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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?

Community
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iam_agf
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  • 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.) – rschwieb Dec 12 '16 at 17:43

1 Answers1

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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.

Stahl
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    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? – iam_agf Dec 12 '16 at 09:27
  • @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. – Alex Wertheim Dec 12 '16 at 09:29
  • Thanks a lot. Now I can continue with the other part of the proof. (I have to wait one minute to accept the answer) – iam_agf Dec 12 '16 at 09:30