high-powered
relator. II. ..... T. W. On the theorem of Ivasev-Musatov III. 53
(1986) 143–192 .... Inverse monoids with a natural semilat- tice
ordering. 70 (1995) 146–182 ...... of permutation groups, and group
rings of locally finite ..... 52 (1986) 1–19 ...
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Result for query "keyword(s)=theorem author= title="Obviously the
inverse Fourier transform is defined by replacing $\mF$ with $\mF^\ast
$ in \eqref formula_Fourier. ...... Theorem~ tells us that we need
only to quotient the ring; Theorem~, or the purely ...... On a theorem
of Ivasev-Musatov III, ...... The next corollary follows immediately
from Theorem~\thmref 19. ...
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Arturo Magidin
<cbr...@cbrownsystems.com> wrote:
> On Jun 11, 9:40 am, Arturo Magidin <magi...@member.ams.org> wrote:
>
> <snip>
>
> > The key is: POLYNOMIALS ARE NOT FUNCTIONS. POLYNOMIAL FUNCTIONS ARE
> > NOT THE SAME THING AS POLYNOMIALS.
>
> What this interchange makes me reflect on is the pedantic value of
> teaching group theory before approaching ring theory; it eases the
> task of UN-learning certain things about the meaning of "using a
> notation".
You don't really need the full abstraction of "ring theory" and "group
theory" for the purpose of understanding the construction of the
complex numbers as a quotient of the ring R[X]. One can talk about R
[X], with R the reals, without really discussing much about rings
other than the bare-bones definition; I learned it that way three
courses before we actually got to 'abstract algebra', as an
undergraduate. Once we did get to abstract algebra, we had this
construction in hand as an instantiation of the more general concept.
I would have suggested (and was attempting) that approach, leaving the
abstract notions behind, but the OP is too busy not listening to make
any headway.
--
Arturo Magidin
Martin Michael Musatov
This morning I picked up my bible and read two verses:
(This is a real book. I bought it. It is in front of me.
The back of it reads: Zondervan
ISBN 0-310-92391-3)
The first verse was Daniel 12:1
"At that time Michael, the great prince who protects your people,
will arise. There will be a time of distress
such as has not happened from the begin-
ning of nations unti8l then. But at that time
your people--everyone whose name is
found written in the book--will be deliv-
ered."
The second one was Revelations 12:9
"The great drag-
on was hurled down--that ancient serpent
called the devil, or Satan, who leads the
whole world astray. He was hurled to the
earth, and his angels with him."
Prince of Persia | Learn The BibleRather, Michael was the angel of Daniel's people. He was the prince of Israel. .... Similar Bible References. Jesus and Michael the Archangel ...
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Daniel 10:13 "But the prince of the kingdom of Persia was withstandingKing James Bible But the prince of the kingdom of Persia withstood me one and twenty days: but, lo, Michael, one of the chief princes, came to help me; ...
bible.cc/daniel/10-13.htm
Who is Michael the Archangel?What was Michael's purpose? Where is he discussed in the Bible and what mighty ... is] none that holdeth with me in these things, but Michael your prince. ...
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Daniel 10:21 "However, I will tell you what is inscribed in theBut I will tell thee that which is inscribed in the writing of truth: and there is none that holdeth with me against these, but Michael your prince. Bible ...
bible.cc/daniel/10-21.htm
Seventh-Day Adventists: Moses Resurrected cont. and Jesus call ...Feb 23, 2008 ... I'll let the Bible speak for itself and you decide if Christ and Michael are one and the same. Daniel 10:13 states that the prince of the ...
en.allexperts.com/q/Seventh-Day-Adventists-2318/2008/2/Moses-Resurrected-cont-Jesus-1.htm
Daniel 10. The Holy Bible: King James Version.The Holy Bible: King James Version. Daniel 10. ... the Scripture of truth: and there is none that holdeth with me in these things, but Michael your prince. ...
www.bartleby.com/108/27/10.html
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Prince Michael in scriptures « AngelsMar 10, 2008 ... Revelation 12: 7 – Apostle John saw Chief Prince/General Michael war against the dragon as, satan (also known as the Prince of the ...
thelordsangels.wordpress.com/2008/03/10/prince-michael-in-scriptures/
Revelation 12:7 And there was war in heaven, Michael and his angelsDaniel 12:1 "Now at that time Michael, the great prince who stands guard over the sons ... Revelation 12:3 Then another sign appeared in heaven: and behold, ...
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Saint Michael the Archangel
A revelation was given to Daniel, who had been named Belteshazzar. ... I left him (Michael) there with the prince of the kings of Persia, and came to make ...
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It is the identity of Michael, the archangel and prince, mentioned in the last five ..... In Revelation, Michael is portrayed as leading the heavenly hosts, ...
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Michael definition of Michael in the Free Online Encyclopedia. In the Bible and early Jewish literature, Michael is one of the angels of God's presence. He is depicted as a warrior-prince leading the celestial armies ...
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Daniel 12:1 "Now at that time Michael, the great prince who stands King James Bible And at that time shall Michael stand up, the great prince which standeth for the children of thy people: and there shall be a time of ...
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Archangel Michael-The Prince-Messiah To Come Archangel Michael is written all over the Bible and in Dead Sea Scrolls, mentioned as the Prince of the Congregation, The Heavenly Prince Melchizedek(IIQ13) ...
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Mar 27, 2009 ... Difficult Bible Passages About Michael Explained. ...... Here is Michael our prince, the great protector of the church, and the patron of ...
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Revelations Michael Martin | Metro.co.ukMay 12, 2009 ... Latest news about revelations michael martin. Breaking UK & world news from Metro, with celebrity gossip and weird new, and exclusive ...
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AFP: Two lawmakers forced out over expenses scandal
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Iranian Revelations - Michael Ledeen - The Corner on National ...May 14, 2005 ... Iranian Revelations [Michael Ledeen]. I am still in Naples, finishing the research for my book "virgil's golden egg and other neapolitan ...
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Michael the Great Prince -- Who Is He? The spirit creature Michael ...Dec 15, 1984 ... "And during that time Michael will stand up, the great prince who is .... Hence, the conquering Michael referred to in Revelation 12 must be ...
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JSH: Negative Pell's Equation, consequences - alt.math.undergrad ...9 posts - 5 authors - Last post: May 31
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groups.google.com/group/alt.math.undergrad/browse_thread/thread/ef8d9c91ff1b5af6/818d93bb0932769a?lnk=raot
Extra Nos: Godel and Falsifiability of EvolutionI just love it when from time to time you would post something about philosophy of .... Yes, re:sci.math board, I know. Once I get more time which should be fairly soon, ... But this whole world is not about me but it is about Christ's payment for me. LPC. 11:25 PM · Martin Musatov said. ... Member Lutheran Ring ...
extranos.blogspot.com/2008/12/godel-and-falsifiability-of-evolution.html
Posts tagged Comicbook at CinematicalDon't be fooled by the fact that this one opens just like the teaser trailer .... Also, Moviehole.net spoke with the writing team of Marty Musatov and Ethan Erwin .... Fargus on 50 Cent and Chris Klein Will Be 'Caught in the Crossfire' ... Bells Ring · Gummi Bear Surgery -- a How-To · The Math All Girls Are Good At ...
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Shtetl-Optimized » Blog Archive » The complement of Atlas ShruggedAlas, the price of keeping out the likes of Martin Musatov (who at the height of .... fourteen-year old's life: The Lord of the Rings and Atlas Shrugged. ..... mathematics has became “just” theorem-proving, and political philosophy has .... day after day, about your alleged P=NP proof, Jesus Christ, your father, ...
scottaaronson.com/blog/?p=386
Russian Poetry and the October Revolutionmathematics. Maiakovskii said that, having found theprinciple ofa poem, Khlebnikov would leave it ... the memory of a sweet,effeminate, pastoral Christ whom he adores but resists. ..... Borisov-Musatov, one of the Blue Rose painters renowned for ..... In his decrees the authoritative ring of an Abbot's exhortations ...
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by A Pyman - 1990
Result for query "keyword(s)=theorem author= title="On a theorem of Ivasev-Musatov III, ...... Combining this with a theorem of Michael Christ \cite ch, the authors prove; To prove Part 2 of Theorem in the case $1 ...... Small extensions of Witt rings [Pacific Journal of Mathematics] ...... The second part of the theorem is just Proposition 2.1, (iv) above. ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/search/j?firstyear=1994&journaldir=combined&lastyear=2001&query=theorem
Wikipedia:Dead external links/404/m - Wikipedia, the free encyclopedia... United Church of Christ "short course" on the German Reformed Church 404 Not Found .... A paper on the mathematics of change ringing 404 Object Not Found ...... Middle-earth in popular culture, REC Studios' Fellowship of the Ring 404 ...... Students in the Classroom: Not Absent, Just Overlooked': 404 Not Found ...
en.wikipedia.org/wiki/Wikipedia:Dead_external_links/404/m
Result for query "keyword(s)=theorem author= title="This property is proved in \cite BKL, and we could have just; following theorem. the above theorem implies that the ...... On a theorem of Ivasev-Musatov III, ...... Combining this with a theorem of Michael Christ \cite ch, the authors prove ...... Small extensions of Witt rings [Pacific Journal of Mathematics] ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/search/j?firstyear=1994&journaldir=combined&lastyear=2002&query=theorem
From Chekhov to the Revolution: Russian Literature, 1900-1917a repentant nobleman; it had a more realistic ring when used by the ...... Musatov) showed their works next to those of such Muscovites as Serov, .... Salieri, and sang the praises of 'the invincible logic of mathematics.' He ..... In Christ's incarnation he saw the evidence that man can be redeemed ...
www.questia.com/PM.qst?a=o&se=gglsc&d=77723673
by M SLONIM - Cited by 14 - Related articles - All 2 versions
Result for query "keyword(s)=theorem author= title="The theorem just stated is a model result for the conjecture using; $b_ 2$ is to emphasize that the $b_ 2$ in the theorem ...... Rings with sovable quivers [Pacific Journal of Mathematics] ...... On a theorem of Ivasev-Musatov III, ...... Combining this with a theorem of Michael Christ \cite ch, the authors prove ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/search/j?firstyear=1994&journaldir=combined&lastyear=2006&query=theorem
Martin Michael Musatov
Mariano Suárez-Alvarez
> [snip]
> How arbitrary a language is though ought
> to be a concern of a mathemtatician. Fundamentals are important and so
> the primitives in use and how they are used do weigh heavily in
> mathematics.
If by fundamentals you mean "explicit constructions
of objects" then no, they are *not* important and they
are *absolutely* irrelevant to the use of those objects.
It is absolutely unimportant and irrelevant how you
construct a complete totally ordered field. The only
important thing is that such a construction can be
carried out, and that all such objects are isomorphic,
because then one can talk about "the complete totally
ordered field" with the assurance that such a thing
actually exists. Absolutely nothing in maths needs to
know whether such a beast was constructed using
Dedekind cuts, Cauchy sequences or through your pretty
clumsy procedure involving "magnitudes".
-- m
Tim BandTech.com
I know you are an individual Mariano and others will no doubt study
the presentation of a Dedekind cut versus a Cauchy sequence and
attempt perhaps even another new method. Still, I do sense that your
position is fairly equitable among mathematicians, which has been
reinforced by the boundary between physics and mathematics, and
philosophy too. How one resolves these as disparate or tied together
and which one would place beneath another is an individual choice. It
seems no matter what choices one makes the tighter these boundaries
are laid then these do actually intersect each other. The topic of
time is so fundamental yet has lacked a clean representative in
mathematics. Clearly the real value which composes the 3D spatial
representation cannot be symmetrically introduced as an additional
dimension, for then time would be of the same type as space. The
Minkowski metric answers this with a negative distance contribution-
something that mathematicians ought to quibble over but don't since
that is physics right? Rather than concern myself with mimicing these
topics as perfectly as I can I prefer to try and cover new ground,
which is a challenging pursuit. From my own perspective much of
mathematics nomenclature looks disfunctional. It is lost in
uninstantiable abstraction. This subject of quotient ring does have
this feature. The position of the physicist versus the mathematician
is well known. This tangle could be allayed by admitting that we are
not at any final theory in either subject. In other words the way we
look at these ought to be taken with a more open stance- open to their
being wrong, open to a better way, open perhaps to their being no
finality in them.
Your subject has a heavier burden than you are willing to admit to
because you are caught in the accumulation. That is my simple
criticism of both you and the topic of mathematics. It was to be a
thing so fundamental that no quibbling over the construction should be
possible without a direct contradiction somewhere. That was the
axiomatic style. I feel comfortable saying that the ring quotient is
far beyond this level. So far that perhaps it is fraudulent, but I'm
not through it so I cannot say for certain. I have been able to make
similar arguments on other topics such as bivectors where the bivector
x ^ x = 0
becomes dubious through the construction of a unit bivector
a ^ b
whose angle and magnitudes vary but maintains constant value. Upon
reaching the overlapped position the direction is undefined. Hence the
unit bivector is undefined at this position. Thus assigning a zero
value is as impossible as is assigning a sensible value to a division
by zero. The usage of infinity here is in play as it is in Arturo's
insistence on an infinite length polynomial, though in a completely
different form. In this unit bivector form we see instantiable
instances rather that a claim of specificity in terms of an
indeterminate variable. Thus I use Grassman's bivector instance as
support for my argument on instantiable mathematics. It seems that no
such form will be permitted within the indeterminate polynomial
representation that the quotient ring
R[X] / ( X X + 1 )
requires. Thus no disproof by instance can ever exist. Hence the logic
of the construction becomes dubious.
I understand that the probablility of my argument being correct is
slim. So I welcome a refutation. I must have made a mistake somewhere
in my reasoning. It shouldn't be so hard to find this, especially for
one who understands this subject. There are corollaries on human
intelligence here which do have readily available instances.
Unfortunately I am one of those instances. Call this an abstract Godel
style if you like- the problems remain open universally.
Sorry to be so long winded. Please feel free not to respond and I will
do the same.
- Tim
Denis Feldmann
> On Jun 12, 2:43 pm, Mariano Su�rez-Alvarez
> <mariano.suarezalva...@gmail.com> wrote:
>> On Jun 12, 8:33 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>>
>>> [snip]
>>> How arbitrary a language is though ought
>>> to be a concern of a mathemtatician. Fundamentals are important and so
>>> the primitives in use and how they are used do weigh heavily in
>>> mathematics.
>> If by fundamentals you mean "explicit constructions
>> of objects" then no, they are *not* important and they
>> are *absolutely* irrelevant to the use of those objects.
>>
>> It is absolutely unimportant and irrelevant how you
>> construct a complete totally ordered field. The only
>> important thing is that such a construction can be
>> carried out,
[cut a lot of nonsensical answer, then...]
>>Thus I use Grassman's bivector instance as
> support for my argument on instantiable mathematics. It seems that no
> such form will be permitted within the indeterminate polynomial
> representation that the quotient ring
> R[X] / ( X X + 1 )
> requires. Thus no disproof by instance can ever exist. Hence the logic
> of the construction becomes dubious.
>
> I understand that the probablility of my argument being correct is
> slim. So I welcome a refutation.
Certainly not. You dont understand much, you would not welcome a
refutation, but (lucky you) you wouldn't recognize one if it were
shouting "I AM A REFUTAZTION" while dancing the tango, and if by some
slim chance you suspected it *was*, you would put your fingers in your
ears and shout "I cannot hear you"
I must have made a mistake somewhere
> in my reasoning.
Noo, you never made any reasoning (and dont know the meaning of the world)
It shouldn't be so hard to find this, especially for
> one who understands this subject.
One typical "mistake" is saying things like "Arturos 'dimensional
arguyment ust be wrong, because 5<6", while Arturo never used the word
dimension to begin with...
There are corollaries on human
> intelligence here which do have readily available instances.
> Unfortunately I am one of those instances.
You are drifting in insanity again...
Call this an abstract Godel
> style if you like- the problems remain open universally.
>
> Sorry to be so long winded. Please feel free not to respond and I will
> do the same.
Is this a promise? Oh, please, please, keep it...
>
> - Tim
Mariano Suárez-Alvarez
This impression of yours is one we see all the time
with our students: they argue that the work they do
in their exams cannot be *that* wrong since we cannot
point to a specific mistake in it. And the thing is,
it is impossible to point to a specific mistake
neither in their work and in your "reasoning", because
both are not even wrong: they are meaningless.
You plainly do not know the subject about which you
have "criticisms". And I do not have in mind the
high and subtle points of any area of mathematics:
you have shown quite clearly that you do not
understand what a polynomial is. No matter how long
a tirade you are able to write about your criticisms,
using that obsrvation to put your writings in perspective
it is aparent that those criticisms cannot possibly carry
any weight, for they are firmly grounded on your
ignorance.
Ignorance is seldom a unsurmontable problem, for it can
be resolved by learning. But it *is* an obstacle.
-- m
Arturo Magidin
Listen, Tim: you have no clue whatsoever about anything I wrote, as
you have amply demonstrated again and again and again and again. You
have consistently misunderstood it, misquoted it, and misrepresented
it. As such, comments such as:
> The usage of infinity here is in play as it is in Arturo's
> insistence on an infinite length polynomial, though in a completely
> different form.
constitute nothing but rather pathetic (and dishonest) attempts on
your part to try to gain some credibility by associating my name
somehow with your ramblings and your ignorant assertions.
Kindly, do *not* mention my name when you are being an ignoramus. As
such, kindly do *not* mention my name at all whenever you post to
sci.math, ever, because you have never shown an iota of an inkling of
an idea of what I may or may not have said, let alone what I may have
meant.
--
Arturo Magidin
Tim BandTech.com
> On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> Listen, Tim: you have no clue whatsoever about anything I wrote, as
> you have amply demonstrated again and again and again and again. You
> have consistently misunderstood it, misquoted it, and misrepresented
> it. As such, comments such as:
>
> > The usage of infinity here is in play as it is in Arturo's
> > insistence on an infinite length polynomial, though in a completely
> > different form.
>
> constitute nothing but rather pathetic (and dishonest) attempts on
> your part to try to gain some credibility by associating my name
> somehow with your ramblings and your ignorant assertions.
You've presented a polynomial in real coefficients but insist that X
is an indeterminate variable. The format you chose is
( a0, a1, a2, ... )
where a(n) are real. This is indistinguishable from an infinite
dimensional cartesian coordinate. This is so simple that how you
cannot concede the overlap is puzzling. Next, rather than attempt to
remedy any nuance of interpretation you choose to dismiss my argument
without discussing any detail, particularly the detail of this
construction not allowing a standard representation
( a0, a1, ..., an )
due to the fact that products of this form require a result out to
2n-1.
Thus under this form above the polynomials are not rings until the
form is enforced to infinite length, whereupon the issue of a
superinfinite form arises.
This is very different from other usages of infinity. For instance
within calculus we can typically choose a largest n and still get a
result, with a larger choice yielding a more accurate result. Here
what we see is that the math itself is broken under such choices. The
usage of the phrase "polynomial with real coefficients in X" as a ring
is not satisfactory, particularly in the construction of an instance
of a quotient ring. Maintenance of X as an indeterminate variable
helps to cloud this issue further.
- Tim
victor_me...@yahoo.co.uk
> without discussing any detail, particularly the detail of this
> construction not allowing a standard representation
> ( a0, a1, ..., an )
> due to the fact that products of this form require a result out to
> 2n-1.
Gosh! The product of two degree n polynomials might have degree
as large as 2n.! As we see this means that
> what we see is that the math itself is broken under such choices.
or does this simply mean that Dim Golden is talking shite again?
Virgil
<166ee32c-d48b-4aa5...@s38g2000prg.googlegroups.com>,
"Tim BandTech.com" <tttp...@yahoo.com> wrote:
> On Jun 13, 3:28 pm, Arturo Magidin <magi...@member.ams.org> wrote:
> > On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
> >
> > Listen, Tim: you have no clue whatsoever about anything I wrote, as
> > you have amply demonstrated again and again and again and again. You
> > have consistently misunderstood it, misquoted it, and misrepresented
> > it. As such, comments such as:
> >
> > > The usage of infinity here is in play as it is in Arturo's
> > > insistence on an infinite length polynomial, though in a completely
> > > different form.
> >
> > constitute nothing but rather pathetic (and dishonest) attempts on
> > your part to try to gain some credibility by associating my name
> > somehow with your ramblings and your ignorant assertions.
>
> You've presented a polynomial in real coefficients but insist that X
> is an indeterminate variable.The format you chose is
> ( a0, a1, a2, ... )
> where a(n) are real. This is indistinguishable from an infinite
> dimensional cartesian coordinate.
Actually, in that format, X is (0,1,0,0,0,...).
But polynomial rings over a ring R are not required to use that format.
They are equally well represented by the set of functions from
N = {0,1,2,3,...} to the ring R having finite carrier, with suitable
rules for addition and multiplication.
> This is so simple that how you
> cannot concede the overlap is puzzling. Next, rather than attempt to
> remedy any nuance of interpretation you choose to dismiss my argument
> without discussing any detail, particularly the detail of this
> construction not allowing a standard representation
> ( a0, a1, ..., an )
> due to the fact that products of this form require a result out to
> 2n-1.
> Thus under this form above the polynomials are not rings until the
> form is enforced to infinite length, whereupon the issue of a
> superinfinite form arises.
On the contrary, the ring of finite tuples of elements of R with the
appropriate sum and product rules has absolutely no need for any
infinite tuples. The tuple length being unbounded does not imply any
infinite ones.
>
> This is very different from other usages of infinity. For instance
> within calculus we can typically choose a largest n and still get a
> result, with a larger choice yielding a more accurate result. Here
> what we see is that the math itself is broken under such choices. The
> usage of the phrase "polynomial with real coefficients in X" as a ring
> is not satisfactory, particularly in the construction of an instance
> of a quotient ring. Maintenance of X as an indeterminate variable
> helps to cloud this issue further.
>
> - Tim
You evidently have a profound mental hangup about polynomial rings that
you will have to work through on your own.
Work on other things for a while before you come back to it, and it may
resolve itself.
--
Virgil
Arturo Magidin
> On Jun 13, 3:28 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
> > On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > Listen, Tim: you have no clue whatsoever about anything I wrote, as
> > you have amply demonstrated again and again and again and again. You
> > have consistently misunderstood it, misquoted it, and misrepresented
> > it. As such, comments such as:
>
> > > The usage of infinity here is in play as it is in Arturo's
> > > insistence on an infinite length polynomial, though in a completely
> > > different form.
>
> > constitute nothing but rather pathetic (and dishonest) attempts on
> > your part to try to gain some credibility by associating my name
> > somehow with your ramblings and your ignorant assertions.
>
> You've presented a polynomial in real coefficients but insist that X
> is an indeterminate variable.
I just checked all my messages. One time I said "X is an indeterminate
variable", that was on June 5, before I attempted to explain
polynomials from the groun up on June 6; that was the only time I did,
and I have not said "indeterminate variable" since attempting to
explain to you what polynomials are. The clam that I "insist that X is
an indeterminate variable" (when I mentioned it *once*, and that was
before "presenting" polynomials (not the plural)O) is
misrepresentation and misleading, if not outright dishonest.
You have, in fact, demonstrated my assertion above that you have shown
yourself incapable of reporting what I did or did not say without
mispresenting, misunderstanding, or misquoting it.
>The format you chose is
> ( a0, a1, a2, ... )
> where a(n) are real.
That's not a format. I *defined* a ring structure on the set of such
tuples, and then I defined the ring R[X] as consisting of those tuples
that are almost null.
>This is indistinguishable from an infinite
> dimensional cartesian coordinate.
Only if you aren't paying attention. There is a fundamental
differerence between a vector space (which is what the "infinite
dimensional cartesian coordinate" is) and a ring. A vector space has
an action of a field on it, and the only operation between two vectors
that is allowed is addition. The *ring* R[X] does not have a family of
scalar multiplication defined on it (not as I presented it), and there
is an operation of multiplication of elements. These elements are not
vectors, and as such, the claim that they are "indistinguishable" from
the vector space structure simply demonstrates once again that you (i)
are utterly confused; (ii) very ignorant; and (iii) incapable of
understanding what I wrote. The fact that you insist on blaming me for
your legion of shortcomings is just an other shortcoming of yours.
>This is so simple that how you
> cannot concede the overlap is puzzling.
Simple, but wrong. The fact that *you* are incapable of distinguishing
a vector space form a ring is the main reason why things seem puzzling
to you, but don't blame *me* for the fact that *you* are *ignorant*.
What I did say, dear, was:
"the linear algebra meaning
*can*, by a suitable point of view, have *some* bearing, but you need
to know the thing you are looking at before you can take the "suitable
point of view". You aren't there yet, by a long shot. "
(message of June 9th). That is where you are confused: you keep
thinking you are dealing with linear algebra when asking about rings.
That same message I said:
"Polynomials, as almost null
sequences, "live" in the set of all almost null sequences. As a vector
space, this space is *infinite dimensional* over the real numbers, and
"dimension" applies to the vector space as a whole, not to its
elements. Individual vectors do not have "dimension", vector spaces
do. And individual polhynomials do not have "dimension", they have
*degree*, which is something else entirely. "
(So much for "cannot concede the overlap"; there it is, the mention of
the vector space of almost null sequences being an infinite
dimensional *vector space* over the real numbers)
Not my fault that you don't know what you are talking about.
> Next, rather than attempt to
> remedy any nuance of interpretation you choose to dismiss my argument
> without discussing any detail,
I discussed the details; repeatedly. I told you that you are confusing
dimension and degree, that you are confusing vector spaces with rings,
polynomials with polynomial functions, etc.
The fact that you haven't *heard* anything I said does not mean I did
not say it.
>particularly the detail of this
> construction not allowing a standard representation
> ( a0, a1, ..., an )
The "standard representation" of polynomials is not as a finite tuple
of coefficients. Your ignorance is showing, *again*.
> due to the fact that products of this form require a result out to
> 2n-1.
The product of a polynomial of *DEGREE* n and a polynomial of *DEGREE*
m is a polynomial of degree n+m. That is what you are hung up on.
Something that most students pretty much get after 10 seconds. Does it
make you wonder what causes you so much trouble?
> Thus under this form above the polynomials are not rings
And this sentence is complete and utter nonsense again. Nobody claims
"polynomials are rings". See, polynomials are almost null sequences.
Rings are sets with two binary operations. Polynomials aren't rings,
and nobody pretends they are; the fact that you even think that
somebody does is just another example of your utter confusion and
ignorance. The fact that you think I said so is just another example
of your inability to understand what I say, and your insistence on
blaming *me* for *your* stupidity. Which is exactly what I asked you
to stop doing.
I have absolutely no problem with you being an ignoramus; that's your
choice, and you obviously work extremely hard at remaining one. Just
don't insist on blaming *me* for it. You have not understood a
*single* thing I wrote, and each and every time you claim to report
something I say you misrepresent, misreport, misquote, and
misunderstand it. At this point, you are pretty much engaging in
dishonesty each time you do it, as simple incompetence no longer
covers the breath of your transgressions.
I'll repeat it again:
Leland McInnes
> On Jun 13, 3:28 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
> > On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> You've presented a polynomial in real coefficients but insist that X
> is an indeterminate variable.
People have merely been insisting that the X doesn't represent some
element of R, but is another ring element entirely, here represented
as a formal symbol. It is not indeterminate, it just isn't
representable as an element of the real numbers (or the complex
numbers for that matter) so we pick a symbol to denote it. That symbol
is usually X, but could be anything you like really -- just not a
number.
> The format you chose is
> ( a0, a1, a2, ... )
> where a(n) are real. This is indistinguishable from an infinite
> dimensional cartesian coordinate.
Right, except for the requirement that all but finitely many of the
components are zero. I shied away from explaining the reason for this
since the only good explanations I know (that don't involve explaining
it in terms of polynomials of finite but unbounded degree, which were
apparently not a good way to explain it to you) wade us into rather
deeper waters. Let's just say that we want the Abelian group structure
of the resulting ring to be the (countably) infinite direct sum, and
not the direct product, of the Abelian group underlying our base ring
(in the case usualyl discussed in this thread, that means the real
numbers). This is the difference between products and coproducts, and
is, in fact, the dual (in the sense of category theory) of a cartesian
product (and hence cartesian coordinate). Now, if you don't know what
some of that means, don't try to work it out on your own and presume
you're guessing right; instead accept that some material may be beyond
what can be explained in a single USENET post, note that there are
actually good deep fundamental reasons for taking care with the
difference between an infinite vector space and the space of nearly
null infinite tuples, and move on.
> This is so simple that how you
> cannot concede the overlap is puzzling.
That much has been conceded regularly, with appropriate caveats as to
actual technical accuracy given. The point is that we don't just have
the "vector space" but we also have the ring structure applied
thereto, which makes it a different object.
> Next, rather than attempt to
> remedy any nuance of interpretation you choose to dismiss my argument
> without discussing any detail, particularly the detail of this
> construction not allowing a standard representation
> ( a0, a1, ..., an )
> due to the fact that products of this form require a result out to
> 2n-1.
You're just tying yourself in knots here. The standard representation
is as countably infinite nearly null tuples. It is quite correct that
you can't stop at some fixed n -- you have to have them always
infinite all the time. But then that's why polynomial notation is
preferred. You see if the tuple is nearly null then all but finitely
many components are zero, and hence it must have only finitely many
terms as a polynomial. That is, our space is that of *all* polynomials
of finite degree. Note the important point there: the degree of the
polynomials must be finite! Any individual polynomial in the space
(which is a single element of the resulting ring) may have arbitrarily
large degree, but it has to be some finite natural number. We do not
allow polynomials of infinite degree -- and this is equivalent to the
requirement for *nearly null* infinite tuples. Keep that in mind for
later.
> Thus under this form above the polynomials are not rings until the
> form is enforced to infinite length, whereupon the issue of a
> superinfinite form arises.
Well yes, the issue arises, but then you stop and think about it
carefully for half a minute and realise that it isn't a problem.
Before we get to why, let me pick a nit in your use of language: you
say "the polynomials are not rings" which is fundamentally wrong. The
*set* of *all* polynomials is a ring (note the singular). Saying "the
polynomials are not rings" is saying that the individual polynomials
are not each rings (by the failure to denote the space/set/collection
of polynomials, and by the use of the plural "rings") which is just
nonsense. Either you were sloppy with your words (which happens, but
is probably why people call what you say nonsense -- it probably often
is if you read what you say (as others are forced to) and not what you
mean) or you have truly fundamental misunderstandings.
Now, back to why we don't have to worry about
"superinfinite" (whatever that means) issues. Remember when I pointed
out that the nearly null requirement on tuples is equivalent to the
requirement that the polynomials have finite degree? Well given a
polynomial of finite degree n and a polynomial of finite degree m,
then their product will have *finite* degree m+n. Thus, as long as our
polynomials have finite degree, we know more have to worry about
multiplication causing problems than we have to worry about addition
of finite (but arbitrarily large) natural numbers causing problems.
You don't believe there is some fundamental issue with arithmetic
where it has to concern itself with "superfinite" numbers do you? In
short, after momentary inspection, we see that countably infinite
nearly null tuples with the multuiplication we have described are
indeed closed under multiplication. There are no issues. Had you
actually taken the trouble to read what people have posted you would
have seen such closure properties proved and not had to ask
nonsensical questions.
> This is very different from other usages of infinity. For instance
> within calculus we can typically choose a largest n and still get a
> result, with a larger choice yielding a more accurate result. Here
> what we see is that the math itself is broken under such choices.
You aren't making sense anymore. Take a moment and actually read what
people have written previously (spend the 10 seconds required to re-
read the earlier portions of this post) and you'll see that you
"issues" are not issues at all, but are easily and clearly dealt with
in a perfectly rigorous manner. This jumping to the conclusion that
the whole edifice is fundamentally broken before you have even taken
the time to actually understand it is one of the reasons you face so
much hostility here.
> The
> usage of the phrase "polynomial with real coefficients in X" as a ring
> is not satisfactory, particularly in the construction of an instance
> of a quotient ring. Maintenance of X as an indeterminate variable
> helps to cloud this issue further.
The problem is that you don't understand the fundamentals, and
apaprently aren't willing to take the time to learn them first. Had
you bothered to spend that time you would not have to make foolish
statements like this.
Tim BandTech.com
> In article
> <166ee32c-d48b-4aa5-bb55-b1244f44a...@s38g2000prg.googlegroups.com>,
Thanks for the simple advice Virgil. On the one hand I see the
polynomial form as a farily simplistic thing, but on the other hand I
see the requirement of not evaluating X as very challenging. Treating
this expression as elemental when it cannot be evaluated is a
hypocritical step. Entering the quotient ring stage with this basis is
a fraudulent construction, especially if claiming to have constructed
an instantiable basis such as the complex numbers. I do find this
circumstance understandable as mathematicians in their plurality and
over generations cover new ground and fill out any voids in the
possibility space. This is where 'new' work lays. Yet here I suspect
that eventually this area will be withdrawn. The rotational aspects of
the polynomial are already inherently within the real number. So
rather than build a confusing structure up above the polynomial (the
quotient ring) all that needs to be done is to build out the real
number more generally. Then, rather than worry about the need for
infinite length sequences the n-length sequence can hold its own since
products will remain at length n. This math relies upon a modulo
effect which is already present within the real valued number.
Clearly for one individual to invalidate mathematics that has been
propagated en masse is not a socially valid step. Here on this thread
few are willing to go back and actually express any doubt on this
topic. I treat this as evidence and accept the position of rejection
without consideration as further evidence. The extensions of this
situation are puzzling. They go far beyond mathematics.
- Tim
Tim BandTech.com
> On Jun 14, 12:58 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > On Jun 13, 3:28 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
> > > On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > You've presented a polynomial in real coefficients but insist that X
> > is an indeterminate variable.
>
> People have merely been insisting that the X doesn't represent some
> element of R, but is another ring element entirely, here represented
> as a formal symbol.
Here I am engaged in a game of whack-a-mole.
You have just stated that X is another ring element entirely. Yet in
the quotient ring construction on the one hand we might take a ring
instance such as Z which does have fully instantiable elements such as
+ 3, + 5
yet when I ask what the elements of X are there is no answer. Thus
expressions in X are not elemental. I simply give you Z as a fine
counterexample. Thus in specifying a ring quotient usage of R[X] in
the construction takes on dubious meaning. There is nothing elemental
about it. Especially when enforced to be infinite in length there is
something rather anti-elemental about R[X]. Yet upon choosing a real
value or a complex value for X, which I believe is where practical
usage of the polynomial form lays, then all of these concerns
evaporate and the entire construction simply collapses to a real or a
complex value, which then exposes the fact that the entire polynomial,
while capable of introducing complexity of raw values, does nothing
for dimensional structure of the domain X, which is completely
counterintuitive to the information enclosed in its general form. This
is just one of several contradictory contexts of the ring quotient
construction
R[ X ] / ( X X + 1 )
> It is not indeterminate, it just isn't
> representable as an element of the real numbers (or the complex
> numbers for that matter) so we pick a symbol to denote it. That symbol
> is usually X, but could be anything you like really -- just not a
> number.
>
> > The format you chose is
> > ( a0, a1, a2, ... )
> > where a(n) are real. This is indistinguishable from an infinite
> > dimensional cartesian coordinate.
>
> Right, except for the requirement that all but finitely many of the
> components are zero. I shied away from explaining the reason for this
> since the only good explanations I know (that don't involve explaining
> it in terms of polynomials of finite but unbounded degree, which were
> apparently not a good way to explain it to you) wade us into rather
> deeper waters.
I think this is more like a game of trickery. You get a student to eat
a fist step that doesn't quite jibe with the next step. Then you cover
that first over, tunnel over to the next thing and present it as if it
were a crystalline form.
But here you are the only one of the group representing the
established math so openly. So I should not try to whack you with my
whack-a-mole plastic hammer.
I understand that series which converge to zero are nice things to
work with. I am fine with your relaince upon these. But this does not
alter my criticism of the quotient ring as a constructor of the
complex numbers. Especially the quotient ring as a definitional stage
which leaps from the simplicity of modulo discrete math over onto
continuum modulo math is not convincing. I already know you'll say
these words are meaningless.
Yes, I pushed my criticism too far in my last post. You see the change
in responses too. Here Arturo now says that he did concede the
dimensional interpretation whereas before he obfuscates that. I am
open to having truly fundamental misunderstandings. I am also open to
mathematics having constructed truly fundamental misunderstandings.
This latter stage is one which good mathematicians need to open to.
Yet it is just where the treachery lays. By stepping over the edge
you'll make yourself incredible. You are willing to dance closer to
this edge than the others. I respect that. I respect that I will not
convince you to walk the tightrope over to the edge of another
precipice. I would merely like to cleanly understand the holy ground
that you've decided to inhabit, plastic hammers holstered but ready
for action.
>
> Now, back to why we don't have to worry about
> "superinfinite" (whatever that means) issues. Remember when I pointed
> out that the nearly null requirement on tuples is equivalent to the
> requirement that the polynomials have finite degree? Well given a
> polynomial of finite degree n and a polynomial of finite degree m,
> then their product will have *finite* degree m+n. Thus, as long as our
> polynomials have finite degree, we know more have to worry about
> multiplication causing problems than we have to worry about addition
> of finite (but arbitrarily large) natural numbers causing problems.
> You don't believe there is some fundamental issue with arithmetic
> where it has to concern itself with "superfinite" numbers do you?
When all practical usage of polynomials collapse these informationally
dense structures down to one informational unit of their claimed
content then yes, I do see a problem here. You have carefully stated
that you will work in null terminated sequences. These taken under
product will propagate upward in length. I don't see that an honest
answer can dismiss this awareness either way. Like working with
infinity within calculus only its limit can be taken. I believe that
infinity is something to be careful with. It can be used, but should
be used carefully. I think that dismissal of this feature especially
to a concerned student is not wise. Simply owning this feature is
enough and then moving on from there rather than being in denial is
appropriate. The future usages of the math being constructed may cause
conflict here. Aren't you open to this? Especially within the concept
of ring whereby the operators are abstracted we should be capable of
building some continuous iterate product or some such form where the
result remains within the original domain, otherwise we are not in a
ring system. You have chosen to work in an unistantiated form R[X] and
if we raise this form to
R[X] ^ m
then we have constructed the same format but at a level of greater
complexity. Since we are engaged in generalities to the fullest you
can take this one as a sneering example. You happily limit the length
of the series whilst you insist that the series be infinite in length.
Please do not take this criticism personally. This is the
establishment mathematics you are upholding. I seriously doubt if one
man alone would ever arrive at this format of construction. This is
the work of one man standing on another man's shoulders. Yes, we'd not
get very far without this behavior. Yet where this behavior can get us
is still dubious. Twisting a zero into an infinity as a child plays
with a loop of yarn should we assume the child's deep understanding?
This situation is far more complicated yet the claim of mathematics as
fundamental construction should be upheld. Scrutiny on this area does
not resolve to the simplistic underpinning that some believe. Instead
we have waivering nuances.
> In
> short, after momentary inspection, we see that countably infinite
> nearly null tuples with the multuiplication we have described are
> indeed closed under multiplication. There are no issues. Had you
> actually taken the trouble to read what people have posted you would
> have seen such closure properties proved and not had to ask
> nonsensical questions.
>
> > This is very different from other usages of infinity. For instance
> > within calculus we can typically choose a largest n and still get a
> > result, with a larger choice yielding a more accurate result. Here
> > what we see is that the math itself is broken under such choices.
>
> You aren't making sense anymore. Take a moment and actually read what
Sorry, but this is still making sense to me. If in calculus we were to
specify a large n and then this large n lead to an even larger m
within a computation we would not arrive at a computable result. Yet
the math which you purvey does contain this behavior. Products reach
ever greater complexity, hence the requirement of not just a length n
sequence, but an infinite length sequence. It's so simple to see this.
No, this point does not prevent you from going onward, but it is a
point of valid awareness. Yet you attempt to stifle this. How many
students ponder this concern? And you stifle them? You stifle the
future by your own fear of leaving the problem open to better
solutions.
- Tim
victor_me...@yahoo.co.uk
> Here I am engaged in a game of whack-a-mole.
> I think this is more like a game of trickery. You get a student to eat
> a fist step that doesn't quite jibe with the next step. Then you cover
> that first over, tunnel over to the next thing and present it as if it
> were a crystalline form.
As illucid as ever, Dim.
> These taken under
> product will propagate upward in length.
Crikey, the product of two n-digit integers might be longer than
n digits, so integers must be suspect too!
Arturo Magidin
The reason you find it challenging is your steadfast refusal to try to
*learn* something.
One cannot construct edifices on the basis of ignorance. Yet, that is
what you insist on trying to do. There is nothing complicated,
hypocritical, social, or flawed here, except your continued belief
that you can make a coherent complaint based ont he fact that you
don't know what you are talking about and you refuse to learn the
material.
--
Arturo Magidin
Arturo Magidin
> Here Arturo now says that he did concede the
> dimensional interpretation whereas before he obfuscates that.
I said it before twice, I'll say it a third time: since you cannot
accurately report what I did or did not say, please stop trying to.
All you are doing is *lying your ass off* whenever you attempt to
claim I did or did not say something.
> I am
> open to having truly fundamental misunderstandings.
Bullshit. You say the words, but you don't mean them. You have kept
that "truly fundamental misunderstanding" and that truly fundamental
ignorance steadfastly throughout the entire thread, and compound it by
lying and misrepresenting what others say. You aren't open to
anything.
Don't claim to report what I did or did not say, because, quite
simply, you have shown you are incapable of doing so without lying.
--
Arturo Magidin
Leland McInnes
If you can just get your mind around it, trust me, it won't be
challenging at all.
> Treating
> this expression as elemental when it cannot be evaluated is a
> hypocritical step.
In what sense is it hypocritical exactly? I'm pretty sure your problem
is that you don't understand it. I'm going to have another go at
explaining it in a different way in another post, but please believe
me when I say that my understanding of it is good, and I've tried in
vain to see how any of the complaints you make could apply to my
understanding (really, I have tried) and they just don't. That means,
to me, that you don't understand it properly.
> Entering the quotient ring stage with this basis is
> a fraudulent construction, especially if claiming to have constructed
> an instantiable basis such as the complex numbers.
What, to you, makes something "instantiable" exactly? What is more
instantiable about the symbol "i" than the symbol "X"?
> I do find this
> circumstance understandable as mathematicians in their plurality and
> over generations cover new ground and fill out any voids in the
> possibility space. This is where 'new' work lays. Yet here I suspect
> that eventually this area will be withdrawn.
I'm very confident that it won't be. Look, polynomial rings are used
regularly in all manner of modern mathematics, and are particularly
foundational in modern algebraic geometry and algebraic number theory.
Not only have these fields made significant theoretical progress,
they've produced very solid practical results. Polynomial rings are
quite rigorous, well defined, and lead to solid practical results.
They aren't going anywhere.
> The rotational aspects of
> the polynomial are already inherently within the real number. So
> rather than build a confusing structure up above the polynomial (the
> quotient ring) all that needs to be done is to build out the real
> number more generally. Then, rather than worry about the need for
> infinite length sequences the n-length sequence can hold its own since
> products will remain at length n. This math relies upon a modulo
> effect which is already present within the real valued number.
Right, well please demonstrate how all of this building up can be used
practically in algebraic number theory and algebraic geometry. I
assure you, you'll simply end up reconstructing the same thing with
different terminology.
> Clearly for one individual to invalidate mathematics that has been
> propagated en masse is not a socially valid step. Here on this thread
> few are willing to go back and actually express any doubt on this
> topic. I treat this as evidence and accept the position of rejection
> without consideration as further evidence. The extensions of this
> situation are puzzling. They go far beyond mathematics.
I have very little doubt that there are any issues with polynomial
rings as rigorous algebraic constructions -- I've studied them, worked
with them, made mistakes with them, caught my own errors in using
them, and spent plenty of time going back through the basics carefully
to find my errors in working with them.
Tim BandTech.com
> On Jun 15, 9:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
>
> > Thanks for the simple advice Virgil. On the one hand I see the
> > polynomial form as a farily simplistic thing, but on the other hand I
> > see the requirement of not evaluating X as very challenging.
>
> If you can just get your mind around it, trust me, it won't be
> challenging at all.
>
> > Treating
> > this expression as elemental when it cannot be evaluated is a
> > hypocritical step.
>
> In what sense is it hypocritical exactly?
On the one hand you've insisted that X will never be evaluated.
On the other hand you're going to wind up evaluating X. For instance
in the buildout of the complex numbers from your infinite dimensional
X-form.
> I'm pretty sure your problem
> is that you don't understand it. I'm going to have another go at
> explaining it in a different way in another post, but please believe
> me when I say that my understanding of it is good, and I've tried in
> vain to see how any of the complaints you make could apply to my
> understanding (really, I have tried) and they just don't. That means,
> to me, that you don't understand it properly.
>
> > Entering the quotient ring stage with this basis is
> > a fraudulent construction, especially if claiming to have constructed
> > an instantiable basis such as the complex numbers.
>
> What, to you, makes something "instantiable" exactly? What is more
> instantiable about the symbol "i" than the symbol "X"?
Nice question. I'm not so sure that there is much better about the
symbol i.
The math which I spoke of that replaces the X format has no need of
the i format either.
The motivation of deriving i from a real basis seems pretty far
fetched. From the perspective of ring definition the sum and product
operations are fundamental and their good behavior is enough. We need
to address
X X = - 1 .
Here we see an assignment of a real value to one of these mysterious
domains of unassignables in X^n. This statement above I believe to be
invalid by your own usage of X. You've insisted that these X do not
perform this way. A real value is being assigned here to XX. You've
already insisted that this need never happen. These X are somehow
sacred cows, or scare crows, all in a line forming some fence that
need never be touched in a field of virtual hay which can somehow
still contain real values.
Complex analysis tags along with real analysis fairly well. Also
complex math is involved in a fair amount of physics and engineering
computation. I do accept the utility of this rotational form. That
rotational form is also existent within the real value, though it's
apparency in the real form is diminished beneath what most will
perceive. I accept the junction of these two maths; the real numbers
and the complex numbers. There need not be so much magic in getting to
the next dimension as is going on in the ring quotient. Rather, the
fundamental definitions of the real numbers (as I recall there are
already six or seven versions of them) can be generalized.
It would be great to see your quotient ring go to work. I want to
understand how this happens. Just how you can breach the hump of
claiming X to not carry any instantiable value and then turn around
and substitute real values in is puzzling to me. I will try to keep
open minded. I am ready for your presentation. Still, it amazes me and
amuses me how much you all are willing to eat that hay from the field
and obey the bizzarre fence that has been laid.
>
> > I do find this
> > circumstance understandable as mathematicians in their plurality and
> > over generations cover new ground and fill out any voids in the
> > possibility space. This is where 'new' work lays. Yet here I suspect
> > that eventually this area will be withdrawn.
>
> I'm very confident that it won't be. Look, polynomial rings are used
> regularly in all manner of modern mathematics, and are particularly
> foundational in modern algebraic geometry and algebraic number theory.
> Not only have these fields made significant theoretical progress,
> they've produced very solid practical results. Polynomial rings are
> quite rigorous, well defined, and lead to solid practical results.
> They aren't going anywhere.
>
> > The rotational aspects of
> > the polynomial are already inherently within the real number. So
> > rather than build a confusing structure up above the polynomial (the
> > quotient ring) all that needs to be done is to build out the real
> > number more generally. Then, rather than worry about the need for
> > infinite length sequences the n-length sequence can hold its own since
> > products will remain at length n. This math relies upon a modulo
> > effect which is already present within the real valued number.
>
> Right, well please demonstrate how all of this building up can be used
> practically in algebraic number theory and algebraic geometry. I
> assure you, you'll simply end up reconstructing the same thing with
> different terminology.
If this is true then all the better. There will be no 'X' at the
bottom of the construction which has no substitutable quality. Yes, I
do believe that this is what I'm talking about. There will be no need
of infinite length progressions. Further the complex numbers take
their place as a natural extension from the real numbers inherently
within the construction. At the basis of the terminology that I
suggest is merely a generalization of the real number and its own
discrete symbollic signage. Thus the chasm between continous and
discrete terminology is layed out on a different map, yet this map
does match up to much of existing mathematics and especially the
development of what might be called multidimensional geometry. The
naming of all these genera of mathematics is a disgusting thing and it
has taken me a long time to arrive at the subject of abstract algebra.
It does seem to be the correct spot to focus on.
- Tim
Arturo Magidin
> On Jun 15, 12:39 pm, Leland McInnes <leland.mcin...@gmail.com> wrote:
>
> > On Jun 15, 9:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > Thanks for the simple advice Virgil. On the one hand I see the
> > > polynomial form as a farily simplistic thing, but on the other hand I
> > > see the requirement of not evaluating X as very challenging.
>
> > If you can just get your mind around it, trust me, it won't be
> > challenging at all.
>
> > > Treating
> > > this expression as elemental when it cannot be evaluated is a
> > > hypocritical step.
>
> > In what sense is it hypocritical exactly?
>
> On the one hand you've insisted that X will never be evaluated.
> On the other hand you're going to wind up evaluating X.
Given that you have no clue or idea how the construction will procede,
on what possible authority could you make such a claim?
> For instance
> in the buildout of the complex numbers from your infinite dimensional
> X-form.
No; the construction of the complex numbers as a quotient of the ring R
[X] never requires an "evaluation" of the polynomial X. Which you
would learn if you were to listen instead of pontificate from a
position of ignorance.
It's really amazing how actual knowledge can help in understanding.
You might want to try it one day, but please forgive me if I don't
hold my breath until you do.
--
Arturo Magidin
Leland McInnes
> On Jun 14, 10:44 pm, Leland McInnes <leland.mcin...@gmail.com> wrote:
>
> > On Jun 14, 12:58 pm, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > On Jun 13, 3:28 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
> > > > On Jun 13, 9:02 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > You've presented a polynomial in real coefficients but insist that X
> > > is an indeterminate variable.
>
> > People have merely been insisting that the X doesn't represent some
> > element of R, but is another ring element entirely, here represented
> > as a formal symbol.
>
> Here I am engaged in a game of whack-a-mole.
Trust me, I feel the same way: each time I try and explain something
you've apparently misunderstood, you come back with a reply that shows
misunderstandings of 5 or 6 new things. Your background really is
lacking, and that makes the explanations I and others give often seem
weird to you because we are often using language very specifically,
and you don't seem to know what the various specifics mean,
misinterpret them, and than take those misinterpretations and build a
whole new edifice of misunderstanding out of them.
> You have just stated that X is another ring element entirely.
Yes, and it seems you've misunderstood what I meant entirely, so let's
nip this in the bud and try and explain it all carefully because it
might provide us with another way to explain to you what the X
actually means. Bear with me as I'm about to go over something that
you already know, but pay attention because it is going to provide the
groundwork for an analogy later.
Let's start with a concrete example you do understand: the real
numbers. Now, suppose we want to consider the complex numbers. How do
we do that? Well we take the ring of real numbers, and add to it an
element "i". Now, "i" is not a real number, it's an object from
outside the ring of real numbers. It is so elusive that, in fact, we
can't use a number to denote it, instead we have to just use a symbol
(usually "i"). Now, to make the real numbers plus this extra symbol
"i" into a ring we need to have addition and multiplication defined.
Addition and multiplication of real numbers is fine (since it was
already a ring) but what happens if we try and do addition and
multiplication that involve "i"? We want to have a+i be meaningful for
any real number a. Now we can't actually evaluate a+i to any real
number, so we'll just have to say that a+i is another element of the
ring (thus, we started by adding just i, but to make sure we get a
ring we've now added a+i for real number a). We also want b*i to be
meaningful, so we'll just have to say all of those are elements of the
ring too. But then once we have all those b*i we'll have to allow
addition with each of them too so we'll have to add a+b*i for each
pair of real number a and i to to thing ring. But wait, we only
considered what happens when we add or multiply real numbers with
"i"; what happens when we add or multiply "i" with itself? Well we
want i+i to be meaningful, so we better add that to the ting as
well ... except that we have the distributive axiom for rings so i+i =
i(1+1) = 2*i, and we already added 2*i to the ring, so we're okay. You
can verify that the rest of the possibilities for adding "i" to itself
work out similarly. Now what of multiplying "i" with itself? Well to
get the complex numbers we have a special requirement for this extra
element "i" that we are adding: specifically we require that i*i = -1.
That means we don't need to add any new elements for multiples of "i"
with itself since -1 is already in the ring, and further multiples
will either be a real number again, or a real number times "i", which
we already added to the ring. You may want to check that the
distributivity axioms for rings (along with knowledge that the real
number are already a ring) ensure that these cases mean that there is
nothing else we need to add to make the real numbers, extended by this
extra (non real number) element "i" into a ring. The result is that we
need all objects of the form a +b*i where a and b are real numbers and
"i" is an extra (non real number!) element we added to the ring of
real numbers.
We could go through the same sort of process to construct the
quaternions by adding extra elements non real number elements "i",
"j", and "k" to the real numbers, and then adding all the extra
elements we'll need to make this larger set into a ring (with the
appropriate properties applied to "i", "j", and "k".
Note that in both these cases "i" is nothing more than a symbol used
to denote a new ring element we wish to add (so as to create a new
ring) to the real numbers. That is, to build a bigger ring we add a
new element to the ring, assume that element has particular properties
with regard to how it multiplies with itself (and with any other new
elements we wish to add), and then add all the extra elements we need
to make sure the resulting new ring is closed under addition and
multiplication.
Now, what say we want to add a new element a bit like "i" to the ring
of real numbers, but we don't want to presume it has specific
properties as to how it multiplies with itself? This is reasonable,
since by not assuming special properties we get the "fully general"
case. So, all we a re going to do is pick a symbol to denote this new
element we want to create, add it to the ring of real numbers, and
then add in all the extra elements we'll need to ensure that the real
numbers along with this new element form a ring. I'm going to pick the
symbol "X" to denote this new element. We then go through the process
just as we did with "i" for the complex numbers: we'll need a + X, and
also b*X, and then a + b*X. Now what about adding "X" to itself? Just
as with the complex number example, we'll have X + X = X(1+1) = 2*X
and so on, meaning we have already added all the possible sums of
"X"'s with those b*X. What about multiplying X with itself? Well we
don't have any requirement as to what happens when we multiply X with
itself, because we wanted to maintain the greatest generality we
could, so X*X is just X*X -- we'll have to add that to the ring. But
then we'll have to go back and add the elements a + b*X*X for each
pair of real numbers a and b; but wait -- we can also form the sum a
+b*X + c+d*X*X = (a+c) + b*X + d*X*X. Now (a+c) is just another real
number, so we're okay there, but the rest... well, we'll just have to
add elements of the form a + b*X + c*X*X for each triple of real
numbers a,b,c as well. But then what about X*X*X? We'll have to add
that to the ring as well. And then we'll have to allow elements of
them form a + b*X + c*X*X + d*X*X*X as well ... and we'll just keep
going on. Now, before you panic of infinities introduced by this "keep
going on", we'll deal with that issue later in this post; forget about
it for now, because it will only distract you from my point.
Note that all we've done is exactly as we would do to create the
complex numbers, or the quaternions, except we didn't place any
restrictions on how the new element we are introducing is to behave
with respect to multiplication with itself. If you want to look at it
as a more general version of the complex numbers that might be a start
-- it is merely an analogy, but it might help you get your head around
it. When I say "X" is another ring element entirely I mean it in the
same sense that "i" is another ring element entirely when we go about
constructing the complex numbers from the real numbers. It is
introduced from outside the ring of real numbers and thus can't be
written as any real number, doesn't stand for any real number, and
can't be evaluated as any real number. It is just "something else".
The "X" is just like that: it is somethign from outside the ring of
real numbers, doesn't stand for any real number, and can't be
evaluated as any real number. It is an extra element for which we have
no other name than "X". Does this make it clearer?
> Yet in
> the quotient ring construction on the one hand we might take a ring
> instance such as Z which does have fully instantiable elements such as
> + 3, + 5
> yet when I ask what the elements of X are there is no answer.
I'm not sure whether I'm nipicking your language, or picking a
misunderstanding here, but "X" doesn't have "elements", it *is* an
element. The ring Z[X] of all polynomials with *integer* coefficients
has elements. "X" is one of those elements. So is 2+3*X+7*X*X*X*X.
> Thus
> expressions in X are not elemental. I simply give you Z as a fine
> counterexample.
I'm still not sure what you mean by "instantiable". If you mean: can I
give you a set of things with "X" many elements in them ... well no.
But then you can't give me a set of things with "-7" things in it, nor
can you show me a line that is exactly "sqrt(2)" long, not can you
point out to me an example of "2+3*i" apples. Thus if I am taking your
meaning of "instantiable" right, then neither the integers, nor the
real numbers, nor the complex numbers are "instantiable". You'll have
to define "instantiable" for me before I can answer this properly
though.
> Thus in specifying a ring quotient usage of R[X] in
> the construction takes on dubious meaning.
It has an explicit meaning, but it is an abstract one. Then again, so
is the notion of "number" to begin with -- try reading some philosophy
of mathematics. Certainly limits in calculus are equally abstract. So
are Dedekind cuts, or Cauchy sequences. So are irrational numbers. If
you don't like taking an abstract generalisation and seeing what you
can get from it (and as I said in another post, the results gained
from polynomial rings in various fields of mathematics are significant
and remarkable) then you don't really want to do mathematics;
mathematics really is the art of abstraction. That's fine, but stop
trying to pretend that all of this isn't rigorous when it is -- it's
just abstract.
> There is nothing elemental
> about it. Especially when enforced to be infinite in length there is
> something rather anti-elemental about R[X].
On the contrary, when viewed the right way it is indeed quite
fundamental -- it's the most general way we can add a single "new"
element to R and close it up under multiplication and addition to get
a ring. The complex numbers are a special case of this. All your
different polysigned numbers can be thought of as special cases of
this (vary the requirements you put on the element you add and you can
get all your polysigned numbers). A whole host of other constructions
are equally special cases -- we opt for the most general since we can
build all these other special cases out of it in terms of quotients:
the complex numbers, your polysigned numbers, and an infinite variety
of other cases which you haven't considered.
> Yet upon choosing a real
> value or a complex value for X, which I believe is where practical
> usage of the polynomial form lays,
No, it really doesn't.
> then all of these concerns
> evaporate and the entire construction simply collapses to a real or a
> complex value, which then exposes the fact that the entire polynomial,
> while capable of introducing complexity of raw values, does nothing
> for dimensional structure of the domain X, which is completely
> counterintuitive to the information enclosed in its general form.
Well yes, that would be true if the only way to do anything was to
consider X to be a some real number. That isn't the only way to do
things, however, so this criticism is quite moot.
> This
> is just one of several contradictory contexts of the ring quotient
> construction
> R[ X ] / ( X X + 1 )
I'll repeat again: there are no contradictions if you actually bother
to understand the construction. We have even gotten as far as getting
you to fully understand R[X], let alone quotient rings, so can you
leave the speculative criticism aside until you actually understand
what you're talking about.
> > Right, except for the requirement that all but finitely many of the
> > components are zero. I shied away from explaining the reason for this
> > since the only good explanations I know (that don't involve explaining
> > it in terms of polynomials of finite but unbounded degree, which were
> > apparently not a good way to explain it to you) wade us into rather
> > deeper waters.
>
> I think this is more like a game of trickery. You get a student to eat
> a fist step that doesn't quite jibe with the next step. Then you cover
> that first over, tunnel over to the next thing and present it as if it
> were a crystalline form.
No one is trying to trick you Tim. There are libraries with lots of
books on these topics. There are a whole host of free online resources
for these topics. Feel free to actually read any of those instead of
listening to us. Perhaps read some of those and ask about anything
that doesn't quite match with what we say. You'll likely get
explanations for why the different things do jibe, it just isn't
necessarily obvious at first.
As to the issue of moving to the second thing before the first is
clear: that's your doing, not ours. You are the one who jumps ahead
and brings up issues about things that have not yet been addressed in
our explanations and demands answers. As I said, it feels like Whack-a-
mole to me too, since as soon as I explain one point, you are
demanding answers on something else down at the third of fourth step
and getting way ahead of yourself. Am I to simply ignore your demands
for explanation, or do things somewhat out of order out of necessity
of keeping up with your demands?
> I understand that series which converge to zero are nice things to
> work with. I am fine with your relaince upon these.
Language use nitpick time: series that converge to zero are something
quite different from nearly null infinite tuples. A series is a sum
(usually infinite) and to say it converges to zero is to say that the
sequence of partial sums has limit zero. That has nothing to do with
nearly null infinite tuples. Either you are misunderstanding something
here, or you are being very sloppy with language. It's this sort of
thing that I believe infuriates Arturo because he will take you at
your word on "series which converge to zero", see that it has nothing
to do with what he was talking about at all, and call your comment
nonsense. I'm going to give you the benefit of the doubt and assume
you meant something else but were sloppy with your language.
> But this does not
> alter my criticism of the quotient ring as a constructor of the
> complex numbers. Especially the quotient ring as a definitional stage
> which leaps from the simplicity of modulo discrete math over onto
> continuum modulo math is not convincing. I already know you'll say
> these words are meaningless.
Polynomial rings and quotient rings as constructors of the complex
numbers are not necessarily the most natural way to go about it. If
all you ever wanted to do was construct the complex numbers then
polynomial rings and quotient rings thereof really aren't worth the
trouble. The key is that they aren't just used to construct the
complex numbers. You seem worried about how complicated this
construction is (although it is simpler than you are making it for
yourself, honest). You seem to want to build up to the specific things
you are interested in -- and that's okay if you're only interested in
those specific things. Polynomial rings are far more general (and are
abstract so as to allow that). Thus instead of building up to the few
specific cases you happen to think of, you can at once encompass all
the possible cases and narrow to whatever specific case you care to.
Importantly you can narrow to specific cases you might not have
otherwise thought of. Equally, given the powerful general framework,
you can prove things for the general cases and have those results
immediately apply to all the specific cases rather than having to
prove the same thing over and over again. So yes it is a little
complicated, and yes it is abstract, but the complication comes from
the abstraction, and the abstraction provides vast sweeping generality
that allows you to do far more, and see further, than special cases
allow. This is why you struggled to see that P4 was just RxC, but
others using polynomial rings and quotient rings can make the
connection simply and easily: the greater generality allows for much
greater flexibility and deeper insight.
Now, as to your complaint about "continuum modulo math": no one has
tried to explain what an ideal is yet, let alone try and explain (in
anything but the most vague handwaving way) what a quotient ring is
and how it works. Your assumptions about what must be going on are
wrong, and that's why it isn't very convincing to you. If you actually
understood how it did work it would be convincing. I won't bother
trying to explain, however, as I'll simply get accused of trying to
explain the second thing while pretending the first is crystalline. We
don't even have polynomial rings halfway sorted, so there's no point
in discussing the rest.
Let me just ask a small question here: was the issue I picked on here
a misunderstanding on your part, or an unfortunate use of language on
your part. You haven't made that clear at all, and it would help if I
actually knew.
> > Now, back to why we don't have to worry about
> > "superinfinite" (whatever that means) issues. Remember when I pointed
> > out that the nearly null requirement on tuples is equivalent to the
> > requirement that the polynomials have finite degree? Well given a
> > polynomial of finite degree n and a polynomial of finite degree m,
> > then their product will have *finite* degree m+n. Thus, as long as our
> > polynomials have finite degree, we know more have to worry about
> > multiplication causing problems than we have to worry about addition
> > of finite (but arbitrarily large) natural numbers causing problems.
> > You don't believe there is some fundamental issue with arithmetic
> > where it has to concern itself with "superfinite" numbers do you?
>
> When all practical usage of polynomials collapse these informationally
> dense structures down to one informational unit of their claimed
> content then yes, I do see a problem here.
Apparently I need to ask you a lot more questions because it is less
and less clear to me what on earth you think is going on. Can you
explain to me clearly and concisely how you think these "dense
structures" are "collapsed down to one informational unit"? For each
and every polynomial we care about every non-zero term. Yes we talk
about the degree of the polynomial, but that is simply extracting one
property out so as to prove that in general certain properties are
preserved. For practical purposes we usually care about all the
individual terms. For certain purposes these terms aren't important --
the proof above just happens to be one of them.
> You have carefully stated
> that you will work in null terminated sequences. These taken under
> product will propagate upward in length. I don't see that an honest
> answer can dismiss this awareness either way.
And no one did, except for technical issues with your sloppy use of
language again. You say the "length" of the sequence will increase --
the length of the sequences are always the same: countably infinite.
Thus Arturo simply complains you are talking nonsense, because you are
if read literally. Being more generous in interpretation: the position
of the last non-zero term will, indeed, appear later if you multiply
two sequences together; you actually have to say that though (or
define beforehand specifically what you mean by length, and explain
why you are using it in a non-standard way).
Now, the question is: why is this necessarily a problem? At a glance
it might be, or it might not be. You can't know until you try it out.
I have tried it out, and it isn't a problem. Can you explain very very
clearly and very specifically why this apparently "breaks" anything?
Let me instead ask you a series of questions, and hopefully we can pin
down where you think the problems arise.
How many terms does the result of multiplying the 3 term polynomial 1
+ X + 2*X^2 with the 2 term polynomial 2 + 4*X^6 have?
Is the answer to the previous question a finite number?
How many terms (at most!) do we expect the result of multiplying a n
term polynomial with an m term polynomial to have (where n and m are
finite numbers)?
Is the answer to the previous question a finite number?
If the multiple of any two polynomials has only finitely many terms as
long as the two polynomials being multiplied have finitely many terms,
is it possible to ever get a polynomial that doesn't have finitely
many terms by multiplying two polynomials with finitely many terms?
If the answer to the previous question is yes, please give an example
of two polynomials with finitely many terms whose product doesn't have
finitely many terms.
Suppose I want to find the result of multiplying together n different
polynomials where n is a finite number and all the polynomials have
only finitely many terms. For purposes of induction let's assume that
the result of multiplying n-1 polynomials that all have finitely many
terms is a polynomial with finitely many terms. Why can I not multiply
together the first n-1 polynomials to get a single polynomial with
finitely many terms, and be left with finding the multiple of two
polynomials, each with only finitely many terms?
Do the previous questions give a valid proof by induction that any
finite product of polynomials with finitely many terms gives a
polynomial with only finitely many terms?
If we take the set of polynomials with finitely many terms, and *all*
finite products thereof, are there any polynomials in that set that
don't have finitely many terms?
If the answer to the above question is no, in what way does the ring
formed by taking all polynomials with finitely many terms and all
finite products thereof have any problem with multiplication?
> Like working with
> infinity within calculus only its limit can be taken. I believe that
> infinity is something to be careful with.
Yes, so do I. The thing is, I have been careful with it, and I've
convinced myself that no entanglements arise when using infinity in
the way that I've described.
> It can be used, but should
> be used carefully.
See above.
> I think that dismissal of this feature especially
> to a concerned student is not wise. Simply owning this feature is
> enough and then moving on from there rather than being in denial is
> appropriate.
But that's what I'm doing. The only point I'm arguing here is your
claim that this somehow falsifies, or otherwise shatters the
consistency of polynomial rings. It doesn't. Can we move on now?
> The future usages of the math being constructed may cause
> conflict here. Aren't you open to this? Especially within the concept
> of ring whereby the operators are abstracted we should be capable of
> building some continuous iterate product or some such form where the
> result remains within the original domain, otherwise we are not in a
> ring system. You have chosen to work in an unistantiated form R[X] and
> if we raise this form to
> R[X] ^ m
> then we have constructed the same format but at a level of greater
> complexity.
I'll have to ask what you mean by R[X]^m. Are we talking about the
Cartesian product of m copies of R[X], or are we talking about
polynomials that are m-fold multiples of polynomials in R[X] or are we
talking about something else? Neither construction presents any
difficulties. I might (reasonably) question your desire to bother with
the former, but given any halfway decent argument as to why one might
want to consider it, I would be happy to do so. I'm not sure I see the
problem.
> Since we are engaged in generalities to the fullest you
> can take this one as a sneering example. You happily limit the length
> of the series whilst you insist that the series be infinite in length.
No. The requirement is that there be infinitely many components to the
vector, but that all but finitely many of those components must be
zero. At this point it is much easier to describe in terms of
polynomials: we simply require that the polynomial have finitely many
terms. Exactly how many terms that is is not bounded -- any natural
number is possible -- but it must be a *finite* number.
> Please do not take this criticism personally. This is the
> establishment mathematics you are upholding. I seriously doubt if one
> man alone would ever arrive at this format of construction. This is
> the work of one man standing on another man's shoulders. Yes, we'd not
> get very far without this behavior. Yet where this behavior can get us
> is still dubious. Twisting a zero into an infinity as a child plays
> with a loop of yarn should we assume the child's deep understanding?
> This situation is far more complicated yet the claim of mathematics as
> fundamental construction should be upheld. Scrutiny on this area does
> not resolve to the simplistic underpinning that some believe. Instead
> we have waivering nuances.
No, what we have a a disparate range of attempts to explain the
situation to you, given at different levels of formality, by different
people, all in the space of a few USENET posts, and with you lacking
the background to make the explanations relatively easy. The result is
that you don't get a single clear consistent development from first
principles up. If you add to that the fact that you are rarely if ever
willing to stick it out with the first principles (and instead want to
leap ahead before you've made sense of the beginning), and the result
is, yes, a little all over the map. Certainly my explanations, trying
to hit the different misunderstandings (which occur all over the map)
on the head, has rambled across the place. But then if you want a
clear and consistent development from first principles you should do
as several people here have suggested and actually invest in a good
textbook, or take a course.
> > You aren't making sense anymore. Take a moment and actually read what
>
> Sorry, but this is still making sense to me. If in calculus we were to
> specify a large n and then this large n lead to an even larger m
> within a computation we would not arrive at a computable result.
Okay, let's try with the questions again:
How many terms must a product of two polynomials have for it to become
uncomputable?
If a product of polynomials has only finitely many terms, can we
compute each of those terms?
If we compute all of the terms appearing in a product of polynomials,
have we computed the product of the polynomials?
Given that any finite product of polynomials with finitely many terms
must itself have only finitely many terms, in what way is it not
computable?
Finally:
Given that the real numbers contain uncomputable numbers, how do you
reconcile your need for computability with your desire to work with
real numbers?
> Yet
> the math which you purvey does contain this behavior. Products reach
> ever greater complexity, hence the requirement of not just a length n
> sequence, but an infinite length sequence. It's so simple to see this.
Well yes, we have infinite length sequences. I believe this was stated
at the very outset. Why is this a problem? You can complain that it is
unnecessarily complex, but see my earlier point about abstraction and
generality. Certainly it doesn't lead to any inconsistency or
contradiction. If it does, then please provide a nice complete proof
thereof.
> No, this point does not prevent you from going onward, but it is a
> point of valid awareness.
Sure, whatever. No one objected to the notion that we will require
infinite length tuples (or polynomials of arbitrarily high degree).
People objected to your claim that this somehow made the mathematics
inconsistent, or false, or wrong. It doesn't. So: it happens, it
doesn't cause any problems, can we move on now?
Tim BandTech.com
Wow. You are really bitter.
That's OK. So is coffee.
As I think I've laid it out mathematics does require a level of
scrutiny. Rather than take the academic path, whereby the successful
will be those who obey the teacher's directives the best, I suggest
that here on Usenet you witness a superior form of learning. There are
no grades and participatioin is fully voluntary. Yes, you get a lot of
idiosynchratic interaction. This is a statement on the human race. Can
there be any doubt that your own internal representation of the
mathematics in dispute is unique to another person's? The masters of
the subject will no doubt quibble over one another's presentation of a
topic in their own book styles. Are such nuances relevant? Are they
all perfectly equivalent? Even within a consistent form such as those
accepting the label 'abstract algebra' qualities do arise. You have
placed full value in your own presentation. I respect that. However my
own scrutiny must be empowered. Rather than validate this necessary
feature of mathematical behavior you deny it. You are the breeder of a
past bread. Your type makes a fixed religion of a topic that is best
left ultimately open at its base. This then enlivens the subject.
Rather than moles discussing dead bread and cheese these cultures can
develop further. I have instantiable proof that this subject remains
open. The lack of sensibility of the language in use here is not
entirely my own. Surely you must see that. Unfortunately there is a
form of repulsion here. I am a bad dog. You come back to tell me this
over and over. And I come back and roll over and play dead, but then I
will get up again and perhaps surprise you. Your own lack of
concession is more proof than my own words are. I've seen this many
times here. Those who concede at least a little bit tend to be more
credible. Anyway, mathematical minds are unusual at the least. I am
happy to have any interaction, including your own hatefulness. This is
par for the course of humanity. The funniest thing to me is to think
of all of the brilliant people who have rejected a subject for its
flaws and their failure within the system because the system refuses
to be challenged. You exemplify this trait. I happily accept my
failure within your system. The mime and the meme of the time form a
theme...
If I had only rejection for your system then my position would be
compromised. Instead I do offer a replacement. This replacement is not
perfectly congruent, which is a good thing because you deal in an
elemental form which has no content.
- Tim
Tonico
******************************************************************
As if somebody still needed some evidence that you have no half an
idea of what you're ranting about, the above shows you've never
honored a university's maths class room with your presence.
Which, of course, is just fine: some study, some don't.
You've turned out to be one of the most bizarre and annoying cranks
this NG has ever had the luck to receive: you know bananas, you've
been corrected, and in many things, over and over by people who know
much more than you do about this stuff, and you insist in your
voluntary ignorance and display a rare kind of stupidity which comes
along with an almost unbelievable stubborn attitude trying to correct
the ones that know!
Stay on course, and please do meet and talk to Musatov, Inverse 19,
James Harris, WM and the rest of nice cuckoos we enjoy of here.
Regards
Tonio
Leland McInnes
> On Jun 15, 12:39 pm, Leland McInnes <leland.mcin...@gmail.com> wrote:
>
> > On Jun 15, 9:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > Thanks for the simple advice Virgil. On the one hand I see the
> > > polynomial form as a farily simplistic thing, but on the other hand I
> > > see the requirement of not evaluating X as very challenging.
>
> > If you can just get your mind around it, trust me, it won't be
> > challenging at all.
>
> > > Treating
> > > this expression as elemental when it cannot be evaluated is a
> > > hypocritical step.
>
> > In what sense is it hypocritical exactly?
>
> On the one hand you've insisted that X will never be evaluated.
> On the other hand you're going to wind up evaluating X. For instance
> in the buildout of the complex numbers from your infinite dimensional
> X-form.
Well, you assume that we'll end up evaluating X, but that assumption
is based on your own wild guesses as to how this will all work, which
is getting way ahead of what anyone has explained to you. The fact is,
we won't have to evaluate X to create a ring isomorphic to the complex
numbers. As surprising and unbelievable as that may sound to you, it
is the case.
> > What, to you, makes something "instantiable" exactly? What is more
> > instantiable about the symbol "i" than the symbol "X"?
>
> Nice question. I'm not so sure that there is much better about the
> symbol i.
> The math which I spoke of that replaces the X format has no need of
> the i format either.
No, it simply pushes the problem somewhere else again. What is more
"instantiable" about the symbol "#" than the symbol "i"? That is, what
does "#i" mean in an "instantiable" way? What does "instantiable" even
mean exactly as far as you are concerned?
> We need
> to address
> X X = - 1 .
> Here we see an assignment of a real value to one of these mysterious
> domains of unassignables in X^n. This statement above I believe to be
> invalid by your own usage of X.
Yup. To get the isomorphism you need to use quotients and what you'll
get is *an isomorphism* of rings. Look that up and see what exactly it
means.
> You've insisted that these X do not
> perform this way. A real value is being assigned here to XX. You've
> already insisted that this need never happen. These X are somehow
> sacred cows, or scare crows, all in a line forming some fence that
> need never be touched in a field of virtual hay which can somehow
> still contain real values.
No, they don't. Look, you need to actually understand quotients or
rings (and for that you need to understand ideals, and a whole host of
other things I'm beginning to suspect you don't entirely grasp) before
you can go pontificating on what is going to go on. Right now your
making wild speculations, assuming that your right, and announcing
that therefore this whole field of mathematics is nonsense. I'm pretty
sure I've warned you several times that this isn't a good plan. Odds
are that if your wild speculations as to how you think things will
work results i bad things, it's probably because your wild
speculations about how things will work are wrong.
I think the real problem here is that you are very very predisposed to
assuming that there must be some problem in polynomial rings, quotient
rings, and commutative algebra in general. Each and every time there
is the least hint of a misunderstanding you leap straight to the
conclusion that the whole field is rotten. That, to me, is very bad
faith, and says that you aren't actually interested in learning this
at all -- you just want us to give you excuses to ignore it. If that's
the case I see little point in wasting my time giving explanations.
Either work in good faith, or admit you aren't interested in actually
learning anything.
> It would be great to see your quotient ring go to work. I want to
> understand how this happens.
See, I'm not all sure you do anymore. It seems to me that you aren't
at all interested in learning about this because you don't bother let
people explain it, but instead yell about how the whole system is
corrupt. Why exactly should I be helping you if you aren't really
interested in learning, and won't actually listen?
> > Right, well please demonstrate how all of this building up can be used
> > practically in algebraic number theory and algebraic geometry. I
> > assure you, you'll simply end up reconstructing the same thing with
> > different terminology.
>
> If this is true then all the better. There will be no 'X' at the
> bottom of the construction which has no substitutable quality.
No. That's what I'm trying to tell you: there will be. To reconstruct
all the material of practical value in algebraic geometry and
algebraic number theory you will end up creating just such an object.
KMF
> <leland.mcin...@gmail.com> wrote:
> > On Jun 15, 9:37 am, "Tim BandTech.com"
> <tttppp...@yahoo.com> wrote:
> >
> >
> > > Thanks for the simple advice Virgil. On the one
> hand I see the polynomial form as a farily simplistic thing, but on the other hand I
> > > see the requirement of not evaluating X as very
> challenging.
> >
Because a polynomial is different from a polynomial
_function_. If you mix these two, you will surely be
confused.
> > If you can just get your mind around it, trust me,
> it won't be challenging at all.
> >
> > > Treating
> > > this expression as elemental when it cannot be
> evaluated is a hypocritical step.
> >
> > In what sense is it hypocritical exactly?
>
> On the one hand you've insisted that X will never be
> evaluated.
> On the other hand you're going to wind up evaluating
> X.
It is not evaluated when X is an element of R[X],
but it will be evaluated when we consider a polynomial
function. You treat things differently in different
contexts/situations. An abstract, general function
f(x) cannot be evaluated, but once you have a specific
function, say, f(x)=x^2 -- a poly. function -- then you
can evaluate it. You may define an algebra of functions,
with f(x)+g(x)=(f+g)(x) , without evaluating.
What's wrong with that?.
This is false. This operation is taking place at a
_coset_ level. XX and -1 are _cosets_ . Cosets
satisfy the property:
[A+B]=[B]+[B]
[-A]=-[A] (in the Abelian case)
[0] is the id. element in the quotient ring, i.e.,
[C]+[0]=[C]
The coset [XX+1] is equivalent, under this relation,
to the coset, 0, to [0]. Then:
[XX+1]=[XX]+[1]=[0] so
[XX]=[-1]+[0]=[-1]
This is the solution to the mystery of X^2=-1
So this is an identity at a _coset_ level.
Under this construction of the complexes,
the numbers are actually cosets. We don't usually
use coset notation because it is cumbersome.
But in this construction, each term a+ib is
a coset representative.
i is a representative for the coset [X] , the
equiv. class of polys. in R[X] , with remainder
x , when dividing by x^2+1
This statement above
> I believe to be
> invalid by your own usage of X. You've insisted that
> these X do not
> perform this way. A real value is being assigned here
> to XX.
See above.
You've
> already insisted that this need never happen. These X
> are somehow
> sacred cows, or scare crows, all in a line forming
> some fence that
> need never be touched in a field of virtual hay which
> can somehow
> still contain real values.
Of course, you are here excluding the possibility that
you have not fully understood things, and you have
refused to take a middle-of-the-road approach of
reading a book, doing some of the exercises, and then
getting back to us with questions after being informed.
Your misunderstanding of the meaning of XX=-1 shows
that you could benefit from hitting the books.
>
> Complex analysis tags along with real analysis fairly
> well. Also
> complex math is involved in a fair amount of physics
> and engineering
> computation. I do accept the utility of this
> rotational form. That
> rotational form is also existent within the real
> value, though it's
> apparency in the real form is diminished beneath what
> most will
> perceive.
In math, you need to define things more clearly.
I don't think anyone can help you unless you clarify
for us what you mean with expressions like:
"existent within the real value" , or "its apparency..."
I accept the junction of these two maths;
> the real numbers
> and the complex numbers. There need not be so much
> magic in getting to the next dimension as is going on in the ring
> quotient.
Again, you are using a term 'dimension'that has different (but specific) meanings in different contexts
loosely. Math is an inherently technical area, and
does not lend itself well to informal talk, nor
informal talk by practitioners that are not aware
of the precise meaning assigned in precise contexts.
So: what do you mean by dimension?. I assume you
are considering some vector space dimension. Why
are you doing this when working with rings?
You are being very loose with your terminology
and phrasing. There is no way around understanding
the precise meaning of each term and definition.
And I agree with others that you have not done
your part of the deal in trying to understand.
Again, why not meet us halfway and read a chapter
on this, do a few exercises, and then get back to us?
Rather, the
> fundamental definitions of the real numbers (as I
> recall there are already six or seven versions of them) can be generalized.
>
> It would be great to see your quotient ring go to
> work. I want to
> understand how this happens. Just how you can breach
> the hump of
> claiming X to not carry any instantiable value and
> then turn around
> and substitute real values in is puzzling to me.
It is the difference between a formal expression and
a function. You can analyze (formal) properties of
an object like R[X] without instantiating. And
you can talk about an element of R[X]
I
> will try to keep
> open minded. I am ready for your presentation. Still,
> it amazes me and amuses me how much you all are willing to eat that hay from the field
> and obey the bizzarre fence that has been laid.
>
Of course, you are completely dismissing off-hand
the possibility that you just don't get it.
> >
> > > I do find this
> > > circumstance understandable as mathematicians in
> their plurality and over generations cover new ground and fill out any voids in the
> > > possibility space. This is where 'new' work lays.
> Yet here I suspect that eventually this area will be withdrawn.
Seriously?. You have been dealing with quotients
for what, a month?. And you think your opinion
is well-informed?.
> >
> > I'm very confident that it won't be. Look,
> polynomial rings are used
> > regularly in all manner of modern mathematics, and
> are particularly
> > foundational in modern algebraic geometry and
> algebraic number theory.
> > Not only have these fields made significant
> theoretical progress,
> > they've produced very solid practical results.
> Polynomial rings are
> > quite rigorous, well defined, and lead to solid
> practical results.
> > They aren't going anywhere.
> >
> > > The rotational aspects of
> > > the polynomial are already inherently within the
> real number. So
> > > rather than build a confusing structure up above
> the polynomial (the
> > > quotient ring)
You are being loose and imprecise. This is built
above the polynomial _ring_
>all that needs to be done
For whom?. Why?. We cannot read your mind. You
seem to be making assumptions that are clear only
to you.
is to
> build out the real
> > > number more generally. Then, rather than worry
> about the need for
> > > infinite length sequences
why do you worry about them?. Many have made
an effort to explain to you. Why not respond
in kind and explain clearly what your problem is,
assuming as little as possible?. I think many
here are not clear on what your doubt is about.
Specifically pinpoint the contradictions
that you see.
the n-length sequence
> can hold its own since
> > > products will remain at length n. This math
> relies upon a modulo
> > > effect which is already present within the real
> valued number.
> >
> > Right, well please demonstrate how all of this
> building up can be used practically in algebraic number theory and algebraic geometry. I
> > assure you, you'll simply end up reconstructing the
> same thing with
> > different terminology.
>
> If this is true then all the better. There will be no
> 'X' at the bottom of the construction which has no substitutable quality.
So when you describe something axiomatically, what
should you do?. You can define an object that satisfies
, say, the group axioms, and derive some properties of
them in a purely formal way. Should we abandon
axiomatic descriptions?
Yes, I
> do believe that this is what I'm talking about. There
> will be no need of infinite length progressions.
Further the complex numbers take
> their place as a natural extension from the real
> numbers inherently within the construction.
??
At the basis of the
> terminology that I
> suggest is merely a generalization of the real number
> and its own
> discrete symbollic signage. Thus the chasm between
> continous and
> discrete terminology is layed out on a different map,
> yet this map
> does match up to much of existing mathematics and
> especially the
> development of what might be called multidimensional
> geometry.
??. I have no idea of what you mean by all this.
The
> naming of all these genera of mathematics is a
> disgusting thing and it has taken me a long time to arrive at the subject of abstract algebra.
> It does seem to be the correct spot to focus on.
>
I think that more could be done, like, say the
way the IEEE goes about setting standards.
> - Tim
Arturo Magidin
> On Jun 15, 12:00 pm, Arturo Magidin <magi...@member.ams.org> wrote:
>
>
>
> > On Jun 15, 10:31 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > Here Arturo now says that he did concede the
> > > dimensional interpretation whereas before he obfuscates that.
>
> > I said it before twice, I'll say it a third time: since you cannot
> > accurately report what I did or did not say, please stop trying to.
> > All you are doing is *lying your ass off* whenever you attempt to
> > claim I did or did not say something.
>
> > > I am
> > > open to having truly fundamental misunderstandings.
>
> > Bullshit. You say the words, but you don't mean them. You have kept
> > that "truly fundamental misunderstanding" and that truly fundamental
> > ignorance steadfastly throughout the entire thread, and compound it by
> > lying and misrepresenting what others say. You aren't open to
> > anything.
>
> > Don't claim to report what I did or did not say, because, quite
> > simply, you have shown you are incapable of doing so without lying.
>
> > --
> > Arturo Magidin
>
> Wow. You are really bitter.
It's not "bitterness"; it's annoyance. Willful ignorance really gets
my goat, and you are a rather clear and extreme example of such.
That you compound your willful ignorance with misrepresentation,
misquotation, and misunderstanding of what I wrote, and have the
chutzpah of trying to attribute to me your stupidities, well, that
doesn't make me bitter either; it just annoys me all the more.
So, kindly, make it clear that everything you write comes from your
own unfathomable pit of ignorance and stupidity, and stop trying to
lay your idiocies at other people's doorstep. They are all your tiny
little children, and you might as well claim them as your own.
--
Arturo Magidin
KMF
By people who have bothered to informed themselves
of definitions and how things actually work. You
want to do disinformed speculation, and that does
not work in an area like mathematics.
I think people are upset because you have not
bothered to learn what you are talking about before
criticizing it.
Terminology in mathematics is precise for a good
reason you seem not to understand.
It is self-serving for you to blame the mathematical
community without making an effort to inform yourself
before talking.
Rather than take the academic path, whereby
> the successful will be those who obey the teacher's directives
bullshit.
the best, I suggest
> that here on Usenet you witness a superior form of
> learning.
By pure speculation?. Inform yourself about what
you are criticizing, before criticizing it and you
will receive a better response. Then you will have
the superior form of learning: by having a response
to your _well-informed_ question on _mathematics_
There are
> no grades and participatioin is fully voluntary. Yes,
> you get a lot of idiosynchratic interaction. This is a statement on the human race. Can
> there be any doubt that your own internal
> representation of the mathematics in dispute is unique to another person's?
> The masters of
> the subject will no doubt quibble over one another's
> presentation of a
> topic in their own book styles. Are such nuances
> relevant? Are they
> all perfectly equivalent? Even within a consistent
> form such as those
> accepting the label 'abstract algebra' qualities do
> arise. You have
> placed full value in your own presentation. I respect
> that. However my
> own scrutiny must be empowered. Rather than validate
> this necessary
> feature of mathematical behavior you deny it. You are
> the breeder of a
> past bread.
You are ignorant and yet continue to talk. It is
absurd to believe that there is a plot by the whole
mathematical community to ignore these holes. You
repeatedly ignore the alternative explanation that
you are confused and have not bothered to inform
yourself before talking. How self-serving: your
disagreement is automatically attributed to a system
that is trying to preserve itself. No need for
self-questioning (and growing up) on your part
to consider your own role.
Your type makes a fixed religion of a
> topic that is best
> left ultimately open at its base. This then enlivens
> the subject.
> Rather than moles discussing dead bread and cheese
> these cultures can
> develop further. I have instantiable proof that this
> subject remains
> open. The lack of sensibility of the language in use
> here is not
> entirely my own. Surely you must see that.
> Unfortunately there is a
> form of repulsion here. I am a bad dog.
Yes. You are the pure, innocent victim against
the dark empire, Luke, er, I mean, Tim. When
there is a disagreement, there will _never_ be
fault on your part.
You come back
> to tell me this
> over and over. And I come back and roll over and play
> dead, but then I
> will get up again and perhaps surprise you.
Read a book, do some problems, reflect on the
topic , and then, _after_ having done this, get
back, and I am sure most will be surprised.
Your own
> lack of
> concession is more proof than my own words are. I've
> seen this many
> times here. Those who concede at least a little bit
> tend to be more
> credible.
When have you conceded anything?.
Anyway, mathematical minds are unusual at
> the least. I am
> happy to have any interaction, including your own
> hatefulness.
Your treatment is insulting too, to many.
You dismiss people who put pride in their work
as followers. Would you like if anyone who disgreed
with you to call you this?
This is
> par for the course of humanity. The funniest thing to
> me is to think of all of the brilliant people who have rejected a subject for its
> flaws and their failure within the system because the
> system refuses to be challenged.
This is true. But it is also true that there
are many who criticize without informing themselves.
You exemplify this trait.
Ditto for the above.
>I happily accept my failure within your system.
Of course, noble (self-righteous) warrior. When
you fail is the big bad system. It is surely not
_your_ fault.
Tim BandTech.com
>
> Well yes, that would be true if the only way to do anything was to
> consider X to be a some real number. That isn't the only way to do
> things, however, so this criticism is quite moot.
>
> > This
> > is just one of several contradictory contexts of the ring quotient
> > construction
> > R[ X ] / ( X X + 1 )
>
> I'll repeat again: there are no contradictions if you actually bother
> to understand the construction. We have even gotten as far as getting
> you to fully understand R[X], let alone quotient rings, so can you
> leave the speculative criticism aside until you actually understand
> what you're talking about.
>
> > > Right, except for the requirement that all but finitely many of the
> > > components are zero. I shied away from explaining the reason for this
> > > since the only good explanations I know (that don't involve explaining
> > > it in terms of polynomials of finite but unbounded degree, which were
> > > apparently not a good way to explain it to you) wade us into rather
> > > deeper waters.
>
> > I think this is more like a game of trickery. You get a student to eat
> > a fist step that doesn't quite jibe with the next step. Then you cover
> > that first over, tunnel over to the next thing and present it as if it
>
> No one is trying to trick you Tim. There are libraries with lots of
> books on these topics. There are a whole host of free online resources
> for these topics. Feel free to actually read any of those instead of
> listening to us. Perhaps read some of those and ask about anything
> that doesn't quite match with what we say. You'll likely get
> explanations for why the different things do jibe, it just isn't
> necessarily obvious at first.
>
> As to the issue of moving to the second thing before the first is
> clear: that's your doing, not ours. You are the one who jumps ahead
> and brings up issues about things that have not yet been addressed in
> our explanations and demands answers. As I said, it feels like Whack-a-
> mole to me too, since as soon as I explain one point, you are
> demanding answers on something else down at the third of fourth step
> and getting way ahead of yourself. Am I to simply ignore your demands
> for explanation, or do things somewhat out of order out of necessity
> of keeping up with your demands?
>
> > I understand that series which converge to zero are nice things to
> > work with. I am fine with your relaince upon these.
>
> Language use nitpick time: series that converge to zero are something
> quite different from nearly null infinite tuples. A series is a sum
> (usually infinite) and to say it converges to zero is to say that the
> sequence of partial sums has limit zero. That has nothing to do with
> nearly null infinite tuples. Either you are misunderstanding something
> here, or you are being very sloppy with language. It's this sort of
> thing that I believe infuriates Arturo because he will take you at
> your word on "series which converge to zero", see that it has nothing
> to do with what he was talking about at all, and call your comment
> nonsense. I'm going to give you the benefit of the doubt and assume
> you meant something else but were sloppy with your language.
Yes, I'd like to get to the ideal. That seems crucial.
A ring is defined as containing elements. Such elements have clear
elemental instances in the reals, the integers, or the complex
numbers, whereas for R[X] there are no such elemental forms. Instead
we see a form
(a0,a1,a2,...,an,...)
which is not elemental. Already this form carries a variable of
undefined domain and an infinite length of terms. This is far from
elemental. In that the form
(a1, a2, a3 )
is simpler than the infinite length above then there exists a more
elemental form than the infinite series. In fact, the elements in the
infinite series are reals. Thus there is a hint of an elemental form
in the representation and such an elemental form is instantiable such
as
- 2.35
which then can be generalized to a symbolic form
a
which is then developed up to a very complicated form at the
polynomial level especially since no complete instance can be
specified due to the infinite length. Consider whether a derived form
is more elemental rather than less elemental than the starting
element. This would be a wise consideration: if a choice of form
exists then a choice which is more elemental would always choose the
simpler form. Like atomic theory we would make a mistake by labelling
a molecule an atom.
Still, here we will not resolve our difference so I announce Uncle in
the interest of moving onward toward the ideal.
>
> > > Now, back to why we don't have to worry about
> > > "superinfinite" (whatever that means) issues. Remember when I pointed
> > > out that the nearly null requirement on tuples is equivalent to the
> > > requirement that the polynomials have finite degree? Well given a
> > > polynomial of finite degree n and a polynomial of finite degree m,
> > > then their product will have *finite* degree m+n. Thus, as long as our
> > > polynomials have finite degree, we know more have to worry about
> > > multiplication causing problems than we have to worry about addition
> > > of finite (but arbitrarily large) natural numbers causing problems.
> > > You don't believe there is some fundamental issue with arithmetic
> > > where it has to concern itself with "superfinite" numbers do you?
>
> > When all practical usage of polynomials collapse these informationally
> > dense structures down to one informational unit of their claimed
> > content then yes, I do see a problem here.
>
> Apparently I need to ask you a lot more questions because it is less
> and less clear to me what on earth you think is going on. Can you
> explain to me clearly and concisely how you think these "dense
> structures" are "collapsed down to one informational unit"?
You seemed happy enough with Arturo's presentation where he takes
(a, 0, 0, ... )
to be simply a. It's all the same point isn't it? So I must bow and
accept your insistence that the infinite length series is as
fundamental as a singular element of that series in order to move on.
My own perspective really is quite defensible here. But you squeeze
and I say 'Uncle'.
> For each
> and every polynomial we care about every non-zero term. Yes we talk
> about the degree of the polynomial, but that is simply extracting one
> property out so as to prove that in general certain properties are
> preserved. For practical purposes we usually care about all the
> individual terms. For certain purposes these terms aren't important --
> the proof above just happens to be one of them.
Alright, here you have contradicted yourself within this one
paragraph. First you say
"For each and every polynomial we care about every non-zero term."
Then you say
"For certain purposes these terms aren't important"
Again we are on the same silly point of contention and we have fully
amplified it ad nauseum. Your attritive method is worn thin here where
you have contradicted yourself within one paragraph of one post while
the issue has been worn to the bone. I say 'Uncle'... with exposed
bone and I am still alive...
>
> > You have carefully stated
> > that you will work in null terminated sequences. These taken under
> > product will propagate upward in length. I don't see that an honest
> > answer can dismiss this awareness either way.
>
> And no one did, except for technical issues with your sloppy use of
> language again. You say the "length" of the sequence will increase --
> the length of the sequences are always the same: countably infinite.
> Thus Arturo simply complains you are talking nonsense, because you are
> if read literally. Being more generous in interpretation: the position
> of the last non-zero term will, indeed, appear later if you multiply
> two sequences together; you actually have to say that though (or
> define beforehand specifically what you mean by length, and explain
> why you are using it in a non-standard way).
>
> Now, the question is: why is this necessarily a problem? At a glance
> it might be, or it might not be. You can't know until you try it out.
> I have tried it out, and it isn't a problem. Can you explain very very
> clearly and very specifically why this apparently "breaks" anything?
Because of the insistence on infinite length sequences the product of
two such sequences whose terms compute out to length 2n will not be
computable. Call this an informational conflict if you like. I have
sited other forms of mathematics where the usage of large n does not
yield this quagmire. However, I concede that for the special case of
finite length series the product of two such sequences will be finite.
However in the case of general product of say one million product
operations that the conflict of interest here, or rather the relevance
of this detail requires accomodation. Because rings are concerned with
the properties of such operators then this attribute on the polynomial
form should be amplified rather than ignored. As you say it might be
or it might not be, and in the most highly generalized form it is, but
you have restricted your study and so we can go on, Uncle.
>
> Let me instead ask you a series of questions, and hopefully we can pin
> down where you think the problems arise.
>
> How many terms does the result of multiplying the 3 term polynomial 1
> + X + 2*X^2 with the 2 term polynomial 2 + 4*X^6 have?
( 1 + X + 2 X^2 )( 2 + 4 X^6 )
= 2 X^0 + 2 X^1 + 4 X^2 + 4 X^6 + 4 X^7 + 8 X^8 .
The result has six nonzero terms.
>
> Is the answer to the previous question a finite number?
Yes, Uncle.
>
> How many terms (at most!) do we expect the result of multiplying a n
> term polynomial with an m term polynomial to have (where n and m are
> finite numbers)?
You must mean nonzero terms here. There should be at most m+n nonzero
terms.
>
> Is the answer to the previous question a finite number?
Yes.
>
> If the multiple of any two polynomials has only finitely many terms as
> long as the two polynomials being multiplied have finitely many terms,
> is it possible to ever get a polynomial that doesn't have finitely
> many terms by multiplying two polynomials with finitely many terms?
Polynomials are defined to have an infinite number of terms, however
the number of nonzero terms obeys the m+n principle only for the
simplest product form that you have selected. Since rings are
concerned with these operators generally then the consideration of the
quantity of product operations ought to be considered within this
specific study. Thus for instance a product series form would fill
this study out nicely, rather than repeating the same old single
product operation. Also it should be noted that my own usage of
'nonzero terms' is weak since the expression
(0,0,0,0,4.5,0,0,0,5.6,0,0,0,...)
contains two nonzero terms. The degree then matters more than the
quantity of terms. Polynomials in two nonzero terms may exist where
the polynomial is infinite in length. Any position may be occupied by
these two nonzero terms by symmetry and so the expressibility of the
form is not general. We can have a two nonzero term polynomial
represented by
(0,0,0,...,0, 4.5,0,0,0,0,...,0,5.6,0,0,0,...) .
Performing operations on this two nonzero term polynomial is not
possible yet by symmetry this form is no different under the laws of
polynomials than the form
(4.5,5.6,0,0,0,...).
While one permits computation the other does not, yet the forms are
identical since X is merely a placeholder of nondescript nature. Each
sequence carries the same quantity of information. This is simply
index shuffling, however because the possible length of the series is
infinite then the upper bound of products on the first form is not
instantiable. We have chosen position n1 and n2 as the positions of
content. Until they are specified then the two nonzero term polynomial
in general is not computable. These considerations go beyond the
simple minded form that you have chosen and may become serious
considerations, but this development here is only my first go at this
problem. You can ignore this information in pursuit of reaching the
ideal and I say Uncle again.
>
> If the answer to the previous question is yes, please give an example
> of two polynomials with finitely many terms whose product doesn't have
> finitely many terms.
( 0,0,0,...,0,1.1,0,0,0,...,0,1.2,0,0,0,0...)( 1.3,1,4,0,0,0,0,...)
There are no polynomials with finitely many terms. Polynomials are
defined to have infinitely many terms. Above is a polynomial product
of two 'elements' each with two nonzero terms which is not computable,
yet this product is known to have finitely many non-zero terms, but it
does not and never will have finitely many terms.
>
> Suppose I want to find the result of multiplying together n different
> polynomials where n is a finite number and all the polynomials have
> only finitely many terms. For purposes of induction let's assume that
> the result of multiplying n-1 polynomials that all have finitely many
> terms is a polynomial with finitely many terms. Why can I not multiply
> together the first n-1 polynomials to get a single polynomial with
> finitely many terms, and be left with finding the multiple of two
> polynomials, each with only finitely many terms?
Again per the definition of polynomial there are no finite length
polynomials. Because the nonzero terms may appear at position n1, n2,
etc. within the general form until those positions are specified the
computation cannot take place. This is a matter of instantiation and
elementary forms, under which the infinite length polynomial is self
conflicted.
It seems more that you are trying to consider forms where the
breakdown as
(a x + b)( c x + d )( e x + f)
are in play. But because x is merely a placeholder this form is no
better than the original polynomial format. Indeed, it seems that
these x ought to be considered as real values here for under
superposition the ring declares its operators to be functioning within
the same domain. However there is a direct conflict with the
polynomial definition here, where x shall not be evaluated at a real
value. This in a nutshell would be the most direct route to the
conflict. I am sorry it has taken me so long to say it so simply. The
ring definition is providing operator theory and yet the polynomial
form provides its own operator theory. These two are in conflict with
each other. To define a superposition operation within a definition
which provides a superposition operation is a logical conflict. Only
just waking up here. Thanks Leland.
>
> Do the previous questions give a valid proof by induction that any
> finite product of polynomials with finitely many terms gives a
> polynomial with only finitely many terms?
No. This subject is an invalid construction as the ring definition
provides the superposition operator and thus to provide a
superposition operator within as a declared ring is not consistent,
especially one which refuses to be placed in the same domain as its
realvalued elements dictate.
>
> If we take the set of polynomials with finitely many terms, and *all*
> finite products thereof, are there any polynomials in that set that
> don't have finitely many terms?
For finitely many nonzero terms it is true that for a singular product
operation there will be finitely many, though their positions are not
necessarily clearly defined. I guess I'll say yes here, but the lack
of attention to the general form and the usage of superposition
operator makes these questions of dubious quality.
>
> If the answer to the above question is no, in what way does the ring
> formed by taking all polynomials with finitely many terms and all
> finite products thereof have any problem with multiplication?
Because the ring itself is providing the quality of superposition and
multiplication the declaration of a form in X which does not allow the
superposition to be taken consistent with the ring operator's quality
yields an invalid construction.
>
> > Like working with
> > infinity within calculus only its limit can be taken. I believe that
> > infinity is something to be careful with.
>
> Yes, so do I. The thing is, I have been careful with it, and I've
> convinced myself that no entanglements arise when using infinity in
> the way that I've described.
These entanglements that I've laid out are sort of second order. I'm
sure I've really got you riled up now. Well, you got me riled up too.
I've got to engage the problem on the superposition operator back out
at ground zero now. Thanks for helping me through this. I know that to
you I am being a dink, and likewise the same back on you. I'm afraid
the anal nature of mathematics brings this forth in human behavior. We
are caught in a language that you've accused me of misusing and all
that I can do is reflect the misuse backward onto yourself and your
compatriots.
My freestanding argument now goes that the ring R[X] is an inherently
invalid construction by the very usage of superposition within the
polynomial, for that superposition does inherently allow the
substitution of real valued X. Thus the insistence on X as
unevaluatable is invalid. By superposing a real value with X as in the
form
(ax+b)(cx+d)
the ring behavior dictates that X be real. Thus the polynomial form
collapses to a single real value.
- Tim
>
> > It can be used, but should
> > be used carefully.
>
> See above.
>
> > I think that dismissal of this feature especially
> > to a concerned student is not wise. Simply owning this feature is
> > enough and then moving on from there rather than being in denial is
> > appropriate.
>
> But that's what I'm doing. The only point I'm arguing here is your
> claim that this somehow falsifies, or otherwise shatters the
> consistency of polynomial rings. It doesn't. Can we move on now?
Leland you've been great about conceding those small points. I know
that my language is loose and I'm still trying to bring it into the
strict conformity. Your own language can be criticized as loose as
well. Within my loose language though you have conceded the points
that I have made and so the many criticisms of my own complete and
utter senselessness are essentially defended by your concession. You
are caught in contradictions as I've pointed out in quotes above here
and I do still hold to this construction as a very weak one. Yet still
I say Uncle in favor of moving on to the ideal. I do get the
polynomial form that you purvey. I've gotten it from the beginning,
but admittedly the stricture of language as communication is
frustrating. Ultimately we are talking about something very simple and
all of this verbiage is merely a human level of expansion. I do not
take back my earlier refutations but do concede that my language can
and should be refined. Even any who write a modern text concern
themselves with this. Thank you so much for engaging in such a lengthy
discourse, and while I Uncle my way through it is in doing so that
you've causated a new response. This is s value of this format. I know
you don't like what I say, yet we've really just been saying the same
things over and over and over. Your side has supposedly been highly
refined not just by you but by many like you. My side is new and
different and has a slight vitality. Best of all I think I've made it
concise now finally, though it is buried in this post.
>
> > The future usages of the math being constructed may cause
> > conflict here. Aren't you open to this? Especially within the concept
> > of ring whereby the operators are abstracted we should be capable of
> > building some continuous iterate product or some such form where the
> > result remains within the original domain, otherwise we are not in a
> > ring system. You have chosen to work in an unistantiated form R[X] and
> > if we raise this form to
> > R[X] ^ m
> > then we have constructed the same format but at a level of greater
> > complexity.
>
> I'll have to ask what you mean by R[X]^m. Are we talking about the
> Cartesian product of m copies of R[X], or are we talking about
> polynomials that are m-fold multiples of polynomials in R[X] or are we
> talking about something else? Neither construction presents any
> difficulties. I might (reasonably) question your desire to bother with
> the former, but given any halfway decent argument as to why one might
> want to consider it, I would be happy to do so. I'm not sure I see the
> problem.
It's just the mth power of an existing polynomial.
>
> > Since we are engaged in generalities to the fullest you
> > can take this one as a sneering example. You happily limit the length
> > of the series whilst you insist that the series be infinite in length.
>
> No. The requirement is that there be infinitely many components to the
> vector, but that all but finitely many of those components must be
> zero. At this point it is much easier to describe in terms of
> polynomials: we simply require that the polynomial have finitely many
> terms. Exactly how many terms that is is not bounded -- any natural
> number is possible -- but it must be a *finite* number.
Well, it's worse than this isn't it? I have shown a finite number of
nonzero terms through symmetry to be troubling.
Obviously I'll have to let you respond and tell me how much nonsense I
am full of, Uncle Leland.
I'll accept this and as you can see I am even able to correct your own
terminology.
So when we're talking sloppy let's allow for some symmetry. We all are
due to be corrected. Still I must defer to you ultimately as the
definer and though at this point you've squeezed me down to a skinny
pancake I'm squashing out sideways from the pressure.
Somehow there is a conservation principle in play.
>
> > > You aren't making sense anymore. Take a moment and actually read what
>
> > Sorry, but this is still making sense to me. If in calculus we were to
> > specify a large n and then this large n lead to an even larger m
> > within a computation we would not arrive at a computable result.
>
> Okay, let's try with the questions again:
>
> How many terms must a product of two polynomials have for it to become
> uncomputable?
Only one nonzero term each:
(0,0,0...,1.1,0,0,0,...)(0,0,0,...,2.2,0,0,...)
>
> If a product of polynomials has only finitely many terms, can we
> compute each of those terms?
We can compute the values of the terms but their positions are not
inherently known.
>
> If we compute all of the terms appearing in a product of polynomials,
> have we computed the product of the polynomials?
No.
>
> Given that any finite product of polynomials with finitely many terms
> must itself have only finitely many terms, in what way is it not
> computable?
In the way that I have shown.
>
> Finally:
>
> Given that the real numbers contain uncomputable numbers, how do you
> reconcile your need for computability with your desire to work with
> real numbers?
They are merely a more elemental form that the polynomial form, whose
elemental nature is highly dubious, especially when defined inherently
as a ring and deemed unevaluatable in X.
>
> > Yet
> > the math which you purvey does contain this behavior. Products reach
> > ever greater complexity, hence the requirement of not just a length n
> > sequence, but an infinite length sequence. It's so simple to see this.
>
> Well yes, we have infinite length sequences. I believe this was stated
> at the very outset. Why is this a problem? You can complain that it is
> unnecessarily complex, but see my earlier point about abstraction and
> generality. Certainly it doesn't lead to any inconsistency or
> contradiction. If it does, then please provide a nice complete proof
> thereof.
Well, I have abstracted and generalized here a bit. I suppose there is
a crux in claiming that
(0,0,0,0,...,0,2.2,0,0,0,...)
is symmetrical to
(2.2,0,0,0,0,...).
The informational content is the same. Only the position of the
information has been changed. I'm not quite sure if a divion will
suffice... I'm pretty sure it will since information will not be
destroyed thanks to the reliance upon zero valued terms.
>
> > No, this point does not prevent you from going onward, but it is a
> > point of valid awareness.
Right, so let's move on to the ideal. I follow you perfectly.
Polynomials are really simple the way that you've presented them. I
know that I've contaminated that buy remaining in your simple
framework I am fine going on.
>
> Sure, whatever. No one objected to the notion that we will require
> infinite length tuples (or polynomials of arbitrarily high degree).
> People objected to your claim that this somehow made the mathematics
> inconsistent, or false, or wrong. It doesn't. So: it happens, it
> doesn't cause any problems, can we move on now?
Yes Uncle Leland, and seriously, I do appreciate you taking all of
this time to defend what would seem not to need any defense at all. I
likely have not convinced you and I don't necesarily expect to
convince anyone. We humans do not deserve so much credit as we give
ourselves, myself included, but yourself included too, by symmetry. I
am a very bad and longwinded dog...
- Tim