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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!

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  • $\begingroup$ I do not know you definition of Dedekind domain, but maybe you will find useful to identify those two classes of rings of your question to that of noetherian domains whose localizations at maximal ideals are discrete valuation rings. $\endgroup$
    – SlavaM
    Oct 26, 2015 at 16:44

1 Answer 1

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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 $R=\bigcap_{m\in\operatorname{Max}R} R_m$.)

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