Question about a module

Question

Please help me to prove the following result using the definition below:

Let $S$ be a ring , $N$ a right $S$-module,

Show that $N$ is a submodule of $S$

Thanks in advance

Answer 1

Extend $\operatorname{Id_N}:N\rightarrow N$ to $M$ by $f$ and take $K=\operatorname{Ker}f$, let $x\in \operatorname{Ker} f\cap N$, $f(x)=\operatorname{Id}_N(x)=0=x$ implies that $\operatorname{Ker}f\cap N=0$.

Let $x\in M$, $x=x-f(x)+f(x)$, $f(x-f(x))=f(x)-f(f(x))$, since $f(x)\in N$, $f(f(x))=f(x))$, we deduce that $f(x-f(x))=0$.

Answer 2

Recall that a module $N$ is said to be injective if for each injection $f:N \to M$ we have that an arbitrary map $h:N \to Q$ can be extended to $M$ so that $\tilde{h}:M \to Q$ so that $\tilde{h} \circ f=h$.

In particular, consider the short exact sequence $$0 \to N \to M \to M/N \to 0$$

since the first arrow (let's call it $i:N \to M$) is an injection, consider $id:N \to N$. We thus have that for the injection $N \to M$, there exists a map $r$ so that $r \circ i=id$, so by splitting lemma we have that this is a split exact sequence.