Nilpotency of the Jacobson radical of an Artinian ring without Axiom of Choice

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?

Solution 1:

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