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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

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    $\begingroup$ I‘m not sure whether the proofs I have in mind would satisfy you. It seems to me that you quickly get to the statement “an integrally closed Noetherian domain with a unique non-zero prime ideal is in fact principal” and from there you have to do some real work, and you would be proving one direction of an equivalence between definitions. I think Serre does this in a very low-tech way at the beginning of his Local Fields, although I don't like that proof very much. Are willing to use some commutative algebra? Basic dimension theory really helps. $\endgroup$ Aug 29, 2012 at 23:59
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    $\begingroup$ Related question: math.stackexchange.com/questions/183467/… may only be edited for 5 minutes(click on this box to dismiss) $\endgroup$ Aug 30, 2012 at 0:33
  • $\begingroup$ See also math.stackexchange.com/questions/95789 $\endgroup$ Aug 22, 2019 at 17:06

5 Answers 5

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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

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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 ring 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 principal.

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 $\mathfrak{m}\mathfrak{m}^{-1} = A$ and therefore $\mathfrak{m}$ is invertible. Hence $\mathfrak{m}$ is a projective $A$-module. Since $A$ is a local ring, $\mathfrak{m}$ is free $A$-module. Hence $\mathfrak{m}$ is principal. 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

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: Let $I$ be a non-zero ideal of $A$ such that $I \neq A$. 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 principal. Hence $\mathfrak{m}$ 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 $I = \mathfrak{m}^n$. Since $I$ is principal, $I$ is also principal. QED

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  • $\begingroup$ Thank you very, very much! $\endgroup$ Aug 30, 2012 at 13:11
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The following proof is similar to the previous ones but we only assume very basic knowledge of commutative algebra.

Lemma 0 Let $A$ be a Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Let $I$ be a non-zero ideal of $A$. Then there exists $n > 0$ such that $\mathfrak{m}^n \subset I$.

Proof: Suppose the assertion is false. Let $\mathfrak{I}$ be the set of non-zero ideal of $A$ such that the statement is false. Let $I$ be a maximal element of $\mathfrak{I}$. Since $I$ is not a prime ideal, there exists $a, b \in A$ such that $ab \in I$ and $a \in A - I, b \in A - I$. Let $J_1 = I + aA, J_2 = I + bA$. Since $I$ is a maximal element of $\mathfrak{I}$, there exists $n_1, n_2 > 0$ such that $\mathfrak{m}^{n_1} \subset J_1$, $\mathfrak{m}^{n_2} \subset J_2$, Since $J_1J_2 \subset I$, this is a contradiction. QED

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 Lemma 1, there exists n > 0 such that $\mathfrak{m}^n \subset aA$. Let $n$ be minimal satisfying this condition. Let $b \in \mathfrak{m}^{n-1} - aA$. Then $\mathfrak{m}b \subset aA$. 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 an integrally closed Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then every non-zero ideal is invertible.

Proof: Suppose the assertionis is false. Let $I$ be a maximal non-zero non-invertible ideal. Then $I \subset \mathfrak{m}$. Hence $I \subset I\mathfrak{m}^{-1} \subset A$. Suppose $I = I\mathfrak{m}^{-1}$. Then every element of $\mathfrak{m}^{-1}$ is integral over $A$ by Lemma 1.5. Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$. By Lemma 1, this is a contradiction. Hence $I \neq I\mathfrak{m}^{-1}$. Hence $I\mathfrak{m}^{-1}$ is invertible. Hence $I$ is invertible. 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: Suppose the assertionis is false. Let $I \neq A$ be a maximal non-zero ideal which is not power of $\mathfrak{m}$. Since $I \subset \mathfrak{m}$, $I \subset I\mathfrak{m}^{-1} \subset A$. $I \neq I\mathfrak{m}^{-1}$ by the same argument of the proof of Lemma 3. Hence $I\mathfrak{m}^{-1}$ is a power of $\mathfrak{m}$. Hence $I$ is a power of $\mathfrak{m}$. This is a contradiction. 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: Suppose $\mathfrak{m} = \mathfrak{m}^2$. Since $\mathfrak{m}$ is invertible, $\mathfrak{m} = A$. This is a contradiction. Hence $\mathfrak{m} \neq \mathfrak{m}^2$. 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

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Lemma 1 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $\bigcap_n \mathfrak{m}^n = 0$.

Proof: Let $\mathfrak{m} = tA$. Let $x \in \bigcap_n \mathfrak{m}^n$. Suppose $x \neq 0$. There exists $y_n \in A$ for every $n$ such that $x = t^ny_n$. Then $t^ny_n = t^{n+1}y_{n+1}$. Hence $y_n = ty_{n+1}$. Hence $y_nA \subset y_{n+1}A$. Since $A$ is Noetherian, there exists $k$ such that $y_kA = y_{k+1}A$. Hence there exists $u \in A$ such that $y_{k+1} = uy_k$. Since $y_k = ty_{k+1}$, $y_k = uty_k$. Hence $(1 - ut)y_k = 0$. Since $t \in \mathfrak{m}$, $1 - ut$ is invertible. Hence $y_k = 0$. Hence $x = t^ky_k = 0$. This is a contradiction. QED

Lemma 2 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $A$ is a discrete valuation ring.

Proof: Suppose $\mathfrak{m} = tA$. By Lemma 1, $\bigcap_n \mathfrak{m}^n = 0$. Let $I$ be a non-zero ideal of $A$. There exists $n$ such that $I \subset \mathfrak{m}^n$ but not $I \subset \mathfrak{m}^{n+1}$. Since $\mathfrak{m}^n = t^nA$, $It^{-n} \subset A$. Suppose $It^{-n} \neq A$. Then $It^{-n} \subset \mathfrak{m}$. Hence $I \subset \mathfrak{m}^{n+1}$. This is a contradictin. Hence $I = t^nA$. QED

Lemma 3 Let $A$ be an integral domain. Let $I$ be an ideal of $A$. Suppose $I$ is invertble. Then $I$ is a finitely generated projective $A$-module.

Proof: Since $II^{-1} = A$, there exist $a_1,\dots,a_n \in I$ and $b_1,\dots,b_n \in I^{-1}$ such that $\sum_i a_ib_i = 1$. Let $f_i:I\rightarrow A$ be the $A$-linear map defined by $f_i(x) = b_ix$. Let $L$ be a free $A$-module with a basis $e_1,\dots,e_n$. Let $g:L \rightarrow I$ be the $A$-linear map defined by $g(e_i) = a_i$. Let $f:I \rightarrow L$ be the $A$-linear map defined by $f(x) = \sum_i f_i(x)e_i = \sum_i b_ixe_i$. Since $gf(x) = \sum_i g(b_ixe_i) = \sum_i b_ia_ix = x$ for every $x \in I$, $gf = 1$. Hence $I$ is isomorphic to a direct summand of $L$. Hence $I$ is a finitely generated projective $A$-module. QED

Lemma 4 Let $A$ be a local ring. Let $M$ be a finitely generated projective $A$-module. Then $M$ is a finitely generated free $A$-module.

Proof: Let $\mathfrak{m}$ be the maximal ideal of $A$. Let $k = A/\mathfrak{m}$. Since $M$ is finitely generated, dim$_k M\otimes_A k$ is finite. Let $a_1,\dots,a_n$ be elements of $M$ such that $\{a_1\otimes 1,\dots,a_n\otimes 1\}$ is a basis of $M\otimes_A k$ over $k$. By Nakayama's lemma, $a_1,\dots,a_n$ generates $M$ over $A$. Let $L$ be a free $A$-module with a basis $\{e_1,\dots,e_n\}$. Let $f:L\rightarrow M$ be the $A$-linear map such that $f(e_i) = a_i (i = 1,\dots,n)$. Let $K$ be the kernel of $f$. Then we get the following exact sequence.

$0 \rightarrow K \rightarrow L \rightarrow M \rightarrow 0$

Then the following sequence is exact by the well known theorem of homological algebra.

Tor$_1(M, k) \rightarrow K\otimes_A k \rightarrow L\otimes_A k \rightarrow M\otimes_A k \rightarrow 0$

Since $M$ is projective, Tor$_1(M, k) = 0$. Since $L\otimes_A k \rightarrow M\otimes_A k$ is an isomorphism, $K\otimes_A k = 0$. Since $M$ is projective, $K$ is a direct summand of $L$. Hence $K$ is finitely generated. Hence $K = 0$ by Nakayama's lemma. QED

Lemma 5 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 6 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 7 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 6. Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$. This is a contradiction by Lemma 5. Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$ and therefore $\mathfrak{m}$ is invertible. QED

Lemma 8 Let $A$ be an integral domain. Let $L$ be a finitely generated free A-module. Let $M$ be a finitely generated free A-submodule of L. Then rank$_A M \le$ rank$_A L$.

Proof: Let $K$ be the field of fractions of $A$. Since $K$ is a flat $A$-module, the canonical homomorphism $M\otimes_A K \rightarrow L\otimes_A K$ is injective. Hence we are done. 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 Lemma 7, $\mathfrak{m}$ is invertible. By Lemma 3, $\mathfrak{m}$ is projective over A. By Lemma 4, $\mathfrak{m}$ is a finitely generated free module $A$. By Lemma 8, $\mathfrak{m}$ is principal. Hence $A$ is a discrete valuation ring by Lemma 2. QED

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Lemma 1 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $\bigcap_n \mathfrak{m}^n = 0$.

Proof: Let $\mathfrak{m} = tA$. Let $x \in \bigcap_n \mathfrak{m}^n$. Suppose $x \neq 0$. There exists $y_n \in A$ for every $n$ such that $x = t^ny_n$. Then $t^ny_n = t^{n+1}y_{n+1}$. Hence $y_n = ty_{n+1}$. Hence $y_nA \subset y_{n+1}A$. Since $A$ is Noetherian, there exists $k$ such that $y_kA = y_{k+1}A$. Hence there exists $u \in A$ such that $y_{k+1} = uy_k$. Since $y_k = ty_{k+1}$, $y_k = uty_k$. Hence $(1 - ut)y_k = 0$. Since $t \in \mathfrak{m}$, $1 - ut$ is invertible. Hence $y_k = 0$. Hence $x = t^ky_k = 0$. This is a contradiction. QED

Lemma 2 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $A$ is a discrete valuation ring.

Proof: Suppose $\mathfrak{m} = tA$. By Lemma 1, $\bigcap_n \mathfrak{m}^n = 0$. Let $I$ be a non-zero ideal of $A$. There exists $n$ such that $I \subset \mathfrak{m}^n$ but not $I \subset \mathfrak{m}^{n+1}$. Since $\mathfrak{m}^n = t^nA$, $It^{-n} \subset A$. Suppose $It^{-n} \neq A$. Then $It^{-n} \subset \mathfrak{m}$. Hence $I \subset \mathfrak{m}^{n+1}$. This is a contradictin. Hence $I = t^nA$. QED

Lemma 3 Let $A$ be a local domain. Let $\mathfrak{m}$ be its maximal ideal. Suppose $\mathfrak{m}$ is invertble. Then $\mathfrak{m}$ is principal.

Proof(Serre's Local fields): $\mathfrak{m}\mathfrak{m}^{-1} = A$, there exist $a_1,\dots,a_n \in \mathfrak{m}$ and $b_1,\dots,b_n \in \mathfrak{m}^{-1}$ such that $\sum_i a_ib_i = 1$. If $a_ib_i \in \mathfrak{m}$ for all $i$, $1 \in \mathfrak{m}$. This is a contradiction. Hence there exists $k$ such that $a_kb_k \in K - \mathfrak{m}$. Since $b_k \in \mathfrak{m}^{-1}$, $a_kb_k \in A$. Hence $a_kb_k = u$ is invertible. Hence $a_ku^{-1}b_k = 1$. Let $a = a_ku^{-1}$. Then $a \in \mathfrak{m}$ and $ab_k = 1$. Let $x \in \mathfrak{m}$. $x = xab_k$. Since $b_k \in \mathfrak{m}^{-1}$, $xb_k \in A$. Hence $x \in aA$. Hence $\mathfrak{m} = aA$. QED

Lemma 4 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 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 6 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 5. Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$. This is a contradiction by Lemma 4. Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$ and therefore $\mathfrak{m}$ is invertible. 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 Lemma 6, $\mathfrak{m}$ is invertible. By Lemma 3, $\mathfrak{m}$ is principal. Hence $A$ is a discrete valuation ring by Lemma 2. QED

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