# Example of a non-gcd domain

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 ?

• See this answer for some insight on the standard quadratic integer examples. – Bill Dubuque Jul 22 '17 at 16:38

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.
• 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 – Friedrich Jul 22 '17 at 9:55
• What about $Z/4Z$ which is not an UFD ?? – Maman Oct 25 '17 at 13:33