Localizations of Dedekind Domains are Discrete Valuation Rings 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
 A: 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
A: 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
A: 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
A: 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
A: 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
