Lemma 46.8.3. Let $A$ be a ring. For any $A$-module $M$ set $M^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(M, \mathbf{Q}/\mathbf{Z})$.

1. For any $A$-module $M$ the $A$-module $M^\vee$ is pure injective.

2. An $A$-module $I$ is pure injective if and only if the map $I \to (I^\vee )^\vee$ splits.

3. For any module $M$ there exists a universally exact sequence $0 \to M \to I \to N \to 0$ with $I$ pure injective.

Proof. We will use the properties of the functor $M \mapsto M^\vee$ found in More on Algebra, Section 15.55 without further mention. Part (1) holds because $\mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee ) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N \otimes _ A M, \mathbf{Q}/\mathbf{Z})$ and because $\mathbf{Q}/\mathbf{Z}$ is injective in the category of abelian groups. Hence if $I \to (I^\vee )^\vee$ is split, then $I$ is pure injective. We claim that for any $A$-module $M$ the evaluation map $ev : M \to (M^\vee )^\vee$ is universally injective. To see this note that $ev^\vee : ((M^\vee )^\vee )^\vee \to M^\vee$ has a right inverse, namely $ev' : M^\vee \to ((M^\vee )^\vee )^\vee$. Then for any $A$-module $N$ applying the exact faithful functor ${}^\vee$ to the map $N \otimes _ A M \to N \otimes _ A (M^\vee )^\vee$ gives

$\mathop{\mathrm{Hom}}\nolimits _ A(N, ((M^\vee )^\vee )^\vee ) = \Big(N \otimes _ A (M^\vee )^\vee \Big)^\vee \to \Big(N \otimes _ A M\Big)^\vee = \mathop{\mathrm{Hom}}\nolimits _ A(N, M^\vee )$

which is surjective by the existence of the right inverse. The claim follows. The claim implies (3) and the necessity of the condition in (2). $\square$

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).