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Let $A$ be a commutative ring. Suppose $A$ has a composition series as an $A$-module. EDIT Since $A$ has a composition series, $A$ has a maximal ideal.

Let $J$ be the intersection of all the maximal ideals of $A$. Can we prove that $J$ is nilpotent without Axiom of Choice?

EDIT[July 14, 2012] In the following argument, am I using any form of Axiom of Choice?

Let $\Lambda$ be nonempty set of ideals of $A$. Choose $I_0 \in \Lambda$. Let $leng_A A/I_0 = r$. Since $A$ has a composition series, $r$ is finite. If $I_0$ is not maximal in $\Lambda$, we can choose $I_1 \in \Lambda$ such that $I_0 \neq I_1$ and $I_0 \subset I_1$. Repeat this. This procedures must terminate with no more than $r$ times trials. Hence there exists a maximal element in $\Lambda$.

EDIT May I ask 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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I have a hunch you are going to answer your own question. Am I right? –  Marc van Leeuwen Jul 13 '12 at 8:26
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I have a hunch, much like @Marc, that you will write the answer is no less than 10 edits, too. –  Asaf Karagila Jul 13 '12 at 8:29
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I object to closing this question. On the face of it it looks reasonable and I think ridiculing it is completely unwarranted (I'm aware of the meta thread on multiple edits and its history). Can we please take it easy? –  t.b. Jul 13 '12 at 23:31
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There are two votes to close as "not a real question". I strongly disagree: the OP has posted a precisely stated, nontrivial mathematical question. It assumes a certain familiarity with the subject matter (e.g. that a commutative ring is Artinian iff it admits a composition series as a module over itself), which is certainly okay. It seems to me that people are letting personal feelings about the OP influence their votes to close. You need not be interested in this question -- I am not so interested in commutative algebra without AC, for instance -- but that's no reason to close it. –  Pete L. Clark Jul 14 '12 at 5:32
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@Asaf and others: Please stop discussing these meta matters on the main site. Instead please take the discussion to our meta site, e.g. the meta thread on this topic, or create another meta question if need be. It is important to keep the main site free of meta tension, since that may alienate many members. The moderators are awaiting community feedback on this matter, so please join said meta discussion. –  Bill Dubuque Jul 14 '12 at 14:36
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2 Answers

The axiom of choice is equivalent to the statement "every commutative ring with unity has a maximal ideal"--even to the weaker statement "every unique factorization domain has a maximal ideal". This suggests that the answer to your question is "no". Now, if your rings were Noetherian instead of Artinian, it's quite possible.

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So which is your answer, yes or no? –  Makoto Kato Jul 13 '12 at 18:26
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I suspect the answer is no, though I've not come up with a counterexample, yet. I'll think on it. –  Cameron Buie Jul 13 '12 at 18:40
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Dear Cameron Buie, "This suggests that the answer to your question is "no"." I'm afraid I don't think I understand this. Could you elaborate on this? –  Makoto Kato Jul 14 '12 at 11:44
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Definition 1 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module. Suppose $M \neq 0$ and $M$ has no proper $A$-submodule other than $0$. Then we say $M$ is simple.

Definition 2 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module. Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a finite descending sequence of $A$-submodules. If each $M_i/M_{i+1}, i = 0, 1, ..., n - 1$, is simple, this sequence is called a composition series of $M$. The $n$ is called the length of the composition series. We define the length of the $0$-module as $0$.

Definition 3 Let $A$ be a (not necessarily commutative) ring. Let $M$ be a left $A$-module. Suppose $M$ has a composition series, the lengths of each series are the same by Jordan-Holder theorem. We denote it by $leng_A M$ or $leng$ $M$. If $M$ does not have a composition series, we define $leng_A M = \infty$.

Lemma 1 Let $A$ be a ring. Let $M$ be a left $A$-module. Let $M_1 \supset M_2$ be $A$-submodules of $M$. Suppose $M_1/M_2$ is simple. Let $N$ be an $A$-submodule of M. Then $(M_1 + N)/(M_2 + N)$ is simple or $0$.

Proof: Let $f:M_1 \rightarrow (M_1 + N)/(M_2 + N)$ be the restriction of the canonical morphism $M_1 + N \rightarrow (M_1 + N)/(M_2 + N)$. Since $f(M_2) = 0$, $f$ induces a morphism $g:M_1/M_2 \rightarrow (M_1 + N)/(M_2 + N)$. Since $f$ is surjective, $g$ is also surjective. Since $M_1/M_2$ is simple, the assertion follows. QED

Lemma 2 Let $A$ be a ring. Let $M$ be a left $A$-module. Suppose $n = leng$ $M$ is finite. Let $N$ be an $A$-submodule of $M$. Then $leng$ $M/N \leq n$.

Proof: This follows immediately from Lemma 1.

Lemma 3 Let $A$ be a ring. Let $M$ be a left $A$-module. Suppose $n = leng$ $M$ is finite. Let $N$ be a proper $A$-sumodule of $M$. Then $leng$ $N \leq n - 1$.

Proof: We use induction on $n$. If $n = 0$, the assertion is trivial. Hence we assume $n \geq 1$. Let $M = M_0 \supset M_1 \supset ... \supset M_n = 0$ be a composition series. Since $leng$ $M_1 = n - 1$, if $N \subset M_1$, $leng$ $N \leq leng$ $M_1$ by the induction assumption. Hence we can assume that $M_1 \neq N + M_1$. Since $M_1 \subset N + M_1 \subset M$, $M = N + M_1$. Since $N/(N \cap M_1)$ is isomorphic to $M/M_1$, it is simple. Hence it suffices to prove that $leng$ $N \cap M_1 \leq n - 2$. Suppose $N \cap M_1 = M_1$. Then $N = M_1$ or $N = M$. Since $N$ is a proper submodule, $N = M_1$. But this contradicts our assumption. Hence $N \cap M_1 \neq M_1$. By the induction assumption, $leng$ $N \cap M_1 \leq n - 2$. QED

Lemma 4 Let $A$ be a ring. Let $M$ be a left $A$-module. Suppose $leng$ $M$ is finite. Let $N_1 \subset N_2$ be $A$-submodules of $M$. Suppose $leng$ $N_1 = leng$ $N_2$. Then $N_1 = N_2$.

Proof: $leng$ $N_2 = leng$ $N_1 + leng$ $N_2/N_1$. Hence $leng$ $N_2/N_1 = 0$. Hence $N_1 = N_2$. QED

Lemma 5 Let $A$ be a commutative ring. Suppose $leng$ $A$ is finite. Let $\Lambda$ be nonempty set of ideals of $A$. Then there exists a maximal element in $\Lambda$.

Proof: Let $r = sup$ {$leng$ $I$; $I \in \Lambda$}. Since $r$ is finite, there exists $I \in \Lambda$ such that $r = leng$ $I$. By Lemma 4, $I$ is a maximal element of $\Lambda$. QED

Lemma 6 Let $A$ be a commutative ring. Suppose $leng$ $A$ is finite. Let $\Lambda$ be nonempty set of ideals of $A$. Then there exist a minimal element in $\Lambda$.

Proof: Let $r = inf$ {$leng$ $I$; $I \in \Lambda$}. Since $r$ is finite, there exists $I \in \Lambda$ such that $r = leng$ $I$. By Lemma 4, $I$ is a minimal element of $\Lambda$. QED

Lemma 7 Let $A$ be a commutative ring. Suppose $leng$ $A$ is finite. Then every ideal of $A$ is finitely generated.

Proof: Let $I$ be an ideal of $A$. Let $\Lambda$ be the set of finitely generated ideals contained $I$. Since $0 \in \Lambda$, $\Lambda$ is not empty. By Lemma 5, there exitst a maximal element $I_0$ in $\Lambda$. Suppose $I \neq I_0$. There exists $x \in I - I_0$. Let $I_1$ be the ideal generated by $I_0 \cup$ {$x$}. Since $I_1 \in \Lambda$ and $I_1 \neq I_0$, this is a contradiction. Hence $I = I_0$. QED

Lemma 8 Let $A$ be a commutative ring. Suppose $leng$ $A$ is finite. Let $J$ be the intersection of all the maximal ideals of $A$. Then there exists an integer $k \geq 1$ such that $J^k = J^{2k}$.

Proof: Let $\Lambda$ = {$J^k; k = 1, 2, ...$}. By Lemma 6, there exists a minimal $J^k$ in $\Lambda$. Since $J^k = J^{k+1} = ...$, $J^k = J^{2k}$. QED

Proposition Let $A$ be a commutative ring. Suppose $leng$ $A$ is finite. Let $J$ be the intersection of all the maximal ideals of $A$. Then $J$ is nilpotent.

Proof: By Lemma 8, there exists an integer $k \geq 1$ such that $J^k = J^{2k}$. Let $I = J^k$. Then $I = I^2$. By Lemma 7, I is finitely generated. By Nakayama's lemma, there exists $r \in A$, such that $r \equiv 1$ (mod $I$) and $rI = 0$. We claim that r is invertible. If otherwise, $rA$ is a proper ideal. By Lemma 5, there exists a maximal ideal $P$ such that $rA \subset P$. Since $I \subset P$, $0 \equiv 1$ (mod $P$). This is a contradiction. Hence $r$ is invertible. Since $rI = 0$, $I = 0$ as desired. QED

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