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This should be a simple question on basic definitions.

In an integral domain $R$ we will say that $gcd(x,y)$ exists and is equal to some $r\in R$ if $r$ divides $x$ and $y$, and any common divisor $r'$ of $x,y$ divides $r$ (i.e. there exists an $e\in R$ such that $r=r'e$).

Now, is it always true for $x,y,z\in R$ that $gcd(zx,zy)$ necessarily exists if $gcd(x,y)$ does (then it can be easily verified that $gcd(zx,zy)=z\cdot gcd(x,y)$)? This statement seems to be wrong in general: for a field $k$ consider the domain $R=k[x,y,z,t]/(xt-yz)\cong k[x,y,z,yz/x]$. Then $gcd(x,y)=1$, whereas $gcd(zx,zy)$ does not exist, since both $x$ and $z$ are common divisors of $zx$ and $zy$. Is this correct?

Also (as shown by this example) it seems that $gcd(x,yz)$ does not have to be $=1$ if $gcd(x,y)=gcd(x,z)=1$. Is this true?

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    $\begingroup$ I think the answers are yes and yes. By the way: en.wikipedia.org/wiki/GCD_domain $\endgroup$
    – Mark Grant
    Commented Apr 20, 2012 at 11:20
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    $\begingroup$ I agree with Mark. The gcd is not well behaved in arbitrary integral domains. $\endgroup$
    – Martin Brandenburg
    Commented Apr 20, 2012 at 11:43
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    $\begingroup$ If you define $gcd(x,y)=Rx+Ry=(x,y)$ then it's always defined (though no element in R but an ideal) and $(x,y)=1=(x,z)$ implies $x,yz)=1$. $\endgroup$
    – Ralph
    Commented Apr 20, 2012 at 18:38

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