Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951.
Can we prove the following theorem without Axiom of Choice?
Theorem Let $A$ be an integrally closed 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 the following assertions hold.
(1) Every ideal of $A$ is finitely generated.
(2) Every non-zero ideal of $A$ is invertible.
(3) Every non-zero ideal of $A$ has a unique factorization as a product of prime ideals.
EDIT May I ask the reason for the downvotes? Is this the reason for the downvotes?
EDIT What's wrong with trying to prove it without using AC? When you are looking for a computer algorithm for solving a mathematical problem, such a proof may provide a hint. At least, you can be sure that there is a constructive proof.
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.