If $\mathfrak{m}_1\mathfrak{m}_2\cdots \mathfrak{m}_n=0$ then $A$ is an Artinian ring.

Question

I want to prove the following:

Let $A$ be a Noetherian commutative ring with unity, and suppose that zero ideal is a product $\mathfrak{m}_1\mathfrak{m}_2\ldots \mathfrak{m}_n$ of maximal ideals in $A$. Then $A$ is Artinian.

The first step in this theorem is to prove that the quotients $\displaystyle\prod_{i=1}^{j-1}\mathfrak{m}_j/\prod_{i=1}^{j}\mathfrak{m}_j$ are finite-dimensional vector spaces over $A/\mathfrak{m}_j$. Here, there is a natural scalar product, and it is well defined, but first I am not sure why they are finite-dimensional.

The following steps are easy. The proof follows inductively using a short exact sequence. However, I am studying another proof with the same first step.

I have to prove two things:

1. I can refine the sequence $A\supset \mathfrak{m_1}\supset \mathfrak{m}_1\mathfrak{m_2}\supset \cdots\supset \mathfrak{m}_1\mathfrak{m_2}\cdots\mathfrak{m}_n= 0$ to a composition series $A\supset B_1\supset\cdots\supset B_r=0$.
2. If $N\subset A$ is a submodule and $N_i=N\cap B_i$ then some subset of such $N_i$'s is a composition series of $N$. So, any chain of submodules of $A$ has finite length.

I do not know how to prove this. I will appreciate your ideas for both propositions.

Answer 1

Use the Chinese remainder theorem, $A = A/(\prod \mathfrak m_i)\simeq A/(\mathfrak m_1^{n_1})\times\cdots\times A/(\mathfrak m_i^{n_i})$. It's enough to show $A/\mathfrak m_j^{n_j}$ is Artinian. (if you assume $\mathfrak m_i$ are distinct, then $A$ is just a finite product of fields, hence Artinian, without assuming it's already Noetherian.) This follows from $\mathfrak m_j + \mathfrak m_j^{n_j}$ is the only prime ideal, as it consists of only nilpotent elements, so it has Krull dimension $0$.

Note that this is false if $A$ is not assumed to be Noetherian in the first place. Let $A=F[x_1, x_2, \cdots]/\mathfrak m^2$ where $\mathfrak m=(x_1, x_2, \cdots)$. Then $\overline{\mathfrak m}^2=0$, but $A$ is not Artinian: $(\bar x_1, \bar x_2, \bar x_3, \cdots)\supsetneq (\bar x_2, \bar x_3, \cdots)\supsetneq(\bar x_3, \cdots)\supsetneq \cdots$ form an infinite chain of decreasing ideals of $A$. In particular, $\prod_{i=1}^{j-1}\mathfrak{m}_j/\prod_{i=1}^{j}\mathfrak{m}_j$ is not finite-dimensional vector spaces over $A/\mathfrak{m}_j=F$.

Answer 2

We shall not pay attention on refining the chain to be composition chain. We only need to use the factors in the chain. We can prove as follows:

Suppose $A$ is Noetherian. Then it is clear that $A/\mathfrak{m}_1...\mathfrak{m}_n$ is Noetherian as for $A$-module. Consider the following exact sequences: $$0\rightarrow \mathfrak{m}_1...\mathfrak{m}_{n-1}/\mathfrak{m}_1...\mathfrak{m}_n \rightarrow A/\mathfrak{m}_1...\mathfrak{m}_n \rightarrow A/\mathfrak{m}_1...\mathfrak{m}_{n-1}\rightarrow 0$$ We can deduce that both $\mathfrak{m}_1...\mathfrak{m}_{n-1}/\mathfrak{m}_1...\mathfrak{m}_n$ and $A/\mathfrak{m}_1...\mathfrak{m}_{n-1}$ are Noetherian as for $A$-module. Repeating using this methods, one can find that for any $1\le i\le n-1$, $\mathfrak{m}_1...\mathfrak{m}_i/\mathfrak{m}_1...\mathfrak{m}_{i+1}$ is Noetherian as for $A$-module, hence it is also Noetherian as for $A/\mathfrak{m}_{i+1}$-vector space (If $IM=0$, then $A$-module $M$ is equivalent to $A/I$-module $M$.). Hence $\mathfrak{m}_1...\mathfrak{m}_i/\mathfrak{m}_1...\mathfrak{m}_{i+1}$ is Artinian as for $A/\mathfrak{m}_{i+1}$-vecor space, so does as for $A$-module. Inductively back to deduce that $A/\mathfrak{m}_1...\mathfrak{m}_i$ is Artinian as for $A$-module for all $i$, hence $A$ is Artinian.