Can we always write $gcd(x,y)$ as $ax+by$ in UFD?

Question

Let $R$ be a commutative ring with unity. Now assume that $R$ is Unique Factorization Domain, but not necessarily Principal Ideal Domain.

Question: Let $x,y\in R$ be such that their GCD exists in $R$, and let $d=\gcd(x,y)$. Then does there always exists $a,b\in R$ such that $d=ax+by$?

The answer is yes if the ring is PID, but here I am not considering $R$ as a PID, but considering $x,y$ for which GCD exists.

Here $R$ is a UFD, so every element has a factorization. For $x,y\in R$, we say that $d$ is GCD of $x,y$ (provided it exists), if

(1) $d$ divides both $x,y$.

(2)If $c$ divides $x,y$ then $c$ divides $d$.

I think, common divisors of two elements in a UFD always exists (for example, at least $1$). But, GCD does not exists means we can find two divisors, which are maximal but different.

Thanks. it is useful link, and some new terminology I am looking there. — Groups, Jul 02 '15 at 12:01
@Daniel No need to link offsite when there are proofs onsite. — Bill Dubuque, Jul 02 '15 at 13:27
@Bill It was the first link my search turned up, so I took that. — Daniel Fischer, Jul 02 '15 at 13:31
@Fischer: in the link, the hypothesis is that "GCD exists for all pairs", or similar. Here I am considering only two elements which have a GCD. $R$ need not be a Bezout domain here. — Groups, Jul 03 '15 at 03:32
@Groups Any UFD is a GCD domain so every pair of nonzero elements has a gcd. — Eoin, Jul 03 '15 at 04:41
OK. In the link of Daniel Fischer, it says $R$ is PID $\Longleftrightarrow$ [$R$ is UFD] + [$gcd(x,y)=ax+by$ for all $x,y$]. So according to this, there must be a UFD in which $gcd(x,y)$ is not necessarily expressible as $ax+by$. What is this example? I mean, I want example of not only such $R$, but also the elements $x,y$ in it. — Groups, Jul 03 '15 at 06:20
Possible duplicate of Example of GCD in UFD that can't be expressed as linear combination — qwr, Dec 15 '18 at 07:50

score 2 · Answer 1 · edited Jul 07 '15 at 05:45

The answer is NO. We consider the standard example $\mathbb{Z}[x]$, which is a UFD but not a PID. In fact, $(2,x)$ is not a principal ideal. Consider the elements $2$ and $x$ in this ring. The only divisors of $2$ are $1$ and $2$, and among them only $1$ divides $x$. This means $\gcd(2,x)=1$. Then can we write $1$ as $2a+bx$? Clearly, NO. Simply compare the constant coefficients on both sides.

Can we always write $gcd(x,y)$ as $ax+by$ in UFD?

1 Answers1