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

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.

• The vector spaces are finite dimensional because they're Artinian as modules (being quotients of ideals of $A$), and in vector spaces fin dim = Artinian = Noetherian. Oct 15, 2021 at 1:01

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 $$A/\mathfrak{m}_j=F$$.
• Sorry, I forgot a hypothesis. The ring $A$ has to be Noetherian. Now, I edited this. Oct 15, 2021 at 2:15
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.