Lemma 15.55.8. Let $R$ be a ring. For every $R$-module $M$ the $R$-module $J(M)$ is injective.

Proof. Note that $J(M) \cong \prod _{\varphi \in M^\vee } R^\vee$ as an $R$-module. As the product of injective modules is injective, it suffices to show that $R^\vee$ is injective. For this we use that

$\mathop{\mathrm{Hom}}\nolimits _ R(N, R^\vee ) = \mathop{\mathrm{Hom}}\nolimits _ R(N, \mathop{\mathrm{Hom}}\nolimits _{\mathbf{Z}}(R, \mathbf{Q}/\mathbf{Z})) = N^\vee$

and the fact that $(-)^\vee$ is an exact functor by Lemma 15.55.6. $\square$

Comment #113 by Kiran Kedlaya on

In the first sentence, I think $m \in M$ should be $m \in M^\vee$.

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).