seeking comments on expository article on factorization
plc...@gmail.com
in integral domains:
http://math.uga.edu/~pete/factorization.pdf
I would appreciate feedback in the form of comments and constructive
criticism. This project began as supplementary notes for an
undergraduate/basic graduate class on number theory. The current
version is a bit more ambitious, but nevertheless includes a treatment
of the basic material essentially from scratch. I am interested to
know to what extent a general mathematical audience will find it
readable and useful.
On the other hand, at the end of the note are some "open questions",
meaning that I do not know the answers (but I am not a true expert in
the subject). I would be glad to hear any ideas or pointers to the
literature.
P.L. Clark
tc...@lsa.umich.edu
plc...@gmail.com <plc...@gmail.com> wrote:
>I have just finished a draft of an expository article on factorization
>in integral domains:
>
>http://math.uga.edu/~pete/factorization.pdf
I've only skimmed this but it looks very nice.
Here are a few comments that you may or may not want to incorporate into
your treatment.
1. For an example of a domain with non-unique factorization, consider the
ring Q[w,x,y,z]/<wx-yz>. This often strikes students as being abstract
and possibly unilluminating, but for that very reason it can be quite
instructive.
2. There exist proofs of unique factorization (in Z, at least) that proceed
by a direct induction and that do not mention GCD's. It might be
interesting to present one such proof early on and pose the question,
is the existence of GCD's really necessary for unique factorization?
This would motivate some of the later theorems.
3. It may be good to discuss the questions "Is 0 a prime number?" and
"Is -1 a prime number?" since these typically cause confusion among
students.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
[mod note: I was hoping to persuade the audience to discuss the questions
at the end. I am hoping that we don't drift into sci.math territory here.
In particular I would strongly discourage a discussion about whether or
not 0 or -1 are prime numbers here! I know Tim isn't suggesting we discuss
this here but I am hoping no-one takes the bait.]
plc...@gmail.com
wrote:
> In article <go8fba$uj...@crackerjack.ma.ic.ac.uk>,
>
> plcl...@gmail.com <plcl...@gmail.com> wrote:
> >I have just finished a draft of an expository article on factorization
> >in integral domains:
>
> >http://math.uga.edu/~pete/factorization.pdf
>
> I've only skimmed this but it looks very nice.
>
> Here are a few comments that you may or may not want to incorporate into
> your treatment.
Thanks for your comments.
> 1. For an example of a domain with non-unique factorization, consider the
> ring Q[w,x,y,z]/<wx-yz>. This often strikes students as being abstract
> and possibly unilluminating, but for that very reason it can be quite
> instructive.
This is already discussed, although briefly, in passing, and towards
the end: see
the paragraph following Theorem 46 on page 24.
> 2. There exist proofs of unique factorization (in Z, at least) that proceed
> by a direct induction and that do not mention GCD's. It might be
> interesting to present one such proof early on and pose the question,
> is the existence of GCD's really necessary for unique factorization?
> This would motivate some of the later theorems.
Agreed. The article as is makes some references to the "Hasse-
Lindemann-Zermelo"
proof of unique factorization in Z, which is a direct inductive
proof. It's not discussed
directly in the article only for "historical" reasons, i.e., I started
writing it as a supplementary
handout for my number theory course, in which the HLZ proof had
already been presented. I think that an introductory
section doing things explicitly for R = Z is probably merited.
> 3. It may be good to discuss the questions "Is 0 a prime number?" and
> "Is -1 a prime number?" since these typically cause confusion among
> students.
Okay, but we won't discuss it here!
Thanks again,
P.L. Clark
plc...@gmail.com
A colleague of mine pointed out that what was "Question 2" at the end
of the article had a trivial answer. (If an element admits a
factorization into irreducibles, then by the associativity of
multiplication we can achieve this by a sequence of "elementary
factorizations", i.e., by splitting off one irreducible in the assumed
factorization at each stage.) A new draft in which this question has
been removed is available in the same place as before:
http://math.uga.edu/~pete/factorization.pdf
Comments still welcome on the rest of the article and the remaining
two open questions at the end.
P.L. Clark
mjc
This is a comment regarding divisability rather than factorization so
it might OT but...
Most proofs that sqrt(2) is irrational involve evenness or the
fundamental theorem of arithmetic or remainders mod 3 or ...
The only one that I know that has nothing whatsoever to do with
divisibility is the one based on the fact that
x^2 - 2y^2 = 1 has solutions with arbitrarily large integral x and y.
In this proof, the crital fact is that two distinct integers differ by
at least one.
Does anyone know of any others of this type?
Waldek Hebisch
> I have just finished a draft of an expository article on factorization
> in integral domains:
>
> http://math.uga.edu/~pete/factorization.pdf
>
> meaning that I do not know the answers (but I am not a true expert in
> the subject). I would be glad to hear any ideas or pointers to the
> literature.
>
Concerning open question 1: AFAICS the class of rings which admit
multiplicative norm in your sense is exactly class of Krull rings
R such that quotient of multiplicative semigroup of R by group of
units of R is countable.
More generally one can say that the family {N_\alpha} of
weak norms is locally finite <=> for each x \in R, x= 0 or
there is only finitely many \alpha such that N_\alpha(x) \ne 1.
We also can say that the family {N_\alpha} represents division <=>
(if a, b \in R, a \ne 0, b\ne 0, then a divides b <=> for
each \alpha, N_\alpha(a) divides N_\alpha(b).
Now, R admits a locally finite family of weak norms which
represents division <=> R is a Krull ring.
It is easy to see that given a locally finite, countable family {N_\alpha}
of weak norms one can find a single weak norm M which is equivalent in
the sense that {x : for each \alpha N_{\alpha}(x) = 1} = {x : M(x) = 1}.
One unrelated remark: AFAICS definition of greatest common divisor d
of a and b misses condition that d should divide a and b.
--
Waldek Hebisch
heb...@math.uni.wroc.pl
plc...@gmail.com
> plcl...@gmail.com <plcl...@gmail.com> wrote:
> > I have just finished a draft of an expository article on factorization
> > in integral domains:
>
> >http://math.uga.edu/~pete/factorization.pdf
>
> > On the other hand, at the end of the note are some "open questions",
> > meaning that I do not know the answers (but I am not a true expert in
> > the subject). I would be glad to hear any ideas or pointers to the
> > literature.
>
> Concerning open question 1: AFAICS the class of rings which admit
> multiplicative norm in your sense is exactly class of Krull rings
> R such that quotient of multiplicative semigroup of R by group of
> units of R is countable.
This can't be quite right, can it? There are abstract number rings
that are not integrally
closed -- e.g. a nonmaximal order in a number field -- and hence non-
Krull domains which admit a
multiplicative norm.
But it may well be that the condition that you gave is a
characterization of which Krull domains admit
multiplicative norms, which is certainly an important special case. I
will need to think about it further.
> [more technical things snipped, for now]
> One unrelated remark: AFAICS definition of greatest common divisor d
> of a and b misses condition that d should divide a and b.
That is an especially unfortunate oversight. Thanks for pointing it
out.
P.L. Clark
tc...@lsa.umich.edu
mjc <mjc...@acm.org> wrote:
>Most proofs that sqrt(2) is irrational involve evenness or the
>fundamental theorem of arithmetic or remainders mod 3 or ...
>
>The only one that I know that has nothing whatsoever to do with
>divisibility is the one based on the fact that
>x^2 - 2y^2 = 1 has solutions with arbitrarily large integral x and y.
>In this proof, the crital fact is that two distinct integers differ by
>at least one.
>
>Does anyone know of any others of this type?
Here's a proof that sqrt(2) is irrational. Assume not, and let A be the
smallest strictly positive integer such that A*sqrt(2) is an integer.
Let B = A*(sqrt(2) - floor(sqrt(2))). Then B is a strictly positive
integer such that B*sqrt(2) is an integer, and B < A; contradiction.
To me, this proof seems different from the proofs you mentioned. Note
that it works with any non-square integer in place of 2.
There is some discussion of various proofs of the irrationality of sqrt(2)
on Gowers's blog.
http://gowers.wordpress.com/2007/10/04/when-are-two-proofs-essentially-the-same
Gerry Myerson
tc...@lsa.umich.edu wrote:
> Here's a proof that sqrt(2) is irrational. Assume not, and let A be the
> smallest strictly positive integer such that A*sqrt(2) is an integer.
> Let B = A*(sqrt(2) - floor(sqrt(2))). Then B is a strictly positive
> integer such that B*sqrt(2) is an integer, and B < A; contradiction.
I've written an expository paper about this kind of proof
and how far one can push it; Irrationality via well-ordering,
Austral Math Soc Gazette 35 (2008) 121-125,
http://www.austms.org.au/Publ/Gazette/2008/May08/Myerson.pdf
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
mjc
> In article <godkh3$6b...@crackerjack.ma.ic.ac.uk>,
>
> mjc <mjco...@acm.org> wrote:
> >Most proofs that sqrt(2) is irrational involve evenness or the
> >fundamental theorem of arithmetic or remainders mod 3 or ...
>
> >The only one that I know that has nothing whatsoever to do with
> >divisibility is the one based on the fact that
> >x^2 - 2y^2 = 1 has solutions with arbitrarily large integral x and y.
> >In this proof, the crital fact is that two distinct integers differ by
> >at least one.
>
> >Does anyone know of any others of this type?
>
> Here's a proof that sqrt(2) is irrational. Assume not, and let A be the
> smallest strictly positive integer such that A*sqrt(2) is an integer.
> Let B = A*(sqrt(2) - floor(sqrt(2))). Then B is a strictly positive
> integer such that B*sqrt(2) is an integer, and B < A; contradiction.
>
> To me, this proof seems different from the proofs you mentioned. Note
> that it works with any non-square integer in place of 2.
>
> There is some discussion of various proofs of the irrationality of sqrt(2)
> on Gowers's blog.
>
>
> --
> Tim Chow tchow-at-alum-dot-mit-dot-edu
> The range of our projectiles---even ... the artillery---however great, will
> never exceed four of those miles of which as many thousand separate us from
> the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
Of course! I should have remembered this one.
I know it in this form:
If sqrt(2) = m/n then
m/n = (m/n)*(sqrt(2)-1)/(sqrt(2)-1)
= sqrt(2)*(sqrt(2)-1)/(m/n-1)
= (2-sqrt(2))/(m/n-1)
= (2-m/n)/(m/n-1)
= (2n-m)/(m-n)
which has smaller denominator.
I like your version better - it is less cluttered.
Thanks,
Martin Cohen
Waldek Hebisch
> On Mar 1, 4:29 am, Waldek Hebisch <hebi...@math.uni.wroc.pl> wrote:
> > plcl...@gmail.com <plcl...@gmail.com> wrote:
> > > I have just finished a draft of an expository article on factorization
> > > in integral domains:
> >
> > >http://math.uga.edu/~pete/factorization.pdf
> >
> > > On the other hand, at the end of the note are some "open questions",
> > > meaning that I do not know the answers (but I am not a true expert in
> > > the subject). I would be glad to hear any ideas or pointers to the
> > > literature.
> >
> > Concerning open question 1: AFAICS the class of rings which admit
> > multiplicative norm in your sense is exactly class of Krull rings
> > R such that quotient of multiplicative semigroup of R by group of
> > units of R is countable.
>
> This can't be quite right, can it? There are abstract number rings
> that are not integrally
> closed -- e.g. a nonmaximal order in a number field -- and hence non-
> Krull domains which admit a
> multiplicative norm.
>
> But it may well be that the condition that you gave is a
> characterization of which Krull domains admit
> multiplicative norms, which is certainly an important special case. I
> will need to think about it further.
>
I have misunderstood your definition of multiplicative norm (at
first I thought that implies a stronger condition). Actually,
all Krull rings (and in particular all Dedeking rings) admit
norm in your sense. Namely, in Krull ring you have unique
factorization into simple divisors x = p_1^{n_1}...p_k^{n_k}.
Just take N(x) = 2^{n_1+...+n_k}.
One can restate condition of existence of norm as follows.
Let S be multiplicative semigroup of R divided by group
of units. Now, R has norm <-> S admits additive homomorphism
\phi into N such that \phi(s) = 0 if and only if s is unit
element of S.
The condition above is satisfied if S is a free abelian semigroup
(that is when R has unique factorization) and when S is a
subsemigroup of free semigroup (which holds for Krull rings).
However, it is clear that one can build rather strange commutative
semigroups which satisfy property above (in particular one can have
a lot of relations). Of course related question is if arbitrary
semigroup can appear as S above. Existence of norm implies
cancellation property for S, but ATM I do not see any more
obvious restrictions. In fact I think that in many cases
given a commutative semigroup S one can build a ring R such
that S is multiplicative semigroups of R divided by units
(for this I would try to modify construction of formal power
series rings).
--
Waldek Hebisch
heb...@math.uni.wroc.pl
Bill Dubuque
Khurana's example [1] (= your Theorem 31), of elts in a
quadratic order with no gcd, is more clearly viewed as
a consequence of a failure of Euclid's Lemma. That such
failure yields nonexistent gcds or nonprincipal ideals is
something that deserves to be better known - I emphasize
it often on sci.math and elsewhere. Here are the details.
LEMMA 1 (ac,bc) exists => (a,b) exists and (ac,bc) = (a,b)c
EUCLID'S LEMMA a|bc and (a,b)=1 => a|c, if (ac,bc) exists
PROOF a|ac,bc => a|(ac,bc) = (a,b)c = c via Lemma 1. QED
So if a,b,c don't satisfy the implication in Euclid's Lemma,
namely if a|bc and (a,b) = 1 and not a|c, then one
immediately deduces that the gcd (ac,bc) fails to exist.
Khurana's example [1] is just the very special case of this
where a,b,c = p,1+w,1-w in the quadratic ring Z[w], ww = -d.
Such elementary observations on arithmetic in quadratic orders
are at least a century old. Likewise, Lemma 1 and the gcd / lcm
lemmas in Khurana's paper are quite old and well-known. A recent
reference is Section 1.6, Factorization in Commutative Monoids,
in Robert Gilmer's book: Commutative Semigroup Rings, 1984.
More remarks to follow when I find some spare time (I only just
discovered this thread now when it was mentioned on sci.math).
I have added sci.math. Please keep it in any followups since
I don't read sci.math.research as frequently as I do sci.math.
Sci.math readers can read the entire thread via Google Groups [2].
--Bill Dubuque
[1] D. Khurana, On GCD and LCM in domains: A Conjecture of Gauss.
Resonance 8 (2003), 72-79.
http://www.ias.ac.in/resonance/June2003/pdf/June2003Classroom.pdf
[2] http://google.com/groups?threadm=go8fba$uji$1%40crackerjack.ma.ic.ac.uk