3

I am looking for examples which shows that the inclusion $$\text{GCD}\subsetneq \text{Integrally Closed}$$ is strict.

I've found one pretty esoteric counterexample:

Let $R$ be the algebraic closure of $\Bbb C(x)$ and let $S$ be the integral closure of $\Bbb C[x]$ in $R$. Let $M$ be a maximal ideal of $S$, and let $\overline{\Bbb Q}$ be the algebraic closure of $\Bbb Q$. Then $\overline{\Bbb Q}+MS_M$ is integrally closed but not a GCD domain. (Example 2.10 here.)

Does anyone know of a more elementary example, or at least one which is more well-known?

user26857
  • 52,094
pancini
  • 19,216

1 Answers1

4

Any Dedekind domain which is not a P.I.D. will be a counter-example.

Indeed, it can be shown that any noetherian gcd domain is a U.F.D., and a Dedekind domain which is a U.F.D. is actually principal.

Now Stark-Heegner's theorem gives the list of the $9$ imaginary quadratic fields with a principal ring of integers. $\mathbf Q(\sqrt{-5})$ is not in the list; so its ring of integers, which is $$\mathbf Z[\sqrt{-5}]\quad\text{since }-5\not\equiv 1\bmod 4,$$is a counter-example: it is integrally closed, but not a gcd domain.

Bernard
  • 175,478