# Let $M_1$, $M_2$ be Artinian modules over $R$. Then $M_1\times M_2$ is Artinian.

Using exact sequences, it's fairly easy to prove the converse, but I can't figure out how to prove this statement.

Suppose we have a descending chain $N_1\supset N_2\supset\cdots$ of $R$-submodules of $M_1\times M_2$. Then if $N_i'=\pi_1(N_i)$, $N_i''=\pi_2(N_i)$ are the projections for all $i$, then I see that $\exists\,n_0$ s.t. $n\geq n_0 \implies N_n'=N_{n_0}', N_n''=N_{n_0}''$, so, looking at $N_{n_0}\supset N_{n_0 +1}\supset\cdots$, we have a descending chain of submodules with the same projection. Unfortunately I can't figure out how to get a contradiction from here.

I might be applying a topological intuition where it doesn't really apply.

• Prove that if you have an exact sequence $0 \to E' \to E \xrightarrow{\pi} E'' \to 0$ and submodules $F \subset G \subset E$ such that $F \cap E' = G \cap E'$ and $\pi(F) = \pi(G)$ then $F = G$. – Hoot Jun 27 '16 at 16:48
• Given a short exact sequence $0\to M' \to M \to M'' \to 0$ of $R$-modules, if $M'$ and $M''$ are Artin, then so is $M$. If you can prove this result, then you can apply this result to the short exact sequence $0 \to M_1 \to M_1 \oplus M_2 \to M_2 \to 0$ (Note that $M_1 \oplus M_2 \cong M_1 \times M_2$ – user348338 Jun 28 '16 at 2:49
• I mean I already got it. I'll post an answer then. – Monstrous Moonshine Jun 28 '16 at 3:40

## 1 Answer

Consider the short exact sequence $0\xrightarrow{}M_1\xrightarrow{}M_1\oplus M_2\xrightarrow{\pi}(M_1\oplus M_2)/M_1\xrightarrow{}0$, then $M_1$ and $(M_1\oplus M_2)/M_1$ are both Artinian.

Let $N_1\supset N_2\supset\cdots$ be the descending chain, let $N_i' = N_i\cap M_1$, let $N_i'' = (N_i+M_2)/M_2$, then it is easy to see that $\{N_i'\}_{i=1}^\infty$ and $\{N_i''\}_{i=1}^\infty$ are descending sequences of $R$-submodules of $M_1$ and $(M_1\oplus M_2)/M_1$, respectively. Fix $n_0$ s.t. $\{N_i'\}_{i=n_0}^\infty$ and $\{N_i''\}_{i=n_0}^\infty$ are constant. Then we use the following lemma:

Lemma (due to Hoot): If $0 \to E' \to E \xrightarrow{\pi} E'' \to 0$ is an exact sequence of $R$-modules, and $F\subset G\subset E$ are submodules s.t. $F \cap E' = G \cap E'$ and $\pi(F) = \pi(G)$, then $F=G$.

Proof: By exactness, we set $E'\subset E$ and $E''=E/E'$. Suppose $x\in G$, then $\overline{x}\in\pi(G)=\pi(F)=(F+E')/E'\implies x\in F+E'$. Let $x=x_1+x_2,\,x_1\in F\subset G,\,x_2\in E'$. Then $x_2=x-x_1\in E'\cap G=E'\cap F\implies x\in F$.

Now, $0\xrightarrow{}M_1\xrightarrow{}M_1\oplus M_2\xrightarrow{\pi}(M_1\oplus M_2)/M_1\xrightarrow{}0$ is exact, and for $i\geq n_0$, $N_i\subset N_{n_0}\subset M_1\oplus M_2$, $N_i\cap M_1 = N_i' = N_{n_0}' = N_{n_0}\cap M_1$, $\,\pi(N_i) = N_i'' = N_{n_0}'' = \pi(N_{n_0})$, so $N_i=N_{N_0}\,\forall\,i\geq n_0$. Then $\{N_i\}_{i=1}^\infty$ is eventually constant, and $M_1\oplus M_2$ is therefore Artinian.