# Is one direction of the equivalence $I$ a direct factor of every $R$-module containing it iff I is injective trivial?

Let $R$ be a commutative ring with identity. I have been working through the various definitions of injective modules tyring to show every equivalence. I have gotten stuck on two parts one I think which requires the use of Baer's criteria and showing the one direction of a definition involving the following problem:

Let $I$ be an $R$-module.

Suppose that $I$ is a direct factor of every $R$-module containing it and suppose $V' \rightarrow V \rightarrow V''$ is an exact sequence of $R$-modules. How do you show $Hom_R(V'',I) \rightarrow Hom_R(V,I) \rightarrow Hom_R(V',I)$ is exact?

The opposite direction involved constructing the sequence $0 \rightarrow I \rightarrow M \rightarrow M/I \rightarrow 0$ if $I \subset M$ and showing it splits. I was wondering if we needed to do something so elaborate or if there was an easy way to prove the question.

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Yes, it's trivial. Embed $I$ into an injective $J$ and show that a direct summand of an injective is injective by showing that a direct summand of a short exact sequence is short exact. –  commenter Sep 26 '11 at 9:01
thanks it is really easy to think about it the way you just stated. –  user7980 Sep 26 '11 at 9:02
Embed $I$ into an injective $J$ and show that a direct summand of an injective is injective by showing that a direct summand of a short exact sequence is short exact.