A characterization of semisimple module related to anihilators

Question

I'm trying to solve the following question:

Let $R$ be a ring and let $M$ be an $R$-module. Prove that $M$ is semisimple iff $ann(m)$ is the intersection of finitely many maximal left ideals of $R$ for all $m\in M\backslash\{0\}$. Hint: Try it first for $M$ cyclic.

For the $\Rightarrow$ part, I argued in this way:

Since $M$ is semisimiple, then $M=\oplus_{i\in I}M_i$ where $\{M_i\}$ is a family of simple modules. Choose $m\in M\backslash\{0\}$ and suppose $m\in M_k, k\in I$. As $M_k$ is simple, $M_k\simeq R/ann(m)$. So, $ann(m)$ is a maximal left ideal of $R$. Thus, $ann(m)$ is the intersection (created with only element) of maximal left ideals of $R$.

But I couldn't go that long in the $\Leftarrow$ part and I'd like to seek some hint. Thanks in advance!

Answer 1

Let $m$ be an arbitrary nonzero element of $M$. Let $ann(m)= I_1\cap \cdots \cap I_r$ with $I_r$ maximal left ideal of $R$. Then the canonical map $$ R\cdot m \cong R/ ann(m) \to R/I_1 \times \cdots R/I_r=:N $$ is an injective homomorphism of left $R$-modules. Since $N$ is clearly semisimple, $R\cdot m$ is semisimple as a left $R$-module. Thus $M$ is a sum of semisimple submodules, and so $M$ is itself semisimple.

Answer 2

To complete your argument of $\implies$, this is what you need:

Let $m\in M$ be nonzero. Then in $\oplus_{i\in I}M_i$, $m$ is only nonzero on finitely many coordinates, say in the finite subset $F\subseteq I$.

For each term $0\neq m_i\in M_i$, you argue as above that $ann(m_i)$ is a maximal left ideal. Then we contend that $ann(m)=\bigcap_{i\in F} ann(m_i)$.

The containment $\supseteq$ is obvious, and the other containment $\subseteq$ is almost as obvious due to the definition of the direct sum:

If $rm=0$, then $r(\sum_{i\in F} m_i)=\sum_{i\in F} rm_i=0$, but by the uniqueness of representation of elements in the direct sum, this means $rm_i=0$ for all $i\in F$.

G.S. Zhou has already handled the other direction nicely and I won't bother making another solution to it.