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Show that the following are equivalent for a ring:

(1) any $R$-module is projective.

(2) any $R$-module is injective

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Hint: think in terms of short exact sequences. – Qiaochu Yuan Jan 11 '13 at 3:08
What definition of injective and projective do you know? Do you know any equivalent conditions? – Tobias Kildetoft Jan 11 '13 at 17:35
I don't think this equivalence has ever been pointed out to me, but I love the symmetry! I can easily come up with a proof that works in any abelian category with both enough projectives and enough injectives. Is the above true for all abelian categories? Or can someone come up with a counterexample? – Piotr Pstrągowski Jan 11 '13 at 17:53
For projective modules I know they are equivalent to $M$ is a summand of a free module.Or any short exact sequence $L \rightarrow M \rightarrow Q$, where Q is the projective module, then this is a split sequence. – Alex Jan 11 '13 at 18:58
and for injective modules $D$, I think that any short exact sequence $D \rightarrow M \rightarrow N$ is split. Besides these two, I know nothing more. – Alex Jan 11 '13 at 19:01
up vote 2 down vote accepted

Here is a proof that if $ R $ is a ring such that any $ R $-module is projective, then any $ R$-module is injective (it will hopefully then be clear how to do the other direction).

We wish to show that if $ A$ is some arbitrary $ R $-module, then any short exact sequence $0 \to A\to B \to C \to 0$ splits. But by assumption $ C $ is projective (since all $R$-modules are), which means that the sequence does indeed split as we wanted.

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