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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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