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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.

EDIT Why worry about the axiom of choice?

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Take an arbitrary statement of non-trivial commutative algebra and ask if it holds in ZF - what's the point? Do you know that in ZF there might be nontrivial rings with no maximal ideals at all? So why should one expect that any "spectral" methods take over? –  Martin Brandenburg Jul 12 '12 at 13:39
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I completely agree with Martin. It is very easy to generate a huge amount of these questions all across the board in mathematics. Especially since almost all your questions in this topic are either sparse ("reprove all basic calculus for me.") or you end up writing an answer on your own, it would seem to me that you would really benefit from having your own blog where you can post these things in a complete and self-contained format. If you intend to keep posting, every few days, another question of this form and then answer it yourself... then there is something wrong, in my eyes. –  Asaf Karagila Jul 12 '12 at 13:57
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Write a paper and put it on arXiv. –  Asaf Karagila Jul 12 '12 at 14:06
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Why do you think I am uninteresting in this type of questions? In fact I love these sort of questions as they are interesting (research-wise) for me, and to some extent keep the AC related questions in the mathematical interest so it is not becomes a void topic that only a few weirdos care about. What I do feel hard to ignore is someone practically turning this website to his blog posting questions and answers that he solved, and making about gazillion edits in the process instead of preparing the answers and questions in advance to post them once, and then make final improvements later. –  Asaf Karagila Jul 12 '12 at 15:07
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The use of boldface and triple question marks looks like you are yelling/whining/crying. If you want to ask a question in a civilized way and not look like a pouting child, drop the histrionics. (No, I did not downvote). –  Arturo Magidin Jul 12 '12 at 20:24
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2 Answers 2

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.

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let $A$ be an algebra over field $K$ such that every element of $A$ is algebraic over $K$ show that $A$ is Dedekind finite

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A great revision to a great answer. –  user26857 Feb 14 '13 at 21:52
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