I am trying to prove the following implication, and can't seem to find my way around all the equivalent definitions of Dedekind domains and DVRs:
I have a ring $R$ with the following properties:
1) $R$ is Noetherian.
2) $R$ is integrally closed.
3) Every nonzero prime ideal in $R$ is maximal.
I wish to show that every localization of $R$ at a maximal ideal is a principal ideal domain.
Does anyone know a direct argument proving this (i.e. not passing through the myriad of equivalent definitions of Dedekind domains and DVRs)? Alternatively, I would be thankful if someone could provide me with a "road map" to proving this claim in a a way which would convince someone (namely, me) without knowledge of Dedekind domains and DVRs.
Thanks a lot!
Roy
Best Answer
In my previous answer, we used a fact that an invertible ideal is projective and a fact that a finitely generated projective module over a local ring is free. Here is a proof without using these facts.
Lemma 1 Let $A$ be a Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Let $K$ be the field of fractions of $A$. Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} ⊂ A\}$. Then $\mathfrak{m}^{-1} \neq A$.
Proof: Let $a \neq 0$ be an element of $\mathfrak{m}$. By the assumption, Supp$(A/aA) = \{\mathfrak{m}\}$. Since Ass$(A/aA) \subset$ Supp($A/aA)$, Ass$(A/aA) = \{\mathfrak{m}\}$. Hence there exists $b \in A$ such that $b \in A - aA$ and $\mathfrak{m}b \subset aA$. Since $\mathfrak{m}(b/a) \subset A$, $b/a \in \mathfrak{m}^{-1}$. Since $b \in A - aA$, $b/a \in K - A$. QED
Lemma 1.5 Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $M \neq 0$ be a finitely generated $A$-submodule of $K$. Let $x \in K$ be such that $xM \subset M$. Then $x$ is integral over $A$.
Proof: Let $\omega_1,\dots,\omega_n$ be generators of $M$ over $A$. Let $x\omega_i = \sum_j a_{i,j} \omega_j$. Then $x$ is a root of the characteristic polynomial of the matrix $(a_{ij})$. QED
Lemma 2 Let $A$ be an integrally closed Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then $\mathfrak{m}$ is invertible.
Proof: Let $K$ be the field of fractions of $A$. Let $a \neq 0$ be an element of $\mathfrak{m}$. Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} \subset A\}$. Since $\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1} \subset A$, $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$ or $\mathfrak{m}\mathfrak{m}^{-1} = A$. Suppose $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, every element of $\mathfrak{m}^{-1}$ is integral over $A$ by Lemma 1.5. Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$. This is a contradiction by Lemma 1. Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$. QED
Lemma 3 Let $A$ be a Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then $\bigcap_n \mathfrak{m}^n = 0$.
Proof: Let $I = \bigcap_n \mathfrak{m}^n$. Suppose $I \neq 0$. Since dim$(A/I) = 0$, $A/I$ is an Artinian ring. Hence there exists $n$ such that $\mathfrak{m}^n \subset I$. Since $I \subset \mathfrak{m}^n$, $I = \mathfrak{m}^n$. Since $I \subset \mathfrak{m}^{n+1}$, $\mathfrak{m}^n = \mathfrak{m}^{n+1}$. By Nakayama's lemma, $\mathfrak{m}^n = 0$. Hence $I = 0$. This is a contradiction. QED
Lemma 4 Let $A$ be an integrally closed Noetherian local. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Let $I$ be a non-zero ideal of $A$ such that $I \neq A$. Then $I = \mathfrak{m}^n$ for some integer $n > 0$.
Proof: By Lemma 3, there exists $n > 0$ such that $I \subset \mathfrak{m}^n$ and I is not contained in $\mathfrak{m}^{n+1}$. By Lemma 2, $\mathfrak{m}$ is invertible. Since $I \subset \mathfrak{m}^n$, $I\mathfrak{m}^{-n} \subset A$. Suppose $I\mathfrak{m}^{-n} \neq A$. Then $I\mathfrak{m}^{-n} \subset \mathfrak{m}$. Hence $I \subset \mathfrak{m}^{n+1}$. This is a contradiction. Hence $I\mathfrak{m}^{-n} = A$. Hence $I = \mathfrak{m}^n$. QED
Theorem Let $A$ be an integrally closed Noetherian local. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then $A$ is a discrete valuation ring.
Proof: By Nakayama's lemma, $\mathfrak{m} \neq \mathfrak{m}^2$. Let $x \in \mathfrak{m} - \mathfrak{m}^2$. By Lemma 4, $xA = \mathfrak{m}$. Let $I$ be a non-zero ideal of $A$ such that $I \neq A$. By Lemma 4, $I = \mathfrak{m}^n$. Hence $I$ is principal. Hence $A$ is a discrete valuation ring. QED