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Let $M$ be a commutative cancellative monoid. For elements $a,b \in M$ a gcd of $a,b$ is an element $\mathrm{gcd}(a,b)$ with the (universal) property $\forall c \in M (c |\mathrm{gcd}(a,b) \Leftrightarrow c|a \wedge c|b)$.

Assume that $a,b,t \in M$ are elements such that $\mathrm{gcd}(a,b)$ exists. Does then also $\mathrm{gcd}(at,bt)$ exist?

If it exists, then it is necessarily given by $\mathrm{gcd}(a,b)t$. But curiously I cannot show that $\mathrm{gcd}(a,b)t$ satisfies the defining property of $\mathrm{gcd}(at,bt)$ without assuming that $\mathrm{gcd}(at,bt)$ already exists. See my answer here (Lemma 1). Therefore, I suspect that the answer to my question is "no" and actually there are examples where $a,b$ have a gcd, but $at,bt$ do not. I prefer examples of the form $M = R \setminus \{0\}$ for integral domains $R$.

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Martin Brandenburg
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  • If $d=\gcd(a,b)$ and $a=da',\ b=db'$ so that $gcd(a',b')=1,$ plugging things in shows you may assume that, in your question as stated, one may as well assume $\gcd(a,b)=1$ and ask if then $gcd(at,bt)$ exists. Is this version any easier? [more details if desired on this comment] – coffeemath Nov 24 '14 at 12:22

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Consider an example similar to one OP used in a now deleted answer, where $M$ is the commutative monoid generated by $x,y$ with the single relation $x^2=y^2.$ Now let $a=x,\ b=y$ so that $\gcd(a,b)=1.$ Let also $t=x.$ Now here's where I have to rely on the OP's statement that, if $\gcd(at,bt)$ exists, then it is necessarily $\gcd(a,b)t$. [That seem right to me but I don't have that proof yet for myself. Note: Daniel Fisher has kindly included a proof of this in a comment.]

With our definitions of $a,b,t$ this means that $gcd(x^2,yx)=x.$ But then we have $y|x^2$ and $y|yx,$ without also having $y|x.$ [Note that Daniel Fischer pointed out I previously had finished this incorrectly.]

coffeemath
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    Oh, by the way. Let $d = \gcd (a,b)$, and suppose $g = \gcd(at,bt)$ exists. Then $dt \mid at$ and $dt\mid bt$, hence $dt \mid g$. Let $g = udt$. Then $at = \alpha\cdot g = \alpha udt$, and since the monoid is cancellative, $a = \alpha ud$. Similarly we obtain $b = \beta ud$. Hence $ud\mid d$, and $u$ is a unit. – Daniel Fischer Nov 24 '14 at 17:05
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    "That seem right to me but I don't have that proof yet for myself." - I have given the proof in the other thread and even linked it in my question. – Martin Brandenburg Nov 24 '14 at 19:10