# Let $R$ be an integral domain and $r, s$ $\in$ $R$

Let R be an integral domain and r, s $$\in$$ R.

I know that an integral domain is a commutative ring with identity that has no zero-divisors. Hence, R is a commutative ring with identity and has no zero-divisors.

Given this, can the gcd(r, s) have more than one gcd? Also, would the gcd(r, s) be associates of one another?

I am trying to work this out in $$\mathbb{Z}$$[$$x$$] (since I know that $$\mathbb{Z}$$ is an integral domain as $$\mathbb{Z}$$[$$x$$] is a commutative polynomial ring with identity that has no zero-divisors) but I do not know if this is a smart integral domain to work with. Would appreciate some guidance. Thanks in advance

• What you are looking for is called a GCD domain en.wikipedia.org/wiki/GCD_domain In particular for your example, since $\mathbb{Z}$ is a UFD, $\mathbb{Z}[x]$ is also a UFD. Therefore, $\mathbb{Z}[x]$ is a GCD domain based on the inclusion chart in the link. Nov 21, 2019 at 9:03
• @take008 Since $\mathbb{Z}$[$x$] is a GCD domain, does that mean that gcd(r, s) has more than one gcd and thus the gcd(r, s) are all associates of one another? Nov 21, 2019 at 9:14
• Depends on how you are defining $\gcd$. But in the most common definition, yes the $\gcd$ is unique up to units in GCD domains. Nov 21, 2019 at 9:20
• @take008 Thanks for the feedback!! So would the gcd's of (r, s) be associates of one another? Nov 21, 2019 at 9:31
• Let's do this from the definition. "There exists a unique minimal principal ideal $I$ containing the ideal generated by the two elements $r$, $s$." Suppose $a,b\in I$ both generated $I$: $(a)=(b)$. Then $a\in (b)$, so that $a=cb$. Similarly, $da=b$ for some $c,d\in R$. Therefore $a=cda$ and hence $cd=1$ by cancellation law and stuff. So, $c,d$ are units. So yes, they are associates. Nov 21, 2019 at 9:38

Hint: if $$\,d_1,d_2$$ are gcds then by definition $$\, c\mid d_1\!\iff c\mid a,b\iff c\mid d_2\,$$ so $$\,\color{#c00}{d_1\mid d_2\mid d_1}$$
Beware  In rings that are not domains $$\rm\color{#c00}{associates}$$ need not be a unit multiples.