If $\text{Ext}_R^1(A,I) = 0$ for all $A\in ob(_R\text{Mod})$, then $I\in ob(_R\text{Mod})$ is injective.

Question

Let $\text{Ext}^1(A,I)=0$ for all $A\in ob(_R\text{Mod})$, then $I\in ob(_R\text{Mod})$ is injective.

I got stuck by this problem. Any ideas?

Answer 1

Use the long exact sequence of $\operatorname{Ext}$: if $0 \to A \to B \to C \to 0$ is exact, then it induces a long exact sequence (everything is over $R$): $$0 \to \hom(C,I) \to \hom(B,I) \to \hom(A,I) \to \operatorname{Ext}^1(C,I) \to \operatorname{Ext}^1(B,I) \to \operatorname{Ext}(A,I) \to \dots$$

Now for an injection $i : A \to B$, let $C = \operatorname{coker}(i)$ so that you have a short exact sequence $0 \to A \to B \to C \to 0$. Then the long exact sequence above and the fact that $\operatorname{Ext}^1(C,I) = 0$ shows that $\hom(B,I) \to \hom(A,I)$ (induced by $i$) is surjective, in other words for any morphism $f : A \to I$ there exists some $g : B \to I$ such that $g \circ i = f$. This is exactly the definition of an injective module.

Answer 2

By definition $I$ is injective if the functor $Hom(-,I)$ is exact. To show this, let $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ be an exact sequence. The derived functors long exact sequence shows that $0\rightarrow Hom(M'',I)\rightarrow Hom(M,I)\rightarrow Hom(M',I)\rightarrow 0$ is exact.