
Lemma 10.119.17. Let $R$ be a ring. The following are equivalent

1. $R$ is a Dedekind domain,

2. $R$ is a Noetherian domain, and for every maximal ideal $\mathfrak m$ the local ring $R_{\mathfrak m}$ is a discrete valuation ring, and

3. $R$ is a Noetherian, normal domain, and $\dim (R) \leq 1$.

Proof. Assume (1). The argument is nontrivial because we did not assume that $R$ was Noetherian in our definition of a Dedekind domain. Let $\mathfrak p \subset R$ be a prime ideal. Observe that $\mathfrak p \not= \mathfrak p^2$ by uniqueness of the factorizations in the definition. Pick $x \in \mathfrak p$ with $x \not\in \mathfrak p^2$. Let $y \in \mathfrak p$ be a second element (for example $y = 0$). Write $(x, y) = \mathfrak p_1 \ldots \mathfrak p_ r$. Since $(x, y) \subset \mathfrak p$ at least one of the primes $\mathfrak p_ i$ is contained in $\mathfrak p$. But as $x \not\in \mathfrak p^2$ there is at most one. Thus exactly one of $\mathfrak p_1, \ldots , \mathfrak p_ r$ is contained in $\mathfrak p$, say $\mathfrak p_1 \subset \mathfrak p$. We conclude that $(x, y)R_\mathfrak p = \mathfrak p_1R_\mathfrak p$ is prime for every choice of $y$. We claim that $(x)R_\mathfrak p = \mathfrak pR_\mathfrak p$. Namely, pick $y \in \mathfrak p$. By the above applied with $y^2$ we see that $(x, y^2)R_\mathfrak p$ is prime. Hence $y \in (x, y^2)R_\mathfrak p$, i.e., $y = ax + by^2$ in $R_\mathfrak p$. Thus $(1 - by)y = ax \in (x)R_\mathfrak p$, i.e., $y \in (x)R_\mathfrak p$ as desired.

Writing $(x) = \mathfrak p_1 \ldots \mathfrak p_ r$ anew with $\mathfrak p_1 \subset \mathfrak p$ we conclude that $\mathfrak p_1 R_\mathfrak p = \mathfrak p R_\mathfrak p$, i.e., $\mathfrak p_1 = \mathfrak p$. Moreover, $\mathfrak p_1 = \mathfrak p$ is a finitely generated ideal of $R$ by Lemma 10.119.16. We conclude that $R$ is Noetherian by Lemma 10.27.9. Moreover, it follows that $R_\mathfrak m$ is a discrete valuation ring for every prime ideal $\mathfrak p$, see Lemma 10.118.7.

The equivalence of (2) and (3) follows from Lemmas 10.36.10 and 10.118.7. Assume (2) and (3) are satisfied. Let $I \subset R$ be an ideal. We will construct a factorization of $I$. If $I$ is prime, then there is nothing to prove. If not, pick $I \subset \mathfrak p$ with $\mathfrak p \subset R$ maximal. Let $J = \{ x \in R \mid x \mathfrak p \subset I\}$. We claim $J \mathfrak p = I$. It suffices to check this after localization at the maximal ideals $\mathfrak m$ of $R$ (the formation of $J$ commutes with localization and we use Lemma 10.22.1). Then either $\mathfrak p R_\mathfrak m = R_\mathfrak m$ and the result is clear, or $\mathfrak p R_\mathfrak m = \mathfrak m R_\mathfrak m$. In the last case $\mathfrak p R_\mathfrak m = (\pi )$ and the case where $\mathfrak p$ is principal is immediate. By Noetherian induction the ideal $J$ has a factorization and we obtain the desired factorization of $I$. We omit the proof of uniqueness of the factorization. $\square$

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