# Every short exact sequence with a simple module splits

More concretely, if $$R$$ is a ring, and $$M$$ is a simple $$R$$-module, I want to show that any short exact sequence $$0 \rightarrow L \rightarrow M \rightarrow N \rightarrow0$$ splits. To this end, I have already proved the fact that for any simple $$R$$-module $$M$$, there exists a maximal left ideal $$I \triangleleft R$$ such that $$M \cong_R R/I$$. Now my question is this: could I in some way identify $$L \oplus N$$ with $$R/I$$? In that way, I could compose the isomorphism $$\phi : M \rightarrow R/I$$ with some isomorphism $$h : L \oplus N \rightarrow R/I$$ to split the sequence. I'm a bit worried that the short-exactness doesn't seem te get into play here, though.

• It splits because one of the other terms is $0$. – Tobias Kildetoft Apr 7 '19 at 19:00
• Hint kernel of a map from a simple module is either the module itself or trivial – Ignorant Mathematician Apr 7 '19 at 19:01

Let $$0 \rightarrow L \rightarrow M \rightarrow N \rightarrow0$$ be any short exact sequence. Note that im$$f$$ is a submodule of $$M$$. As $$f$$ is injective, we have $$L \cong \text{im}f \in \{\{0_M\}, M\}$$ (since these are the only submodules of $$M$$). Consider the map $$r_1 : M \rightarrow L$$, $$m \mapsto 0_L$$. If $$L \cong \{0_M\}$$ (i.e. $$L = \{0_L\}$$), the for all $$l \in L$$ we have $$r_1 \circ f(l) = r_1(f(l)) = 0_L = l$$, so $$r_1 \circ f$$ = id$$_L$$ and $$r_1$$ is a retraction of $$f$$. Hence the sequence splits. If, on the other hand, im $$f = M$$, $$f$$ is bijective, so $$r_2 := f^{-1}$$ exists. Then $$r_2 \circ f =$$ id$$_L$$, so the sequence splits again.