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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

Alex
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    Hint: think in terms of short exact sequences. – Qiaochu Yuan Jan 11 '13 at 03:08
  • 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
  • You mean put $D\rightarrow M \rightarrow Q$, where D is the injective module and Q is the projective one?@YACP I am not sure what I can derive from here.Would you please explain more explicitly? – Alex Jan 12 '13 at 15:47

1 Answers1

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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.

Tobias Kildetoft
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  • If $A$ is an arbitrary $R$-module, how do you know there exists such a short exact sequence? Also, don't we want to show that $\textit{every}$ short exact sequence splits? – user5826 Apr 23 '20 at 15:53
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    @AlJebr We don't care about existence. We only care what happens when it exists (obviously there are modules $B$ for which no such sequence exists). Also, I did show that every short exact sequence splits. – Tobias Kildetoft Apr 23 '20 at 16:22
  • But you've only shown that short exact sequences which start with $0 \to A \to \cdots$ splits. – user5826 Apr 23 '20 at 17:41
  • @Alj Yes, and $A$ is arbitrary. – Tobias Kildetoft Apr 23 '20 at 21:20