## 22.17 Injective modules over algebras

In this section we discuss injective modules over algebras analogous to More on Algebra, Section 15.55. This section should probably be moved somewhere else.

Let $R$ be a ring and let $A$ be an $R$-algebra, see Section 22.2 for our conventions. For a right $A$-module $M$ we set

which we think of as a left $A$-module by the multiplication $(a f)(x) = f(xa)$. Namely, $((ab)f)(x) = f(xab) = (bf)(xa) = (a(bf))(x)$. Conversely, if $M$ is a left $A$-module, then $M^\vee $ is a right $A$-module. Since $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group (More on Algebra, Lemma 15.54.1), the functor $M \mapsto M^\vee $ is exact (More on Algebra, Lemma 15.55.6). Moreover, the evaluation map $M \to (M^\vee )^\vee $ is injective for all modules $M$ (More on Algebra, Lemma 15.55.7).

We claim that $A^\vee $ is an injective right $A$-module. Namely, given a right $A$-module $N$ we have

and we conclude because the functor $N \mapsto N^\vee $ is exact. The second equality holds because

by Algebra, Lemma 10.12.8. Inside this module $A$-linearity exactly picks out the bilinear maps $\varphi : N \times A \to \mathbf{Q}/\mathbf{Z}$ which have the same value on $x \otimes a$ and $xa \otimes 1$, i.e., come from elements of $N^\vee $.

Finally, for every right $A$-module $M$ we can choose a surjection $\bigoplus _{i \in I} A \to M^\vee $ to get an injection $M \to (M^\vee )^\vee \to \prod _{i \in I} A^\vee $.

We conclude

the category of $A$-modules has enough injectives,

$A^\vee $ is an injective $A$-module, and

every $A$-module injects into a product of copies of $A^\vee $.

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