JSH: Factorization P(x) = 2(x(x+1)/2 + 1)
James Harris
factorizations than the ones I've been giving, where they end up with
units other than 1 or -1, and maybe an example will help some of you,
so here's this post.
Consider P(x) = x^2 + x + 2, which factors nicely as
P(x) = 2(x(x+1)/2 + 1)
which I pick because it's valid in the ring of integers, but NOT
always in the ring of algebraic integers.
That, of course, is because every integer is either even or 1 away
from being even, so it works.
Now then, what if someone wanted to question that fact, so they came
back at me with a *different* factorization that did something
different.
Why would that be relevant?
It wouldn't.
You MUST FOLLOW A PROOF, and not second guess the process.
Now I've given a proof, and mathematicians need to behave like people
who believe in what they're doing.
And make no mistake, if you choose NOT to behave like mathematicians,
then it seems to me that there's no reason for you to remain
mathematicians, even if it is only in name.
After all, fair is fair, and it's inhumane what you people are doing
to me, basically punishing me for making discoveries. I have
rights!!!
James Harris
Dik T. Winter
> Some people have gotten excited by their ability to find *different*
> factorizations than the ones I've been giving, where they end up with
> units other than 1 or -1, and maybe an example will help some of you,
> so here's this post.
>
> Consider P(x) = x^2 + x + 2, which factors nicely as
>
> P(x) = 2(x(x+1)/2 + 1)
>
> which I pick because it's valid in the ring of integers, but NOT
> always in the ring of algebraic integers.
Can you give an example where x^2 + x + 2 is not divisible by 2 when
x is an algebraic integer?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
David Moran
"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message
news:Hnr9...@cwi.nl...
> In article <3c65f87.03110...@posting.google.com> jst...@msn.com
(James Harris) writes:
> > Some people have gotten excited by their ability to find *different*
> > factorizations than the ones I've been giving, where they end up with
> > units other than 1 or -1, and maybe an example will help some of you,
> > so here's this post.
> >
> > Consider P(x) = x^2 + x + 2, which factors nicely as
> >
> > P(x) = 2(x(x+1)/2 + 1)
> >
> > which I pick because it's valid in the ring of integers, but NOT
> > always in the ring of algebraic integers.
>
> Can you give an example where x^2 + x + 2 is not divisible by 2 when
> x is an algebraic integer?
That's impossible because if you do a little mathematical induction, you
find that P(1)=4 and then evaluating for k+1, you get
(k+1)^2+k+1+2
k^2+2k+1+k+1+2
k^2+3k+4
So P(x) is even for all x.
David Moran
Dik T. Winter
> "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message
> news:Hnr9...@cwi.nl...
> > In article <3c65f87.03110...@posting.google.com> jst...@msn.com
> (James Harris) writes:
> > > Some people have gotten excited by their ability to find *different*
> > > factorizations than the ones I've been giving, where they end up with
> > > units other than 1 or -1, and maybe an example will help some of you,
> > > so here's this post.
> > >
> > > Consider P(x) = x^2 + x + 2, which factors nicely as
> > >
> > > P(x) = 2(x(x+1)/2 + 1)
> > >
> > > which I pick because it's valid in the ring of integers, but NOT
> > > always in the ring of algebraic integers.
> >
> > Can you give an example where x^2 + x + 2 is not divisible by 2 when
> > x is an algebraic integer?
>
> That's impossible because if you do a little mathematical induction, you
> find that P(1)=4 and then evaluating for k+1, you get
> (k+1)^2+k+1+2
> k^2+2k+1+k+1+2
> k^2+3k+4
>
> So P(x) is even for all x.
Yes, for the integers. Bit how about the algebraic integers?
David C. Ullrich
>[...]
>
>After all, fair is fair, and it's inhumane what you people are doing
>to me, basically punishing me for making discoveries.
Huh??? The only "punishment" anyone's been giving you is to
say you're wrong and explain why. A minute ago you asked me
"What the fuck does agreement matter?" This is a little puzzling,
since you regard disagreement as "punishment".
>I have
>rights!!!
Yes. You have the right to say what you want. And we all have
the right to say what we want about what you say.
We'd even have the _right_ to say you were wrong if you were
actually _right_! But you're not, and you _don't_ have the right
to agreement from people about things you're wrong about,
sorry.
>James Harris
************************
David C. Ullrich
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>Some people have gotten excited by their ability to find *different*
>factorizations than the ones I've been giving, where they end up with
>units other than 1 or -1, and maybe an example will help some of you,
>so here's this post.
>
>Consider P(x) = x^2 + x + 2, which factors nicely as
>
>P(x) = 2(x(x+1)/2 + 1)
That's not a factorization; that's a rewriting.
>which I pick because it's valid in the ring of integers, but NOT
>always in the ring of algebraic integers.
Nope. For every algebraic integer value of x, P(x) is also an
algebraic integer.
I think what you mean is that not every partial calculation is an
algebraic integer; that is, x(x+1)/2 is not an algebraic integer for
every algebraic integer value of x; even though x(x+1)/2 is always an
integer for every integer value of x.
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Dave Taylor
> > Some people have gotten excited by their ability to find *different*
> > factorizations than the ones I've been giving, where they end up with
> > units other than 1 or -1, and maybe an example will help some of you,
> > so here's this post.
> >
> > Consider P(x) = x^2 + x + 2, which factors nicely as
> >
> > P(x) = 2(x(x+1)/2 + 1)
> >
> > which I pick because it's valid in the ring of integers, but NOT
> > always in the ring of algebraic integers.
>
> Can you give an example where x^2 + x + 2 is not divisible by 2 when
> x is an algebraic integer?
I think James' point is that since x(x+1)/2 is not an algebraic integer for
all algebraic integers x, writing the polynomial in that form implicitly
involves 'going into a larger ring'. On the other hand, if x is an integer,
x(x+1) is always even, so x(x+1)/2 makes sense within the integers.
But then, as so often before, this is a factorization of *numbers* written
as if it were one of *polynomials*.
--
Dave Taylor
"When I want your opinion, I'll ... I'll never want your opinion"
[BtVS]
Arturo Magidin
>In article <3c65f87.03110...@posting.google.com> jst...@msn.com (James Harris) writes:
> > Some people have gotten excited by their ability to find *different*
> > factorizations than the ones I've been giving, where they end up with
> > units other than 1 or -1, and maybe an example will help some of you,
> > so here's this post.
> >
> > Consider P(x) = x^2 + x + 2, which factors nicely as
> >
> > P(x) = 2(x(x+1)/2 + 1)
> >
> > which I pick because it's valid in the ring of integers, but NOT
> > always in the ring of algebraic integers.
>
>Can you give an example where x^2 + x + 2 is not divisible by 2 when
>x is an algebraic integer?
Ah, I think I see now what James is trying to claim; that we can write
it as x(x+1)/2 + 1 times something...
If x=i, then i^2+i+2 = -1+i+2 = 1+i; 1+i is a divisor of 2 and not a
unit (in Z[i], its norm is 2, and if r is an algebraic integer unit,
then it is a unit in the ring of integers of Q(r)); and it is a proper
divisor of 2, since (1-i)(1+i) = 2 and 1-i is also not an algebraic
integer unit.
If it were a multiple of 2, then 1+i would be an associate of 2, which
would make 1-i into a unit, which is impossible.
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
>In article <3c65f87.03110...@posting.google.com>,
>James Harris <jst...@msn.com> wrote:
>>Some people have gotten excited by their ability to find *different*
>>factorizations than the ones I've been giving, where they end up with
>>units other than 1 or -1, and maybe an example will help some of you,
>>so here's this post.
>>
>>Consider P(x) = x^2 + x + 2, which factors nicely as
>>
>>P(x) = 2(x(x+1)/2 + 1)
>
>That's not a factorization; that's a rewriting.
Oh, I get it. You are factoring it as 2 times ((x)(x+1)/2) + 1.
>>which I pick because it's valid in the ring of integers, but NOT
>>always in the ring of algebraic integers.
To be honest, I fail to see your point. Nobody is saying that you
cannot factor your polynomial as
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7).
What people are saying is that this factorization does not yield
algebraic integer factors whenever P(x) is irreducible; you claim it
SHOULD. Are you claiming that your factorization above, which is valid
in algebraic integers when x=0, "should" also be valid for arbitrary
algebraic integer values of x? And if so, why do you think so?
James Harris
> > Some people have gotten excited by their ability to find *different*
> > factorizations than the ones I've been giving, where they end up with
> > units other than 1 or -1, and maybe an example will help some of you,
> > so here's this post.
> >
> > Consider P(x) = x^2 + x + 2, which factors nicely as
> >
> > P(x) = 2(x(x+1)/2 + 1)
> >
> > which I pick because it's valid in the ring of integers, but NOT
> > always in the ring of algebraic integers.
>
> Can you give an example where x^2 + x + 2 is not divisible by 2 when
> x is an algebraic integer?
It's the FACTORIZATION Dik Winter, and if you can't grasp that point,
you'll never get it.
The factorization is valid in integers for any integer x, but not in
the ring of algebraic integers for any algebraic integer x.
My point is that you have to focus on the *factorization* and its
validity in particular rings.
Some factorizations will be valid in one ring, but not another.
James Harris
Dik T. Winter
> "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hnr9...@cwi.nl>...
> > > Consider P(x) = x^2 + x + 2, which factors nicely as
> > >
> > > P(x) = 2(x(x+1)/2 + 1)
> > >
> > > which I pick because it's valid in the ring of integers, but NOT
> > > always in the ring of algebraic integers.
> >
> > Can you give an example where x^2 + x + 2 is not divisible by 2 when
> > x is an algebraic integer?
>
> It's the FACTORIZATION Dik Winter, and if you can't grasp that point,
> you'll never get it.
>
> The factorization is valid in integers for any integer x, but not in
> the ring of algebraic integers for any algebraic integer x.
Oh, there are enough algebraic integers where it is valid in the ring
of algebraic integers. For instance: x = sqrt(2). Arturo gave an
example for which it is indeed not valid.
> My point is that you have to focus on the *factorization* and its
> validity in particular rings.
>
> Some factorizations will be valid in one ring, but not another.
Yes, and your factorisation
P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
is in general not valid in the ring of algebraic integers. So what
are you trying to show?
The Ghost In The Machine
<jst...@msn.com>
wrote
on 2 Nov 2003 17:57:52 -0800
<3c65f87.03110...@posting.google.com>:
> Some people have gotten excited by their ability to find *different*
> factorizations than the ones I've been giving, where they end up with
> units other than 1 or -1, and maybe an example will help some of you,
> so here's this post.
>
> Consider P(x) = x^2 + x + 2, which factors nicely as
>
> P(x) = 2(x(x+1)/2 + 1)
That's not a factorization, just a rewrite (although one could
quibble over the '2'). In this case
P(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2))
is a factorization.
>
> which I pick because it's valid in the ring of integers, but NOT
> always in the ring of algebraic integers.
Erm, it's an abstract formalism/equation/rewrite; how
can it not be valid? Unless 2 = 0, which is only the
case in the field of integers mod 2, which isn't all that
interesting. :-)
>
> That, of course, is because every integer is either even or 1 away
> from being even, so it works.
This is true.
>
> Now then, what if someone wanted to question that fact, so they came
> back at me with a *different* factorization that did something
> different.
>
> Why would that be relevant?
>
> It wouldn't.
>
> You MUST FOLLOW A PROOF, and not second guess the process.
I'm assuming there's a proof to follow somewhere on the
Web; do you have a Web site? I for one want to see the
whole proof, ideally with the corrections others have
mentioned in this newsgroup.
>
> Now I've given a proof, and mathematicians need to behave like people
> who believe in what they're doing.
You've given a sequence of what you purport are logical deductions.
Some of them are logical enough. However, others are jumps over
crevasses which I for one can't follow.
I can only illustrate with one of my own strawman examples, however:
x^3 - 49 is factorizable into (x - r1)(x - r2)(x - r3), but
none of the roots is divisible by 7 -- x divisible by 7 over the
algebraic integers, means that one can find an algebraic
integer y such that x = 7 * y. Since all the r's are products
of (-sqrt(3)/2 + i/2)^n * 7^(2/3) (n=0,1,2), it's clear that
there is no such algebraic integer in this strawman case.
Does your proof suffer from a similar flaw? I hope not.
>
> And make no mistake, if you choose NOT to behave like mathematicians,
> then it seems to me that there's no reason for you to remain
> mathematicians, even if it is only in name.
>
> After all, fair is fair, and it's inhumane what you people are doing
> to me, basically punishing me for making discoveries. I have
> rights!!!
Personally, I prefer to critique your proof, not you.
I can't say anything at all whether you smell good or bad,
run around in your underwear (or not in your underwear),
do weird things with a shishkebab, a chainsaw, and a
Superball, attend church, give to charity, drive around the
city with a large sign saying "I [heart] MATH", etc. :-)
Who really cares? We judge according to what one posts.
It's all we have. :-)
>
>
> James Harris
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
James Harris
> In sci.math, James Harris
> <jst...@msn.com>
> wrote
> on 2 Nov 2003 17:57:52 -0800
> <3c65f87.03110...@posting.google.com>:
> > Some people have gotten excited by their ability to find *different*
> > factorizations than the ones I've been giving, where they end up with
> > units other than 1 or -1, and maybe an example will help some of you,
> > so here's this post.
> >
> > Consider P(x) = x^2 + x + 2, which factors nicely as
> >
> > P(x) = 2(x(x+1)/2 + 1)
>
> That's not a factorization, just a rewrite (although one could
You're showing troubling ignorance here considering how often you've
posted in my threads as if you know basic mathematics.
In my example P(x) is factored into 2 and x(x+1)/2 + 1, and *both* are
indeed factors.
Possibly the symbols are confusing you, so here are some numbers:
Let x=1, then P(1) = 4, and the factors are 2 and 2.
It IS a factorization poster. You're just lost on basics.
> quibble over the '2'). In this case
>
> P(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2))
>
> is a factorization.
Yes, it is a factorization, but it's not the only factorization
poster.
Ok, let's get one thing straight: the discussions here are over
*advanced* topics, so if you don't even know what a factorization is,
you might want to sit back and just read without posting.
And yes, when I see such posts I downgrade a poster.
If I downgrade a poster far enough, I just don't worry about reading
their posts, unless I start seeing evidence that the newsgroup as a
whole is convinced by them in their confusion.
James Harris
James Harris
> In article <3c65f87.03110...@posting.google.com> jst...@msn.com (James Harris) writes:
> > "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hnr9...@cwi.nl>...
> ...
> > > > Consider P(x) = x^2 + x + 2, which factors nicely as
> > > >
> > > > P(x) = 2(x(x+1)/2 + 1)
> > > >
> > > > which I pick because it's valid in the ring of integers, but NOT
> > > > always in the ring of algebraic integers.
> > >
> > > Can you give an example where x^2 + x + 2 is not divisible by 2 when
> > > x is an algebraic integer?
> >
> > It's the FACTORIZATION Dik Winter, and if you can't grasp that point,
> > you'll never get it.
> >
> > The factorization is valid in integers for any integer x, but not in
> > the ring of algebraic integers for any algebraic integer x.
>
> Oh, there are enough algebraic integers where it is valid in the ring
> of algebraic integers. For instance: x = sqrt(2). Arturo gave an
> example for which it is indeed not valid.
>
> > My point is that you have to focus on the *factorization* and its
> > validity in particular rings.
> >
> > Some factorizations will be valid in one ring, but not another.
>
> Yes, and your factorisation
> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
> is in general not valid in the ring of algebraic integers. So what
> are you trying to show?
That's the point. I *prove* that if you have coprimeness between 7
and 22 in the ring in which the factorization is valid, where 7 is NOT
a unit (and neither is 22), then the constant terms of the factors
that result from dividing P(x) by 49 *MUST* be coprime to 7.
That result is then ring independent to the extent that it's not
particular to a particular ring, but to a particular property *of* the
ring.
However, the ring of algebraic integers, which has the desired
coprimeness property, does NOT always have a_1(x)/7 and a_2(x)/7 as
members!
It's a great result, which follows from some fascinatingly basic
algebra.
Get it yet Dik Winter?
James Harris
Arturo Magidin
What you FAIL to realize, though, is that this DOES NOT MEAN that the
factorization is
P(x)/49 = (5a1/7 + 1)(5a2/7 + 1)(5b3 + 22).
What you get is that 49 = w_1(x)*w_2(x)*w_3(x), with w_i varying with
the values of x, and
P(x)/49 = (( 5a1(x)+7)/w_1(x) - 1) + 1)*
(((5a_2(x)+7)/w_2(x) - 1) + 1) *
((( 5b_3(x)+7)/w_3(x) - 22) + 22).
That's IT. You are ASSUMING your conclusion when you claim that
w_1(x)=7 for all x.
Remember: Given ANY function f(x) and ANY constant r, f(x) can be
written as f(x)=g(x) + r, where r is constant, and g(x) is a
function. Given ANY function f(x), and any two values r and s, we can
write f(x) = g(x) + c, where c is a constant, and g(r)=s: just let
g(x) = f(x) - f(r) + s
and let c = f(r)-s.
Then
g(x) + c = f(x)-f(r) + s + f(r) - s = f(x)
and
g(r) = f(r) - f(r) + s = s.
So you can take the function 5a_1(x) + 7, divide it by ANY FUNCTION
r(x) WHICH SATISFIES r(0)=7, and then you can write
(5a_1(x)+7)/r(x) = h_1(x) + 1
where h_(1) = 0.
Likewise, you can divide 5a_2(x)+7 by ANY FUNCTION s(x) which
satisfies s(0)=7, and then you can write it as
(5a_2(x)+7)/s(x) = h_2(x) + 1
And you can divide 5b_3(x) + 22 by ANY FUNCTION t(x) that satisfies
t(0)=1, and then you can write the result as
(5b_3(x)+22)/t(x) = h_3(x) + 22.
So, if you make ANY choice for which r(x)*s(x)*t(x) = 49, the
factorization will work; and if you make the choice in such a way that
r(x)*s(x)*t(x) = 49, r(x) divides a_1(x) and 7, s(x) divides a_2(x)
and 7, and t(x) divides a_3(x) and 7 (all in the ring of algebraic
integers), then the factorization will work, and it will NOT
NECESSARILY BE OF THE FORM YOU CLAIM IT ->MUST<- BE.
You cannot conclude that, because you can ALWAYS write the function as
h_1(x) + 1 with h_1(x) = 0, regardless of the value of
w_1(x). Likewise for w_2 and for w_3. And if you choose them
CORRECTLY, then you get that all of them have values in the algebraic
integers.
Get it, James S. Harris?
Dik T. Winter
...
> > > My point is that you have to focus on the *factorization* and its
> > > validity in particular rings.
> > >
> > > Some factorizations will be valid in one ring, but not another.
> >
> > Yes, and your factorisation
> > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
> > is in general not valid in the ring of algebraic integers. So what
> > are you trying to show?
>
> That's the point. I *prove* that if you have coprimeness between 7
> and 22 in the ring in which the factorization is valid, where 7 is NOT
> a unit (and neither is 22), then the constant terms of the factors
> that result from dividing P(x) by 49 *MUST* be coprime to 7.
If you are talking here about the factorisation
P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)
being valid in that ring (for all x), that is vacuously true. No
proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in
any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22
are units. So we need no proof in that case.
It becomes different about the "MUST". You have made no showing of that.
Why can it not be that for particular x, a division of all three factors
by 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, but
(5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assuming
that because for x = 0 the factors of 49 clearly distribute as 7, 7, 1
this *must* be true for all x. When you stay in the algebraic integers
that simply is not the case. When you talk about divisibility you can
not assume that something that is proper in one case is also proper in
another case.
> Get it yet Dik Winter?
Get it yet James Harris?
James Harris
> In article <3c65f87.03110...@posting.google.com> jst...@msn.com (James Harris) writes:
> > "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hns8J...@cwi.nl>...
> ...
> > > > My point is that you have to focus on the *factorization* and its
> > > > validity in particular rings.
> > > >
> > > > Some factorizations will be valid in one ring, but not another.
> > >
> > > Yes, and your factorisation
> > > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
> > > is in general not valid in the ring of algebraic integers. So what
> > > are you trying to show?
> >
> > That's the point. I *prove* that if you have coprimeness between 7
> > and 22 in the ring in which the factorization is valid, where 7 is NOT
> > a unit (and neither is 22), then the constant terms of the factors
> > that result from dividing P(x) by 49 *MUST* be coprime to 7.
>
> If you are talking here about the factorisation
> P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)
> being valid in that ring (for all x), that is vacuously true. No
> proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in
> any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22
> are units. So we need no proof in that case.
Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as
you're showing your lack of reasonableness.
Now then, consider coprimeness in algebraic integers, and what that
means for 7 and 22, versus talking about coprimeness as if either is a
unit, can you do that Dik Winter?
Assuming you can, now then, my point is that for a factorization valid
IN SUCH A RING, the argument plays out as I've given it, as the
constant terms MUST be coprime to 7, in a way that's rather obvious.
Now you come up with different factorizations, and got all excited as
if finding some other factorization, will invalidate a DIFFERENT
factorization, so I've explained in detail.
Rather than be reasonable, you're being petulant and stubborn.
> It becomes different about the "MUST". You have made no showing of that.
> Why can it not be that for particular x, a division of all three factors
> by 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, but
Because the constant terms of the factors of P(x)/49 are coprime to 7.
Understand Dik Winter?
It's NOT COMPLICATED, but you are behaving as if logic is a disease.
Now then, if you find a *different* factorization, not valid in rings
where 7 is coprime to 22 in the same sense as in algebraic integers,
then you have a different result.
It's math, so I can trace out the mathematical argument, and show you
exactly the point where the factorizations you hacked together go
their own way as they're not valid in rings like algebraic integers.
> (5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assuming
> that because for x = 0 the factors of 49 clearly distribute as 7, 7, 1
> this *must* be true for all x. When you stay in the algebraic integers
> that simply is not the case. When you talk about divisibility you can
> not assume that something that is proper in one case is also proper in
> another case.
Constant terms do NOT changes as variables.
I repeat, constant terms do NOT change as variables.
Given a situation, where one position requires that they do, while
another accepts that they do not, why in the hell would you go for the
position that requires constant terms to change as variables?
> > Get it yet Dik Winter?
>
> Get it yet James Harris?
Yeah, I understand it you mocking nitwit. I'm just wondering why you
REFUSE TO BE LOGICAL. Why do you even bother reading or posting on a
math newsgroup if you display behavior that is anti-thetical to
mathematics?
Now then, given that constant factors do NOT change as variables there
must be a LOGICAL answer for what happens, and I've explained it to
you.
James Harris
Dik T. Winter
> "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hnt0I...@cwi.nl>...
> > In article <3c65f87.03110...@posting.google.com> jst...@msn.com (James Harris) writes:
> > > "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hns8J...@cwi.nl>...
> > ...
> > > > > My point is that you have to focus on the *factorization* and its
> > > > > validity in particular rings.
> > > > >
> > > > > Some factorizations will be valid in one ring, but not another.
> > > >
> > > > Yes, and your factorisation
> > > > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
> > > > is in general not valid in the ring of algebraic integers. So what
> > > > are you trying to show?
> > >
> > > That's the point. I *prove* that if you have coprimeness between 7
> > > and 22 in the ring in which the factorization is valid, where 7 is NOT
> > > a unit (and neither is 22), then the constant terms of the factors
> > > that result from dividing P(x) by 49 *MUST* be coprime to 7.
> >
> > If you are talking here about the factorisation
> > P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)
> > being valid in that ring (for all x), that is vacuously true. No
> > proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in
> > any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22
> > are units. So we need no proof in that case.
>
> Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as
> you're showing your lack of reasonableness.
You are ignoring my statement that in *every* ring that contains 1, 7 and 22,
they are coprime. Whether 7 is a unit or not.
> Now then, consider coprimeness in algebraic integers, and what that
> means for 7 and 22, versus talking about coprimeness as if either is a
> unit, can you do that Dik Winter?
Why? When you read what I write you know what I would say here. Because
7 and 22 are coprime in every ring that contains them, they are also coprime
in the algebraic integers. Or are you not able to raw that conclusion?
> Assuming you can, now then, my point is that for a factorization valid
> IN SUCH A RING, the argument plays out as I've given it, as the
> constant terms MUST be coprime to 7, in a way that's rather obvious.
I am attacking your *MUST*. You have shown nowhere that it *MUST* be
the case.
> > It becomes different about the "MUST". You have made no showing of that.
> > Why can it not be that for particular x, a division of all three factors
> > by 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, but
>
> Because the constant terms of the factors of P(x)/49 are coprime to 7.
*Why?*
> Understand Dik Winter?
No. You have not shown anything about *why* they must be coprime to 7.
You assert it. If for some particular x the factors can be divided by
7^{2/3} leaving algebraic integers, there is no reason to divide two of
the factors by 7 and the third by 1. That 22 is coprime to 7 does
*not* mean that (5 b3(x) + 22) is coprime to 7 in the algebraic integers.
> It's NOT COMPLICATED, but you are behaving as if logic is a disease.
>
> Now then, if you find a *different* factorization, not valid in rings
> where 7 is coprime to 22 in the same sense as in algebraic integers,
> then you have a different result.
Can you show me a ring (with unit) where 7 is *not* coprime to 22? Try
some mathematics instead. In every ring that contains 1, 7 and 22 they
are coprime in the same sense as in the algebraic integers.
> It's math, so I can trace out the mathematical argument, and show you
> exactly the point where the factorizations you hacked together go
> their own way as they're not valid in rings like algebraic integers.
Well, do that for once.
> > (5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assuming
> > that because for x = 0 the factors of 49 clearly distribute as 7, 7, 1
> > this *must* be true for all x. When you stay in the algebraic integers
> > that simply is not the case. When you talk about divisibility you can
> > not assume that something that is proper in one case is also proper in
> > another case.
>
> Constant terms do NOT changes as variables.
They are constant terms *only* if you assume that they ought to be divided
by the same constant value for every x, but that is something you have to
prove. If that is not the case, dividing P(x) by 49 gives the
factorisation
(5 a1(x)/w1(x) + 7/w1(x))(5 a2(x)/w2(x) + 7/w2(x))(5 b3(x)/w3(x) + 22/w3(x))
where the w's depend on the value of x. When x = 0, w1(x) = 7, w2(x) = 7
and w3(x) = 1. For other values of x other values of the w's are available.
That is, if you wish to stay in the algebraic integers. (Note that although
22/w3(x) may very well not be an algebraic integer, (5 b3(x)/w3(x) + 22/w3(x))
is.
> I repeat, constant terms do NOT change as variables.
When you divide through by 49, they are no longer "constant terms"). Unless
you *assume* that the distribution of the factors of 49 must go the same
way for every x.
> Yeah, I understand it you mocking nitwit. I'm just wondering why you
> REFUSE TO BE LOGICAL. Why do you even bother reading or posting on a
> math newsgroup if you display behavior that is anti-thetical to
> mathematics?
Why do you post answers to articles you obviously do not read?
Will Twentyman
> "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hnt0I...@cwi.nl>...
>
>>In article <3c65f87.03110...@posting.google.com> jst...@msn.com (James Harris) writes:
>> > "Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<Hns8J...@cwi.nl>...
>> ...
>> > > > My point is that you have to focus on the *factorization* and its
>> > > > validity in particular rings.
>> > > >
>> > > > Some factorizations will be valid in one ring, but not another.
>> > >
>> > > Yes, and your factorisation
>> > > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)
>> > > is in general not valid in the ring of algebraic integers. So what
>> > > are you trying to show?
>> >
>> > That's the point. I *prove* that if you have coprimeness between 7
>> > and 22 in the ring in which the factorization is valid, where 7 is NOT
>> > a unit (and neither is 22), then the constant terms of the factors
>> > that result from dividing P(x) by 49 *MUST* be coprime to 7.
>>
>>If you are talking here about the factorisation
>> P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)
>>being valid in that ring (for all x), that is vacuously true. No
>>proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in
>>any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22
>>are units. So we need no proof in that case.
>
>
> Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as
> you're showing your lack of reasonableness.
Given a commutative ring with 7 and 22.
Definition of coprime: 7 and 22 are coprime in a ring if there exists p
and q in the ring such that 7p + 22q = 1.
7+7+7 = 21, so 21 is in the ring.
21 is in the ring so -21 is in the ring.
22 + -21 = 1 so 1 is in the ring.
1+1+1 = 3, so 3 is in the ring.
3 is in the ring, so -3 is in the ring.
7*-3 + 22*1 = 1, with both -3 and 1 in the ring, so 7 and 22 are coprime.
Note: whether 7 or 22 are units is *completely* irrelevent to the
computations.
You probably want to say something about whether or not they have any
non-unit common factors, since unit factors are generally ignored.
If we are looking at 7 and 22, we are at least in the integers, since
closure under addition and additive inverses allows us to construct 1.
If the ring has 7 as a unit, it must contain Z[1/7] as a subring.
If the ring has 22 as a unit, it must contain Z[1/22] as a subring.
If the ring has both as units, it must contain Z[1/7][1/22] as a subring.
Regardless of the case, if p is an element of the ring that is a divisor
of both 7 and 22, then p is a divisor of a unit, which means p is a unit.
If 7 and 22 are *not* units, I'm pretty sure they won't have non-unit
divisors, but I'm not sure how to go about showing that. Maybe when
I've got some time later I'll look at it.
What were you objecting to again?
[rest deleted]
--
Will Twentyman
email: wtwentyman at copper dot net
Arturo Magidin
This definition (the standard definition) is stronger than the
definition James uses: a and b are "coprime" if and only if any common
divisor is a unit.
>If 7 and 22 are *not* units, I'm pretty sure they won't have non-unit
>divisors, but I'm not sure how to go about showing that. Maybe when
>I've got some time later I'll look at it.
That's why the usual definition is better:
LEMMA. If R is a ring, and x and y in R are coprime in the sense that
there exist a and b in R such that ax+by = 1, then for any ring that
contains R, x and y are also coprime (in the same sense).
Proof: a and b serve as witnesses in both R and the larger ring. QED
REMARK. If R is a ring, x and y in R are "coprime" in the sense that
any common divisor in R of x and y is a unit in R, then it is possible
for there to be a larger ring S, containing R, where x and y are no
longer coprime (in that sense).
Example: R= Z[sqrt(-5)]; x = 2, y = (1+sqrt(-5)), S=
Z[sqrt(-5),sqrt(2)]; note that y is a multiple of sqrt(2), since
(1+sqrt(-5)) = sqrt(2)*sqrt(3+sqrt(-5)).
PROP. Let R be a ring. If x and y are coprime in R in the sense that
there exists a and b in R such that ax+by=1, then x and y are coprime
in R in the sense that any common divisor in R is a unit in R.
Proof. Let u be a common divisor of x and y. Then it is a divisor of
ax, and it is a divisor of by, so it is a divisor of ax+by=1. Divisors
of 1 are units. So u is a unit. QED
So in ANY ring that contains the integers, 7 and 22 are coprime (under
either definition).
James Harris
Are you just totally stupid? That's irrelevant Dik Winter as the
point is that 7 is NOT a factor of 22, which my saying that it's not a
unit points out.
I'm getting sick of stupid games from stupid people who apparently
have nothing better to do with *their* time.
Now are you or are you not intelligent enough to understand what it
means for 7 NOT to be a factor of 22?
James Harris
Bill Dubuque
>
> So in ANY ring that contains the integers, 7 and 22 are coprime
> (under either definition).
That's true in any ring R since R contains a homomorphic image of Z,
Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
must preserve the relation 22 - 3(7) = 1.
By the way, as I mentioned in a prior post [1], "coprime" is
a highly overloaded term whose meaning depends upon context.
JSH is using one of the most common definitions and it is
incorrect to criticize him for that. Arturo's definition has
also the less ambiguous name "comaximal", and this should
be preferred in contexts where there may be ambiguity.
-Bill Dubuque
Arturo Magidin
Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>
>> So in ANY ring that contains the integers, 7 and 22 are coprime
>> (under either definition).
>
>That's true in any ring R since R contains a homomorphic image of Z,
>Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
>must preserve the relation 22 - 3(7) = 1.
Hmmm... Only if you assume that ring morphisms map 1 to 1, which is
not necessarily a given either. Even assuming rings have a 1, the zero
map is usually considered a valid homomorphism, and your conclusion
would be incorrect there.
>By the way, as I mentioned in a prior post [1], "coprime" is
>a highly overloaded term whose meaning depends upon context.
>JSH is using one of the most common definitions and it is
>incorrect to criticize him for that.
I will quibble that what is "most common" depends on context as
well. Most ring theorists I know would object to using the definition
depending on common divisors, since to them 'prime' refers to ideals,
almost never to elements; and most number theorists would certainly
disagree that the definition via common divisors is 'the most common'
(for the latter the definition ->is<- invariably related to ideals,
never to elements). To me, and particularly given that JSH's work is
taking place in subrings of the ring of all algebraic integers, it
seems that the most common usage from algebraic number theory should
prevail.
It may be "most common" among those who study
domains or other closely related kinds of rings, though.
But in any case, so long as he states explicitly what he means and
sticks to it (neither of which is something he usually does), I have
accepted your correction and rather than criticise him for his use I
simply note the distinction between his use and the one defined via
linear combinations equalling 1 (see below).
>Arturo's definition has
>also the less ambiguous name "comaximal", and this should
>be preferred in contexts where there may be ambiguity.
The definition I use is actually that no prime ideal contains the
principal ideals generated by the elements. It turns out to be
equivalent to comaximal in the presence of a 1 and the Axiom of
Choice, but not in general, as I discussed when James brought up the
example of the ring (or "rng") of even integers. I assume you are
referring to "a and b are coprime if and only if there exists r and s
such that ar+bs = 1", which I did assume throughout above, but which I
would consider a theorem, not the definition.
But, if you'll see James's most recent contributions, his recent
mistake on 'coprime' following his usage has led him to the conclusion
that 'coprime' is "broken", so go figure what will happen next.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Will Twentyman
Very ellegant. Thanks for the result.
Dik T. Winter
Why do you not read what I write, rather than what you think I write?
> I'm getting sick of stupid games from stupid people who apparently
> have nothing better to do with *their* time.
>
> Now are you or are you not intelligent enough to understand what it
> means for 7 NOT to be a factor of 22?
Yup. That means that you think your factorisation of P(x)/49 is valid
in some ring where 7 is not a unit. I am not convinced. I *know* it
is not valid in the algebraic integers, because there is ample proof that
for varying x the factors of 49 distribute differently amongst the three
factors of the polynomial when you wish to stay in the algebraic integers.
Furthermore, I *know* that in the algebraic integers (5 b3(x) + 22) is
*not* coprime (either the standard definition or your definition) to 7
for most values of x. So you should construct your ring such that
both (5 a1(x) + 7) and (5 a2(x) + 7) are divisible by 7 and that
(5 b3(x) + 22) is coprime to 7 *for all x*, and that 7 is not a unit.
I am not satisfied that such a ring does exist.
Moreover, you wish to divide all three factors by a constant. Why do
you assume it must be a constant? Given w1(x), w2(x) and w3(x) such
that their product is 49, and w1(0) = 7, w2(0) = 7 and w3(0) = 1.
Wouldn't a distribution like:
P(x)/49 =
(5 a1(x)/w1(x)+7/w1(x))(5 a2(x)/w2(x)+7/w2(x))(5 b3(x)/w3(x)+22/w3(x))
be feasable? If I calculate correctly, the constant terms in this
factorisation are now 1, 1 and 22, just as you wish. So why can this
not satisfy your requirements? Assuming the w's are constant is just
plain stupid if you wish to stay in a particular ring. The functions
w depend on divisibility considerations within the ring, and so are
not continuous in any sense.
Bill Dubuque
>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>
>>> So in ANY ring that contains the integers, 7 and 22 are coprime
>>> (under either definition).
>>
>> That's true in any ring R since R contains a homomorphic image of Z,
>> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
>> must preserve the relation 22 - 3(7) = 1.
>
> Hmmm... Only if you assume that ring morphisms map 1 to 1, which
> is not necessarily a given either. Even assuming rings have a 1,
> the zero map is usually considered a valid homomorphism,
> and your conclusion would be incorrect there.
For Rings (with 1, as I assume above) ring morphisms must preserve 1,
so the zero map is not a morphism of Rings with 1. If, as you claim,
one considered the zero map as a valid homomorphism of Rings with 1
then basic theorems on rings would fail, e.g. the image of the zero
morphism would fail to be a subring (except if the target ring is 0).
Therefore my above quoted statement is in fact correct as written.
>> By the way, as I mentioned in a prior post [1], "coprime" is
>> a highly overloaded term whose meaning depends upon context.
>> JSH is using one of the most common definitions and it is
>> incorrect to criticize him for that.
> I will quibble that what is "most common" depends on context as well.
> Most ring theorists I know would object to using the definition
> depending on common divisors, since to them 'prime' refers to ideals,
> almost never to elements; and most number theorists would certainly
> disagree that the definition via common divisors is 'the most common'
> (for the latter the definition ->is<- invariably related to ideals,
> never to elements).
This issue is not what _should_ be the proper definition but, rather,
what _is_ past and current usage. The fact of the matter is that due
to the way this and related ideas evolved, the denotation of "coprime"
was - and still is - highly overloaded. Thus it is best to explicitly
specify its meaning in any context where possible ambiguity exists.
> To me, and particularly given that JSH's work is taking place in
> subrings of the ring of all algebraic integers, it seems that the
> most common usage from algebraic number theory should prevail.
But often such subrings are of infinite degree over Q and are not
Dedekind domains. Therefore they lie outside the scope of classical
algebraic number theory. Hence "common usage" may not even apply.
-Bill Dubuque
James Harris
I did read what you wrote.
> > I'm getting sick of stupid games from stupid people who apparently
> > have nothing better to do with *their* time.
> >
> > Now are you or are you not intelligent enough to understand what it
> > means for 7 NOT to be a factor of 22?
>
> Yup. That means that you think your factorisation of P(x)/49 is valid
> in some ring where 7 is not a unit. I am not convinced. I *know* it
That's because you're *refusing* to be rational!!!
Now I've found another way your hacks go away from my argument as the
polynomial I use, with v=-1+49x is simply
P(x) = (v^3+1)5^3 - 3v(5)7^2 + 7^3
which shouldn't look totally unfamiliar to you, but you're a hack,
apparently dedicated to convincing newsgroup readers of what you
believe, without concern about the actual mathematical truth.
Just my problem you and your cabal have been successful enough that I
have to break you.
> is not valid in the algebraic integers, because there is ample proof that
> for varying x the factors of 49 distribute differently amongst the three
> factors of the polynomial when you wish to stay in the algebraic integers.
And my point remains that I've found an error in "core" so your claim
of a proof using core is stupid in context as I've *repeatedly*
explained it to you.
Why do you refuse to be logical Dik Winter? Why do you refuse to
follow a stepped out mathematical proof?
What's wrong with you?
> Furthermore, I *know* that in the algebraic integers (5 b3(x) + 22) is
> *not* coprime (either the standard definition or your definition) to 7
> for most values of x.
You're irrational Dik Winter, desperate to hold on to what you think
you know, as if mathematics need care about your beliefs.
You've been refuted. I've shown yet again how your hack diverges from
my work.
Tell the truth, or go away.
James Harris
Arturo Magidin
Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>>
>>>> So in ANY ring that contains the integers, 7 and 22 are coprime
>>>> (under either definition).
>>>
>>> That's true in any ring R since R contains a homomorphic image of Z,
>>> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
>>> must preserve the relation 22 - 3(7) = 1.
>>
>> Hmmm... Only if you assume that ring morphisms map 1 to 1, which
>> is not necessarily a given either. Even assuming rings have a 1,
>> the zero map is usually considered a valid homomorphism,
>> and your conclusion would be incorrect there.
>
>For Rings (with 1, as I assume above) ring morphisms must preserve 1,
>so the zero map is not a morphism of Rings with 1.
Granted; like I said, "if you assume that ring morphisms map 1 to 1."
On the other hand, you lose some things by assuming that: you
don't get isomorphic copies of the rings in direct products. Some
conventions are better than others, depending on the situation.
> If, as you claim,
>one considered the zero map as a valid homomorphism of Rings with 1
>then basic theorems on rings would fail, e.g. the image of the zero
>morphism would fail to be a subring (except if the target ring is 0).
>Therefore my above quoted statement is in fact correct as written.
If by "Ring" you meant "with 1" and by homomorphism you meant
"preserve 1", then ->of course<- they were correct. But you must grant
that not ->everyone<- means both things, and that there was at least
some sense in my stating so explicitly...
[.snip.]
Bill Dubuque
>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>>>
>>>>> So in ANY ring that contains the integers, 7 and 22 are coprime
>>>>> (under either definition).
>>>>
>>>> That's true in any ring R since R contains a homomorphic image of Z,
>>>> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
>>>> must preserve the relation 22 - 3(7) = 1.
>>>
>>> Hmmm... Only if you assume that ring morphisms map 1 to 1, which
>>> is not necessarily a given either. Even assuming rings have a 1,
>>> the zero map is usually considered a valid homomorphism,
>>> and your conclusion would be incorrect there.
>>
>> For Rings (with 1, as I assume above) ring morphisms must preserve 1,
>> so the zero map is not a morphism of Rings with 1.
>
> Granted; like I said, "if you assume that ring morphisms map 1 to 1."
That's implicit in my statement, but I agree it's worth highlighting.
> On the other hand, you lose some things by assuming that: you
> don't get isomorphic copies of the rings in direct products. Some
> conventions are better than others, depending on the situation.
Indeed. But I think your critique is better targeted at your own:
> LEMMA. If S is a ring, and x and y in S are coprime in the sense that
> there exist a and b in S such that a x + b y = 1, then for any ring
> R that contains S, x and y are also coprime (in the same sense).
COUNTEREXAMPLE Let S = Z x {0} < R = Z x Z, a = (3,0), b = (2,0)
aS + bS = S = aR + bR < R, so a,b are coprime in S but not in R
Your lemma needs S < R is a sub(ring with 1), so 1_S = 1_R.
"R contains S" is not sufficient, as the counterexample shows.
However, my statement remains true here: 3 - 2 = 1 in Z
maps to (3 - 2 = 1) 1_R -> (3,3) - (2,2) = (1,1) in R
-Bill Dubuque
Arturo Magidin
> Arturo Magidin <mag...@math.berkeley.edu> wrote:
> >Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
> >>Arturo Magidin <mag...@math.berkeley.edu> wrote:
> >>>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
> >>>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
> >>>>>
> >>>>> So in ANY ring that contains the integers, 7 and 22 are coprime
> >>>>> (under either definition).
> >>>>
> >>>> That's true in any ring R since R contains a homomorphic image of Z,
> >>>> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
> >>>> must preserve the relation 22 - 3(7) = 1.
> >>>
> >>> Hmmm... Only if you assume that ring morphisms map 1 to 1, which
> >>> is not necessarily a given either. Even assuming rings have a 1,
> >>> the zero map is usually considered a valid homomorphism,
> >>> and your conclusion would be incorrect there.
> >>
> >> For Rings (with 1, as I assume above) ring morphisms must preserve 1,
> >> so the zero map is not a morphism of Rings with 1.
> >
> > Granted; like I said, "if you assume that ring morphisms map 1 to 1."
>
> That's implicit in my statement, but I agree it's worth highlighting.
>
> > On the other hand, you lose some things by assuming that: you
> > don't get isomorphic copies of the rings in direct products. Some
> > conventions are better than others, depending on the situation.
>
> Indeed. But I think your critique is better targeted at your own:
Ouch. Your absolutely right (except, I wasn't critising, just
commenting...) I had a rather nasty double standard there, where on
the one hand I left open the possibility that ring morphisms do not
preserve 1, but on the other I was tacitly assuming that subrings did.
Which of course, makes little sense, as then the image of a morphism
would not necessarily be a subring. I shall be more careful in the
future...
Arturo Magidin, sans .sig.