Regular of Krull Dimension One iff Dedekind Domain

Question

I want a proof for the fact that any Noetherian integral domain with Krull dimension 1 is regular if and only if it is a Dedekind domain.

My try:

In the local case where $(R,m)$ is a Dedekind domain with $m\not =0$, we may choose $a\in m-m^2$, by Nakayama's Lemma. Since $am^{-1}\subseteq R$ we must have equality since otherwise the ideal $am^{-1}$ would fall in $m$, hence $aR=am^{-1}m\subseteq m^2$, contradicting the choice of $a$. Now, from $am^{-1}=R$ we get $aR=am^{-1}m=Rm=m$, i.e. $m$ is principal so that V-dim of $R$ is $1$. But $a$ is a non-zero divisor whence an $R$-sequence generating the maximal ideal $m$, making $R$ a regular ring, hence K-dim of $R$ equals V-dim of $R$ = $1$.

Thanks for any help for the converse!

Answer 1

Let $R$ be a Noetherian integral domain with $\dim R=1$. If $R$ is regular then $R$ is Dedekind.

We only have to show that $R$ is integrally closed. Since $R$ is regular all its localizations are regular (by definition), so all the localizations of $R$ are integrally closed. Now it follows that $R$ is integrally closed. (The last claim follows easily from $R=\bigcap_{m\in\operatorname{Max}R} R_m$.)