B is an injective R-module iff $ Ext^{1}_{R}(A,B) $ vanishes for all A. I know how to prove this statement from left to right, but don't know the opposite direction. Please help.
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Use the long exact sequence for $\operatorname{Ext}$ groups. – Najib Idrissi May 16 '14 at 13:50
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$B$ is injective if and only if the functor $Hom_R(-,B)$ is exact. Take a short exact sequence:
$$ 0 \rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0 $$
And apply the functor $Hom_R(-,B)$ to get an exact sequence:
$$ 0 \rightarrow Hom_R(A'',B) \rightarrow Hom_R(A,B) \rightarrow Hom_R(A',B) \rightarrow Ext_R^1(A'',B) \rightarrow Ext_R^1 (A,B) \cdots $$
By hypothesis $Ext_R^1(A'',B)=0$, then you have an exact sequence:
$$ 0 \rightarrow Hom_R(A'',B) \rightarrow Hom_R(A,B) \rightarrow Hom_R(A',B) \rightarrow 0 $$
Then $B$ is injective.
If you are using the diagram definition of injective, try to prove what I said on the first line.
Marco Armenta
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In fact,in my book, it defines $Ext^{i}{R}(A,B)=R^{i}Hom{R}(A,-)(B) $ After learning some more sections, I found that $ R^{ \ast }Hom(-,B)(A) \cong R^{*}Hom(A,-)(B) $ Then it is clear. Thank you! – user150245 May 17 '14 at 08:26