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From Wikipedia, a GCD domain is an integral domain in which every pair of elements has a GCD.

Let us consider some polynomial quotient ring $R=K[X]/(pq)$ where $K$ is a field and $p$, $q$ are (irreducible) polynomials. Then $R$ is not an integral domain (since $pq=0$ in $R$), so cannot be a GCD domain according to Wikipedia's definition. Yet I am unable to find to elements without GCD. Do they exist?

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Actually, there is no example of such elements: Consider the ring $R=K[X]/(pq)$ where $K$ is a field, $p$ and $q$ are irreducible polynomials.

Claim: Let $A,B\in R$, $a$ and $b$ their respective lifts in $K[X]$, and $g = \gcd(a,b)$. Then the image $G$ of $g$ in $R$ is a gcd of $A$ and $B$.

Proof. First, it is clear that $G$ divides both $A$ and $B$. Indeed, since $a = g\cdot h$ for some $h$, $A=G\cdot H$ where $H$ is the image of $h$ in $R$. Now, let $D$ be any common divisor of $A$ and $B$, that is $A=D\cdot W_a$ and $B = D\cdot W_b$ for some $W_a,W_b\in R$. Denote by $d$, $w_a$ and $w_b$ the respective lifts of $D$, $W_a$ and $W_b$ in $K[X]$. Then $a = d\cdot w_a + \lambda pq$ and $b=d\cdot w_b + \mu pq$. Now, let us write $g = a\cdot u + b\cdot v$ (by Bézout's identity). Then we have $$g = d\cdot (w_a\cdot u + w_b\cdot v) + (\lambda u+\mu v) pq.$$ That is, $G = D(W_a U + W_b V)$ where $U$ and $V$ are the images of $u$ and $v$ in $R$. In other words, $D$ divides $G$.

Remark. Any divisor of $G$ by a unit in $R$ is still a gcd of $A$ and $B$. In particular, the image of $\gcd(a,b,pq)$ is another gcd.

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