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.
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1$\begingroup$ It splits because one of the other terms is $0$. $\endgroup$– Tobias KildetoftCommented Apr 7, 2019 at 19:00
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1$\begingroup$ Hint kernel of a map from a simple module is either the module itself or trivial $\endgroup$– user6Commented Apr 7, 2019 at 19:01
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1 Answer
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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.