Lemma 1
Let $A$ be a commutative algebra over a field $k$.
Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$.
Then every ideal of $A$ is finitely generated.
Proof:
Let $I$ be a non-zero ideal of $A$.
Let $f \in I$ be a non-zero element.
By the assumption, $A/fA$ is a finite $k$-module.
Hence $I/fA$ is also a finite $k$-module.
Hence $I/fA$ is a finite $A$-module.
Since $fA$ is a finite $A$-module, $I$ is also a finite $A$-module.
QED
Lemma 2
Let $A$ be a commutative ring.
Let $P_1, ..., P_{n+1}$ be distinct maximal ideals of $A$.
Then $P_1...P_n \neq P_1...P_{n+1}$.
Proof:
Suppose $P_1...P_n = P_1...P_{n+1}$.
Then $P_1...P_n \subset P_{n+1}$.
Hence $P_i \subset P_{n+1}$ for some $i \leq n$.
This is a contradiction.
QED
Lemma 3
Let $k$ be a field.
Let $A$ be a commutative algebra over a field $k$.
Suppose $A$ is a finite $k$-module.
Then Spec($A$) is finite.
Proof:
Since $A$ is a finite $k$-module, any prime ideal of $A$ is maximal.
Hence the assertion follows from Lemma 2.
QED
Lemma 4
Let $A$ be a commutative algebra over a field $k$.
Suppose $A$ is a finite $k$-module.
By Lemma 3, Spec($A$) is finite.
Let Spec($A$) = {$P_1, ..., P_r$}.
Let $I = P_1 \cap ..., \cap P_r$.
Then $I$ is nilpotent.
Proof:
Since $A$ is a finite $k$-module, every prime ideal is maximal.
Hence every element of $I$ is nilpotent by Lemma 3 of my answer to this question.
By Lemma 1, $I$ is finitely generated.
Hence $I$ is nilpotent.
QED
Lemma 5
Let $A$ be a commutative algebra over a field $k$.
Let $I$ be a non-zero proper ideal of $A$.
Suppose $A/I$ is a finite $k$-module.
Then there exist maximal ideals $P_1, ..., P_r$ such that $P_1...P_r \subset I$.
Proof:
By Lemma 3, Spec($A/I$) is finite.
Let Spec($A/I$) = {$Q_1, ..., Q_s$}.
Let $J = Q_1 \cap ... \cap Q_s$.
Since each $Q_i$ is maximal, $J = Q_1...Q_s$.
By Lemma 4, $J^k = 0$ for some integer $k \geq 1$.
Let $P_i$ be the inverse image of $Q_i$ by the canonical morphism $A \rightarrow A/I$.
Then $(P_1...P_s)^k \subset I$.
QED
Lemma 6
Let $A$ be an integrally close domain containing a field $k$ as a subring.
Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$.
Then every non-zero prime ideal of $A$ is invertible.
Proof:
Let $P$ be a non-zero prime ideal of $A$.
We claim that $P^{-1} \neq A$.
Let $a \in P$ be non-zero.
By Lemma 5, there exist maximal ideals $P_1, ..., P_r$ such that $P_1...P_r \subset aA$.
Choose $r$ such that $r$ is minimal.
Since $P_1...P_r \subset P$, one of $P_i = P$.
Without loss of generality, we can assume $P_1 = P$.
By the minimality of r, $P_2...P_r$ is not contained in $aA$.
Hence there exits $b \in P_2...P_r$ such that $b$ is not contained in $aA$.
Since $bP \subset aA$, $ba^{-1}P \subset A$.
Hence $ba^{-1} \in P^{-1}$.
Since $ba^{-1}$ is not contained in $A$, $P^{-1} \neq A$.
Since $P$ is maximal and $P \subset PP^{-1} \subset A$, $P = PP^{-1}$ or $PP^{-1} = A$.
Suppose $P = PP^{-1}$.
Since $P$ is finitely generated by Lemma 1, every element of $P^{-1}$ is integral over A.
Since $A$ is integrally closed $P^{-1} \subset A$.
This is a contradiction.
QED
Lemma 7
Let $A$ be an integrally close domain containing a field $k$ as a subring.
Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$.
Then every non-zero ideal is invertible.
Proof.
Suppose there exists a non-zero ideal $I$ which is not invertible.
We choose $I$ such that $dim_k A/I$ is minimal.
Since $A \neq I$, there exists a maximal ideal $P$ such that $I \subset P$.
$I \subset IP^{-1} \subset II^{-1} \subset A$.
If $I = IP^{-1}$, since $P$ is finitely generated by Lemma 1, every element of $P^{-1}$ is integral over $A$.
Since $A$ is integrally closed, this cannot happen by the proof of Lemma 6.
Hence $I \neq IP^{-1}$.
By the minimality of $dim_k A/I$, $IP^{-1}$ is invertible.
Hence $I$ is invertible.
This is a contradiction.
QED
Lemma 8
Let $A$ be an integrally close domain containing a field $k$ as a subring.
Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$.
Then every non-zero ideal is a product of prime ideals.
Proof:
Suppose there exists a non-zero ideal $I$ which is not a product of prime ideals.
We choose $I$ such that $dim_k A/I$ is minimal.
Since $I$ is not maximal, there exists a prime ideal $P$ such that $I \subset P$.
Then $IP^{-1} \subset A$ and $IP^{-1} \neq A$.
Suppose $I = IP^{-1}$. Since I is finitely generated by Lemma 1, every element of $P^{-1}$ is integral over $A$. Since $A$ is integrally closed, this cannot happen by the proof of Lemma 6. Hence $I \neq IP^{-1}$.
Since $I \subset IP^{-1}$, $IP^{-1}$ is a product of prime ideals.
Then $I$ is a product of prime ideals.
This is a contradiction.
QED
Proposition
Let $A$ be an integrally close domain containing a field $k$ as a subring.
Suppose $A/fA$ is a finite $k$-module for every non-zero element $f \in A$.
Then every non-zero ideal has a unique factorization as a product of prime ideals.
Proof:
This follows immediately from Lemma 8 and Lemma 6.
