A module $N$ is semisimple $\Longleftrightarrow$ $N$ has no proper essential submodules

Question

Problem: A module $N$ is semisimple $\Longleftrightarrow$ $N$ has no proper essential submodules.

My attempt: If $N$ is semisimple, then every submodule is a direct summand
of $N$ and so not essential unless equal to $N$. Conversely, any $K \subsetneq N$ has a complement $L \subsetneq N$. Then $K \bigoplus L \subseteq N$, so if $N$ has no proper essential submodules, $K$ is a direct summand of $N$.

Please check my proof. Thank all!

Answer 1

Conversely, any $K \subsetneq N$ has a complement $L \subsetneq N$. Then $K \bigoplus L \subseteq N$, so if $N$ has no proper essential submodules, $K$ is a direct summand of $N$.

Saying that "any $K$ has a complement" usually means $K\oplus L=N$, and that would be circular reasoning. What is true is this:

For any $K$, there exists a submodule $L$ such that $K\oplus L\subseteq_e N$.

I wouldn't call that "having a complement", but that is what your argument wants to employ. If that is what you meant, then your line of reasoning would be complete.