Lemma 15.55.3. Let $R$ be a ring. Let $J$ be an $R$-module. The following are equivalent

1. $J$ is injective,

2. $\mathop{\mathrm{Ext}}\nolimits ^1_ R(M, J) = 0$ for every $R$-module $M$.

Proof. Let $0 \to M'' \to M' \to M \to 0$ be a short exact sequence of $R$-modules. Consider the long exact sequence

$\begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, J) \to \mathop{\mathrm{Hom}}\nolimits _ R(M', J) \to \mathop{\mathrm{Hom}}\nolimits _ R(M'', J) \\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, J) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M', J) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M'', J) \to \ldots \end{matrix}$

of Algebra, Lemma 10.71.7. Thus we see that (2) implies (1). Conversely, if $J$ is injective then the $\mathop{\mathrm{Ext}}\nolimits$-group is zero by Homology, Lemma 12.27.2 and Lemma 15.55.2. $\square$

There are also:

• 4 comment(s) on Section 15.55: Injective modules

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).