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Let $A$ be a commutative ring. Let $f$ be any non-zero element of $A$. Suppose that $A/fA$ has a composition series as an $A$-module. Then we say $A$ is a weakly Artinian ring (this may not be a standard terminology).

Can we prove the following theorem without Axiom of Choice?

Theorem Let $A$ be a weakly Artinian integrally closed domain. Then the following assertions hold.

(1) Every ideal of $A$ is finitely generated.

(2) Every non-zero prime ideal is maximal.

(3) Every non-zero ideal of $A$ is invertible.

(4) 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?

EDIT What's wrong with trying to prove it without using AC? A proof without AC is constructive. When you are looking for a computer algorithm for solving a mathematical problem, this type of 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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Are you going to ask a single question for each and every combination of algebraic properties, whether or not it holds without the axiom of choice? –  Asaf Karagila Jul 14 '12 at 5:17
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@BenjaLim: I'm not sure that the good folks on MO will appreciate this behavior of posting a convergent sequence of partial answers. If anything I would expect mathematicians to appreciate the ability to hold back the answer until it was about done. I also don't know if people would appreciate a rain of questions verifying a majority of commutative algebra is constructive for "tame" objects. –  Asaf Karagila Jul 14 '12 at 13:32
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@AsafKragila "Are you going to ask a single question for each and every combination of algebraic properties, whether or not it holds without the axiom of choice?" No. I ask it mostly when the answer is likely to be affirmative and mostly when I think it's interesting and it has important applications. –  Makoto Kato Jul 14 '12 at 22:56
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@Asaf, do you have some kind of objection in principle to asking about the various uses of AC in algebra? To my way of thinking, this is a perfectly reasonable topic of study, particularly in those cases where it seems likely that AC is required. –  JDH Jul 15 '12 at 1:18
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@Makoto Please do not bump questions or answers for the purpose of asking about downvotes. Such questions are only rarely fruitful. If you insist on posing such questions then please do so in comments, so that the thread is not bumped. Better, as always, discuss such meta issues on the meta site - where they belong. –  Bill Dubuque Jul 16 '12 at 6:28

1 Answer 1

We assume tacitly the definitions in my answer to this question.

Definition 1 Let $A$ be a commutative ring. Suppose $A$ has a composition series as an $A$-module. Then we say $A$ is an Artinian ring.

Lemma 1 Let $A$ be an Artinian ring. Let $I$ be an ideal of $A$. Then $A/I$ is an Artinian ring.

Proof: This follows from Lemma 2 of my answer to this question.

Lemma 2 An Artinian ring is weakly Artinian.

Proof: This follows from Lemma 1.

Lemma 3 Let $A$ be a weakly Artinian ring. Let $I$ be a non-zero ideal of A. Then $A/I$ is Artinian.

Proof: Since $I$ is non-zero, $I$ contains a non-zero element $f$. Since $A/fA$ is Artinian, $A/I$ is Artinian by Lemma 1. QED

Lemma 4 Let $A$ be a weakly Artinian ring. Let $I$ be an ideal of $A$. Then $A/I$ is a weakly Artinian ring.

Proof: Clear.

Lemma 5 Let $A$ be a weakly Artinian ring. 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 Lemma 7 of my answer to this question, $I/fA$ is a finite $A$-module. Since $fA$ is a finite $A$-module, $I$ is also a finite $A$-module. QED

Lemma 6 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module. Suppose that $M$ has a composition series. Let $f:M \rightarrow M$ be an injective $A$-homomorphism. Then $f$ is surjective.

Proof: Since $f$ is injective, $leng$ $M = leng$ $f(M)$. By Lemma 4 of my answer to this question, $f(M) = M$. QED

Lemma 7 Let $A$ be an Artinian integral domain. Then $A$ is a field.

Proof: Let $a$ be a non-zero element of $A$. Let $f: A \rightarrow A$ be the map defined by $f(x) = ax$. Since $f$ is injective, it is surjective by Lemma 6. Hence $a$ is invertible. Hence $A$ is a field. QED

Lemma 8 Let $A$ be an Artinian ring. Let $P$ be a prime ideal of $A$. Then $P$ is maximal.

Proof: This follows immediately from Lemma 1 and Lemma 7.

Lemma 9 Let $A$ be a weakly Artinian ring. Let $P$ be a non-zero prime ideal. Then $P$ is maximal.

Proof: By Lemma 3, $A/P$ is an Artinian ring. Since $A/P$ is an integral domain, $A/P$ is a field by Lemma 7. Hence $P$ is maximal. QED

Lemma 10 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module. Suppose that $M$ has a composition series. Let $N$ be an $A$-submodule of $M$. Then $leng$ $M = leng$ $N + leng$ $M/N$.

Proof: By Lemma 2 of my answer to this question, $leng$ $M/N$ is finite. By Lemma 3 of my answer to this question, $leng$ $N$ is finite. Hence $leng$ $M = leng$ $N + leng$ $M/N$. QED

Lemma 11 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module.

Let $N_0 \supset N_1 \supset ... \supset N_r$ be a descending sequence of $A$-submodules of $M$. Suppose that $N_i \neq N_{i+1}$ for $i = 0, 1, ..., r - 1$. Then $r \leq leng$ $M$.

Proof: This follows from Lemma 10.

Lemma 12 Let $A$ be an Artinian ring. Then Spec($A$) is finite.

Proof: This follows from Lemma 2 of my answer to this question and Lemma 8 and Lemma 11.

Lemma 13 Let $A$ be an Artinian ring. By Lemma 12, Spec($A$) is finite. Let Spec($A$) = {$P_1, ..., P_r$}. Let $I = P_1 \cap ..., \cap P_r$. Then $I$ is nilpotent.

Proof: This follows from Lemma 8 and the proposition of my answer to this question.

Lemma 14 Let $A$ be a weakly Artinian ring. Let $I$ be a non-zero proper ideal of $A$. Then there exist maximal ideals $P_1, ..., P_r$ such that $P_1...P_r \subset I$.

Proof: By Lemma 3, $A/I$ is Artinian. By Lemma 13, 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 13, $J^k = 0$ 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 15 Let $A$ be a weakly Artinian integrally closed domain. 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 14, 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 5, every element of $P^{-1}$ is integral over $A$. Since $A$ is integrally closed $P^{-1} \subset A$. This is a contradiction. QED

Lemma 16 Let $A$ be a weakly Artinian integrally closed domain. Then every non-zero ideal is invertible.

Proof. Let $\Lambda$ be the set of non-zero ideals which are not invertible. Suppose $\Lambda$ is not empty. By Lemma 5 of my answer to this question, there exists a maximal element $I$ in $\Lambda$. Since $A \neq I$, by Lemma 5 of my answer to this question, 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 5, every element of $P^{-1}$ is integral over $A$. Since $A$ is integrally closed, this cannot happen by the proof of Lemma 15. Hence $I \neq IP^{-1}$. Since $I$ is a maximal element in $\Lambda$, $IP^{-1}$ is invertible. Hence $I$ is invertible. This is a contradiction. QED

Lemma 17 Let $A$ be a weakly Artinian integrally closed domain. Then every non-zero ideal is a product of prime ideals.

Proof: Let $\Lambda$ be the set of non-zero ideals which are not a product of prime ideals. Suppose $\Lambda$ is not empty. By Lemma 5 of my answer to this question, there exists a maximal element $I \in \Lambda$. Since $I$ is not a maximal ideal, By Lemma 5 of my answer to this question, 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 5, every element of $P^{-1}$ is integral over $A$. Since $A$ is integrally closed, this cannot happen by the proof of Lemma 15. 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 a weakly Artinian integrally closed domain. Then every non-zero ideal has a unique factorization as a product of prime ideals.

Proof: This follows immediately from Lemma 16 and Lemma 17.

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