# Why $R$ is semisimple ring iff every $R$-module is semisimple?

I'm reading An introduction to homological algebra of Rotman, but the proposition 4.5 of the section 4.1 Semisimple rings states this:

The following conditions on a ring $$R$$ are equivalent.

• $$R$$ is semisimple.

• Every left (or right) $$R$$-module $$M$$ is a semisimple module.

• Every left (or right) $$R$$-module $$M$$ is injective.

• Every short exact sequence of left (or right) $$R$$-modules splits.

• Every left (or right) $$R$$-module $$M$$ is projective.

And the proof of the first point to the second doesn't look very clear. This is the proof the book has:

Since $$R$$ is semisimple, it is semisimple as a module over itself; hence, every free left $$R$$-module is a semisimple module. Now $$M$$ is a quotient of a free module, by Theorem $$2.35$$, and so Corollary $$4.2$$ gives $$M$$ semisimple.

I don't understand why the part in boldface is true. Can anyone explain to me the hence part?

• A direct sum of direct sums of simple modules is a direct sum of simple modules. I.e. A direct sum of semisimple modules is again semisimple. (or if your definition is that the module is the sum of its simple submodules, this is easily adapted.) Dec 12, 2016 at 17:43

By definition, a semisimple ring is a ring which is semisimple as a left module over itself. A free left $R$-module is a left module (isomorphic to a module) of the form $\bigoplus_{A} R$ for some index set $A$. Moreover, it is true that if $M_i$ is a collection of semisimple modules, then $\bigoplus_{i\in I}M_i$ is also semisimple. Putting all this together implies the result.
• So the useful fact for that part is that the direct sum of free semisimple $R$-modules (that are basically a sum of copies of $R$) is again a semisiple module? Dec 12, 2016 at 9:27
• @MonsieurGalois: yes. If $M_{i}, i \in I$ are semi-simple $R$-modules, then each $M_{i}$ is a direct sum of simple submodules. These simple submodules are still simple when thought of as submodules of $\bigoplus_{i \in I} M_{i}$, so $\bigoplus_{i \in I} M_{i}$ is semisimple as well. Dec 12, 2016 at 9:29