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There are many facts about integer gcds which can be proved by appealing to unique prime factorization (up to sign). for example $\gcd(a^2,b^2)=\gcd(a,b)^2$. One way to get the machinery (if that is not too strong a word) about primes and unique factorization is to start with the fact that for all integers $a,b$ there are is a linear combination $d=as+bt$ which divides both $a$ and $b$. That is, to show that the integers are a Bezout Domain and then follow the familiar series of results we all learned and perhaps forget (but then relearned when we taught it). An important tool is that the cofactors $s,t$ can be computed via the Euclidean Algorithm. There are other domains with an integer norm where the same path can be followed: $\mathbb{Z}[\sqrt{k}]$ for certain $k$ such as $2$ and $-1$ for example. However, many of the facts about divisibility can be derived without all that machinery. In a recent answer I pointed out that $as+bt=1$ implies after cubing and simplifying that $a^2(as+3bt)s^2+b^2(3as+bt)t^2=1$ so that $\gcd(a,b)=1$ implies $\gcd(a^2,b^2)=1$. That can be fixed up to a proof that $\gcd(a^2,b^2)=\gcd(a,b)^2$. My question is how far one can get assuming only that a commutative ring (with 1) is a Bezout Domain? Lest I be accused of not having a question I'll ask the following (but feel free to tell more).

Is there any nice treatment of the facts about divisibility which follow only from the assumption that we are in a Bezout Domain? And are there any nice Bezout Domains which show some facts which don't follow?

I suppose all the premises would be equations as would the conclusions. Can we say anything about the increase in complexity? In the equation $a^2(as+3bt)s^2+b^2(3as+bt)t^2=1$ we can use $as+bt=1$ to replace $as+3bt$ with $2bt+1$ or $as+bt+2$ although this is still the same value. Must the cofactors for $s',t'$ for $a^2s'+b^2t'=1$ be at least quartic in $a,b,s,t$ or is there a nicer equation? Over the integers the $s'$ and $t'$ I mention are usually much bigger than they need to be. For example $a,b=11,18$ gives $as+bt=11\cdot 5+18\cdot (-3)=1$ which yields $11^2\cdot (-2675)+18^2\cdot 999=1$ But we prefer $11^2\cdot (-83)+18^2\cdot 31=1$ or at least $11^2\cdot 241+18^2\cdot (-90)=1$.

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up vote 6 down vote accepted

I find your question interesting but a little vague. Let me give you a couple of references: perhaps you can use them to sharpen your question. (Or perhaps they're not what you're looking for at all: we'll see...)

First, there are certainly plenty of treatments of factorization from the perspective of Bezout domains (or GCD-domains, etc.): see for instance the Chapters on Factorization, Bezout Domains and Valuation Domains (currently labelled 12,13,14, but beware: this is subject to change!) in

http://math.uga.edu/~pete/integral.pdf

As for properties not shared by all Bezout domains: an integral domain $R$ is called an elementary divisor domain if Smith normal form exists over $R$: that is if every (not necessarily square) matrix with $R$-coefficients can be diagonalized by performing elementary row and column operations. Famously, any PID is an elementary divisor domain.

It is an open question whether every Bezout domain is an elementary divisor domain. Apparently the expected answer among the experts is no, so this gives at least a conjectural answer to your question. For some information about this problem and other facts which may or may not hold for arbitrary Bezout domains, please see the recent paper by my colleague D.J. Lorenzini:

http://www.math.uga.edu/~lorenz/BezoutMay9.pdf

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Thanks, I like your notes. I realize that my question is somewhat different so i will re-ask it. –  Aaron Meyerowitz Dec 11 '10 at 6:16
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Based upon your examples, you may find it more natural to work with the slightly more general class of Prüfer domains (vs. Bezout domains) - i.e. finitely generated ideals$\:\ne 0$ are invertible (vs. principal). Prüfer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. CRT, or Gauss's Lemma for content ideals, or for ideals $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\: $ or $\rm\ (A + B)\ (A \cap B) = A\ B\ $ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my sci.math post for 30.

As a simple example I'll give the natural Prüfer domain proof of a generalization of your example, viz. the ideal-theoretic $\:$ Freshman's Dream $\rm\ \ (A + B)^n = A^n + B^n\:.\: $ This identity is true for both arithmetic of $\:$ GCDs $\:$ and invertible ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\ $ for ideals and $\rm\ (A,A) = A\ $ for GCDs. Combining these properties with the associative, commutative, distributive laws of addition and multiplication we obtain an extremely elementary high-school-level proof of the Freshman's Dream - which is best illustrated for $\rm\: n = 2\:,\:$ viz.

$\rm\quad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4 $

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2 $

$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2 $

So $\rm\ {(A + B)^2 }\ =\ \ A^2 + B^2\ $ if $\rm\ A+B\ $ is cancellative, e.g. if $\rm\ A+B = 1\ $ or if it's invertible.

The same proof generalizes for all $\rm\:n\:$ since, as above

$\rm\quad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n $

$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n $

In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\ $ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. Here too the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain. (Note: one can unify the GCD and ideal cases by employing Divisor Theory).

In fact this yields yet another characterization: a domain is Prufer iff it satisfies the Freshman's Dream for all finitely generated ideals. See said sci.math post for further discussion.

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