Submodules of a product of simple modules

Question

Let $R$ be a ring and $A,B$ two simple $R$-modules. I would like to prove the following:

If $A$ and $B$ are not isomorphic, then the only submodules of $A \times B$ are $\{0\} \times \{0\},A \times \{0\}, \{0\} \times B$ and $A \times B$.

The result is wrong in general (see this question).

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Let $P \leq_R A \times B$. I know that if $π : A \times B \to A$ and $π' : A \times B \to B$ are the canonical projections, then $π(P)$ is either $0$ or $A$, and $π'(P)$ is either $0$ or $B$.

If one of them is $0$, I was able to conclude. But how can I prove that $P=A \times B$ assuming that $π(P)=A$ and $π'(P)=B$? This may be easy, but I didn't really know how to use the fact that $A$ and $B$ are not isomorphic (as $R$-modules). Should I proceed by contradiction?

There are on the site some questions on submodules of semi-simple modules, but I didn't found what I was looking for.

Thank you for your help!

Answer 1

Since $(A \times B)/A \cong B$ is simple, there are no modules properly contained between $A$ and $A \times B$, and the same holds for $B$ and $A \times B$.

Suppose $C$ is a submodule different from the ones you named.

Since $C$ is different from $A$ and $A \times B$, it cannot contain $A$. For the same reason, it cannot contain $B$.

Since $A$ and $B$ are simple, we get $C \cap A = \{ 0 \} = C \cap B$.

Since $C \ne \{ 0 \}$, so that $A + C$ properly contains $A$, we have $A + C = A \times B$, and similarly $C + B = A \times B$.

It follows $$ B \cong (A \times B)/A = (A + C)/A \cong C, \qquad A \cong (A \times B)/B = (C + B)/B \cong C, $$ so that $A \cong B$, against the assumption.

Answer 2

I'll use $\pi_A$ and $\pi_B$ for your $\pi$ and $\pi'$. Let $i: P \to A\oplus B$ be the inclusion map. Let $\hat B = \{(0,b): b \in B\}\subset A\oplus B$ and $\hat A$ be defined similarly.

Under your hypothesis, $\pi_A \circ i : P \to A$ is onto. Its kernel is $P \cap \hat B$. If the kernel is non-zero, then $\hat B \subset P$ and it follows $P=A\oplus B$. If the kernel is zero, $P \cong A$.

Now repeat the argument with $\pi_B \circ i : P \to B$. You get either $\hat A \subset P$, so $P=A\oplus B$ as before, or $P \cong B$. If $P \neq A\oplus B$ then it is iso to both $A$ and $B$ simultaneously; impossible.

Answer 3

Suppose $\{0\}\subsetneq P\subsetneq A\times B$; prove that either $P\oplus(A\times\{0\})$ or $P\oplus(\{0\}\times B)$ is the whole of $A\times B$. Without loss of generality, we can assume the former case holds. Therefore $P\cong B$ is simple.

Let $t=(x,y)\in P$, with $x\ne0$ and $y\ne0$. Then $tR=P$ and the map $tR\to A$ (restriction of the projection) is surjective, because it contains $x\ne0$. Since $tR\cong B$, we conclude that $A\cong B$.