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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

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  • $\begingroup$ 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. $\endgroup$
    – take008
    Nov 21, 2019 at 9:03
  • $\begingroup$ @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? $\endgroup$
    – asuhdude
    Nov 21, 2019 at 9:14
  • $\begingroup$ Depends on how you are defining $\gcd$. But in the most common definition, yes the $\gcd$ is unique up to units in GCD domains. $\endgroup$
    – take008
    Nov 21, 2019 at 9:20
  • $\begingroup$ @take008 Thanks for the feedback!! So would the gcd's of (r, s) be associates of one another? $\endgroup$
    – asuhdude
    Nov 21, 2019 at 9:31
  • $\begingroup$ 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. $\endgroup$
    – take008
    Nov 21, 2019 at 9:38

1 Answer 1

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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.

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