# Local GCD domain and not Bezout domain

I have hard time to find the example of local GCD domain which is not a Bezout domain. Is there any simple example? Thanks

-

Take $R=K[X,Y]_{\mathfrak m}$, where $\mathfrak m=(X,Y)$. This is a local noetherian UFD which is not a PID.

-
@ YACP, Very nice example. I am kind of wondering, is there an example of dimension one. –  Sharma Sharma Jan 20 '13 at 17:01
@SharmaSharma Muhammad Zafrullah answered in the negative to your question. –  user26857 Feb 5 '13 at 0:22

Observation A: Let $D$ be a GCD domain such that $D$ contains no pair of coprime nonzero nonunits. Then $D$ is a valuation domain.

Proof. We show that for every pair of elements $x, y$ of D, $x$ divides $y$ or $y$ divides $x$. Let $x, y$ be two elements in $D$. If either is zero or a unit then $x\mid y$ or $y\mid x$. So, let’s assume that $x$ and $y$ are both nonzero nonunits. Let $d = \operatorname{GCD}(x, y)$. Then $x = ad$, $y = bd$ where $\operatorname{GCD}(a, b) = 1$. By the condition at least one of $a, b$ is a unit. If $a$ is a unit then $a\mid b$ and so $x\mid y$. Similarly if $b$ were a unit $y$ would divide $x$.

Next let $D$ be a one dimensional quasi-local $\operatorname{GCD}$ domain. We show that no two nonzero nonunits of $D$ are coprime. Suppose that $a,b$ are two nonzero nonunits of $D$ such that $\operatorname{GCD}(a,b)=1$. Then $\operatorname{GCD}(a, b^n)=1$ for all $n$, by part (c) of Theorem 49 of [K]. But as $D$ is one dimensional quasi-local by Theorem 108 of [K], $a\mid b^m$ for some positive integer $m$. But then $a\mid 1$, forcing a to be a unit. Thus by Observation A, $D$ is a valuation domain.

[K] I. Kaplansky, Commutative rings, Allyn and Bacon 1970.

-
Thank you Muhammad Zafrullah and YACP. I got it. –  Sharma Sharma Feb 5 '13 at 7:41