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An integral domain $R $ is a gcd-domain if for all $a,b \in R\setminus \{0\} $ there exists $d\in R $ such that

  • $d$ divides both $a $ and $b $
  • for all $d'$ in $R $, if $d' $ is a common divisor of $a $ and $b $ then $d' $ divides $d $

But I can't come up with an example of integral domain which is not a gcd-domain. Can you find one or give me a hint ?

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  • $\begingroup$ See this answer for some insight on the standard quadratic integer examples. $\endgroup$ – Bill Dubuque Jul 22 '17 at 16:38
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All subdomains of $\mathbb{C}$ are integral domains. However, not all subdomains of $\mathbb{C}$ are unique factorization domains.

An example of such a ring is $\mathbb{Z}[\sqrt{-5}]$. In this ring, $6 = 2 \cdot 3$ but also $6 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5}) $. By looking at the norms of $2, 3, (1 + \sqrt{-5})$ and $(1 - \sqrt{-5})$ and using $|ab| = |a|\cdot|b|$, you can see that each of those elements is an irreducible element of $\mathbb{Z}[\sqrt{-5}]$. Therefore, they cannot divide each other.

Now look at $gcd(6, 2 + 2\sqrt{-5})$. Those numbers have $2$ and $(1 + \sqrt{-5})$ as common divisors, but those don't divide each other.

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    $\begingroup$ I see why that ring is not a UFD but I don't understand which elements $a $ and $b $ to choose to get a counterexample to the the property $\endgroup$ – Friedrich Jul 22 '17 at 9:55
  • $\begingroup$ Yes, i just notices that this might not lead to what you are looking for. I'll think about it a bit. $\endgroup$ – cdwe Jul 22 '17 at 9:56
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    $\begingroup$ G.C.D. domains are not necessarily U.F.D.s. $\endgroup$ – Bernard Jul 22 '17 at 9:57
  • $\begingroup$ I think that should work now. $\endgroup$ – cdwe Jul 22 '17 at 10:11
  • $\begingroup$ What about $Z/4Z$ which is not an UFD ?? $\endgroup$ – Maman Oct 25 '17 at 13:33

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