Why we say that the Jacobson radical of a module M is the intersection of all submodules N of M such that M/N is semisimple? I know that the Jacobson radical of a module $M$ is the intersection of all submodules $N$ of $M$ such that $M/N$ is simple. However, why there is another definition saying that the Jacobson radical of a module $M$ is the intersection of all submodules $N$ of $M$ such that $M/N$ is semisimple?
How are these two definitions equivalent?
 A: For a sake of simplicity of notation, let $A= \{ N \le M : M/N \mbox{ is simple} \}$ and $B= \{ N \le M : M/N \mbox{ is semisimple} \}$. Obviously $A \subseteq B$.
Let's prove by double inclusion that
$$\bigcap_{N \in A}N = \bigcap_{N \in B}N$$
The inclusion "$\supseteq$" is a trivial consequence of $A \subseteq B$.
Conversely, for all $x \in \left( \bigcap_{N \in B}N \right)^c$ there exists some $N \in B$ such that $x \notin N$. Denote by $p: M \to M/N$ the canonical projection. We know that every semisimple module is a direct sum of simple modules, so write
$$M/N = \bigoplus_{i \in I} S_i$$
where $\{ S_i \}_{i \in I}$ is some family of simple modules indexed by a set $I$.
For all $i \in I$ consider the canonical projection $\pi_i : M/N \to S_i$ and the composition
$$\pi_i \circ p : M \longrightarrow S_i$$
Obviously $K_i=\ker (\pi_i \circ p)$ is a submodule of $M$ such that $M/K_i \cong S_i$ is simple: in other words $K_i \in A$. You can check that
$$N=\ker p = \bigcap_{i \in I} K_i$$
Since $x \notin N$, there exists some $K_i$ not containing $x$. So that
$$x \notin \bigcap_{N \in A}N$$
This proves the opposite inclusion.
