$N/K$ is direct summand of $M/K$ implies $N$ is direct summand of $M$

Question

Let $K\subset N\subset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.

It is easy to show that

If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.

I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.

Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/K\oplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/K\oplus N'/K=M/K$.

I want to say something like $N\oplus N'=M$ which is not true since $N'\cap N\supset K\neq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=N\oplus B$ but $B$ is not a submodule.

Answer 1

The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=\langle e_3\rangle \subset N=\langle e_2, e_3\rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

Answer 2

Taking $K = N$, it would follow that $N$ is always a direct summand of $M$ - which is false in general (unless $R$ is semisimple, i.e. absolutely projective, i.e. all $R$-modules are projective; for example a field).

Answer 3

In the second part, it is still assumed that $K$ is a direct summand of $M$, hence of $N$ by the first part. (Otherwise the statement would obviously be false, as shown by the two answers of 2018.)

So, $N=K\oplus L$ for some submodule $L$. This, together with your $N/K\oplus N'/K=M/K$, easily implies $L\oplus N'=M$.