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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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    **Please** make any further comment on this topic in a meta thread: this is 200% off-topic here.2012-12-30

1 Answers 1

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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.