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Les $A$ be an integral domain[1] and $S\subset A$. The set $gcd(S)$ is the set of elements $\delta$ satisfying the two following properties :

  • $\delta|s$ for all $s\in S$
  • if $d|s$ for all $s\in S$, then $d|\delta$.

Let $a\in A$. I want to prove the set equality $a gcd(S)=gcd(aS)$.

I am okay with the inclusion $gcd(aS)\subset a\, gcd(S)$. For the reverse inclusion, let $\delta\in gcd(S)$. In want to prove that $a\delta\in gcd(aS)$.

For each $s\in S$, since $\delta|s$, we have $a\delta|s$. So $a\delta$ checks the first property.

Now let $d\in A$ such that $d|as$ for every $s\in S$.

QUESTION: how to prove that $d|a\delta$ ?

[1] Or a unique factorization domain ? What is the most general in which my property is true ?

EDIT : This question is different from that one : Prove that $(ma, mb) = |m|(a, b)\ $ [GCD & LCM Distributive Law] In this question we empathize the fact that the gcd is a set, not a number. The given answers are more general.

Laurent Claessens
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  • Is $S$ assumed to be finite? Then in a UFD surely we have $a\gcd(S)=\gcd(aS)$. Edit: Oh, $S$ need not be assumed to be finite; sorry. – Noiril Jul 28 '23 at 05:36
  • Yes, if needed, we can assume that S is finite. I'm curious to see where a proof can fail when S is infinite... – Laurent Claessens Jul 28 '23 at 05:41
  • See the alternative proof here in the linked dupe, which shows that it holds true in any domain if $,\gcd(aS),$ exists (the proof there is for two arguments but it clearly works for any argument set $S)\ \ $ – Bill Dubuque Jul 28 '23 at 16:17
  • The linked proof shows that your sets have the same elements. Alternatively use the fact that $\gcd S$ is simply the set of all associates of any gcd of $,S,$ (recall gcds are defined only up to associates). – Bill Dubuque Jul 28 '23 at 16:40

2 Answers2

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You can prove the inclusion $a\gcd(S)\subseteq\gcd(aS)$ if you assume $\gcd(aS)\neq\varnothing$; this is true in a UFD.

Let $\delta\in\gcd(S)$ and $d'\in\gcd(aS)$. For every $s\in S$, we have $\delta\mid s$, hence $a\delta\mid as$, from which $a\delta\mid d'$. Let $u\in A$ such that $d'=ua\delta$. By the inclusion $\gcd(aS)\subseteq a\gcd(S)$ (that you already proved), $u\delta\in\gcd(S)$, hence $u\delta\mid\delta$ from which $u\delta a\mid\delta a$, that's $d'\mid\delta a$.

Fabio Lucchini
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Either a UFD or a Bezout domain will be sufficient. These two properties of integral domains do not imply each other. (See this post and note that an integral domain is a PID if and only if it is both a UFD and a Bezout domain.)

In a UFD, two (nonzero) elements $a,b\in R$ with factorizations $$a=u(a)\prod_pp^{n(p,a)}$$ $$b=u(b)\prod_pp^{n(p,b)}$$ satisfies $a\mid b$ if and only if $n(p,a)\le n(p,b)$ for each possible prime factor $p$ of $a$ and $b$. And the set of greatest common divisors of (possibily infinite) set of elements $S$ is $$\gcd S=\left\{u\prod_pp^{\min_{s\in S}n(p,s)}\colon u\in R^\times\right\}.$$

In your proof, $d\mid as$ implies $n(p,d)\le n(p,a)+n(p,s)$. Since $n(p,\delta)=\min_{s\in S}n(p,s)$, we have $n(p,d)\le n(p,a)+n(p,d)$, i.e. $d\mid a\delta$.

Now suppose that $R$ is a Bezout domain. Assume that $S$ is finite. (I don't know if my proof works when $S$ is infinite, a comment on this will be appreciated!) Suppose that $\delta\in\gcd S$ and $d\mid as$ for each $s\in S$. Then by definition of Bezout domains $\delta$ is an $R$-combination of elements of $S$, and so $a\delta$ is an $R$-combination of elements of $aS$. Therefore we have $d\mid a\delta$.

Noiril
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