My question is from T.Y.Lam- A First Course in Noncommutative Rings.

For a ring $R$, if all left $R$-modules are semisimple, then all short exact sequences of left $R$-modules split.

A module $M$ is a semisimple module if every submodule $N$ of $M$ is a direct summand, that is there exists a submodule $N'$ such that $M=N \oplus N'$.

In order to prove the proposition above, let $M$ be a left $R$-module and let us consider $0 \rightarrow A \hookrightarrow B \rightarrow C \rightarrow0$ be a short exact sequence, where $A,B$ and $C$ are semisimple left $R$-modules. What should my strategy be in order to $B\cong A \oplus C$.


2 Answers 2


By the first isomorphism theorem, you always have that $B/A\cong C$.

Since $B$ is semisimple, there exists some submodule $C'$ such that $A\oplus C'=B$. By the second isomorphism theorem $$\frac{B}{A}=\frac{A\oplus C'}{A}\cong\frac{C'}{A\cap C'} =\frac{C'}{\{0\}}\cong C'$$.

So $C\cong C'$, and $A\oplus C\cong B$.

This is an ad-hoc elementary explanation, but ultimately it is going to be important for you to absorb the definition that Mustafa has given where you learn the equivalence of the statements "$0\to A\to B\to C\to 0$ splits" and "$A\oplus C\cong B$".


Theorem: Let be $R$ is a ring and $M,M_1,M_2$ are $R-$module. If $f:M_1\to M$, $g:M\to M_2$ are $R-$homomorphism and $0\to M_1\to M \to M_2\to 0 $ is an exact sequence ,then the following are equivalent:

$1)$ there exists a homomorphism $\alpha:M\to M_1$ where $\alpha \circ f=i_{M_1}$

$2)$ there exists a homomorphism $\beta:M_2\to M$ where $g \circ \beta=i_{M_2}$

$3)$ the exact sequence spilt and $M\cong Im(f) \oplus ker(\alpha) \cong Im(\beta) \oplus ker(g) \cong M_1 \oplus M_2 $


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