# How to prove $E$ is an injective module?

I was stuck with the seemingly simple homework problem:

A $R$-module is injective if and only if every exact sequence $$0\rightarrow E\rightarrow B\rightarrow R/I\rightarrow 0$$ splits.Here $I$ is an ideal of $R$.

The only if direction is straightforward since if $E$ is injective, then every exact sequence of former type must splits. But I do not know how to prove $E$ is injective given the above condition. The natural direction is to use Baer's criterion. However the above exact sequence seems to be quite difficult to apply this criterion. So I am stuck.

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Suppose $I\to R$ is the inclusion map from an left ideal of $R$ and that $f:I\to E$ is a map. We have a diagram

0  --> I --> R --> R/I --> 0
|
|
V
E


So we can construct the push-out, to get a diagram with exact rows

0 --> I --> R --> R/I --> 0
|     |      |
|     |      |
V     V      V
0 --> E --> ? --> R/I ---> 0


where ? is some module. Now use the hypothesis on the bottom row.

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Thank you! This is really clear. – Bombyx mori Dec 13 '12 at 21:47
Can I ask how do I get the arrow from ? to R/I? – Bombyx mori Dec 13 '12 at 21:58
It comes out from the construction of push-outs. – Mariano Suárez-Alvarez Dec 13 '12 at 22:14
I see. Thanks!! – Bombyx mori Dec 14 '12 at 0:38