# Every right ideal of R is injective

How can I prove the following statement:

Every right ideal of $R$ is injective iff $R$ is semisimple.

It's a strange statement. If true, only from the condition satisfied by the ideals we can conclude a property valid for all modules over the ring.

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What's so strange about going from ideals to modules? If $R$ has no nontrivial right modules, then $R$ is a field, so every module is free. Ideals tell you something about $R$ which tell you something about modules over $R$. –  Qiaochu Yuan Jun 20 '12 at 1:23
Further, why are you considering ideals and modules to be different things? Ideals are just submodules of the regular module (ie the ring regarded as a module in the natural way). –  KReiser Jun 20 '12 at 1:26
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## 1 Answer

Recall that one of several equivalent definitions of a module $E$ being injective is that every short exact sequence starting with $E$ splits (exercise: prove this is equivalent to all other definitions you know).

Recall further the definition of a semisimple ring- a ring is semisimple iff all of its modules are a BLANK of BLANKS. In particular, every short exact sequence behaves like BLANK.

Combining these two statements, you should be able to prove what you want.

NOTE: Blanks have been used to help you find the answer on your own!

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