Lemma 46.9.1. Let $A$ be a ring. Let $\mathcal{A}$ be the category of adequate functors on $\textit{Alg}_ A$. The injective objects of $\mathcal{A}$ are exactly the functors $\underline{I}$ where $I$ is a pure injective $A$-module.

**Proof.**
Let $I$ be an injective object of $\mathcal{A}$. Choose an embedding $I \to \underline{M}$ for some $A$-module $M$. As $I$ is injective we see that $\underline{M} = I \oplus F$ for some module-valued functor $F$. Then $M = I(A) \oplus F(A)$ and it follows that $I = \underline{I(A)}$. Thus we see that any injective object is of the form $\underline{I}$ for some $A$-module $I$. It is clear that the module $I$ has to be pure injective since any universally exact sequence $0 \to M \to N \to L \to 0$ gives rise to an exact sequence $0 \to \underline{M} \to \underline{N} \to \underline{L} \to 0$ of $\mathcal{A}$.

Finally, suppose that $I$ is a pure injective $A$-module. Choose an embedding $\underline{I} \to J$ into an injective object of $\mathcal{A}$ (see Lemma 46.4.2). We have seen above that $J = \underline{I'}$ for some $A$-module $I'$ which is pure injective. As $\underline{I} \to \underline{I'}$ is injective the map $I \to I'$ is universally injective. By assumption on $I$ it splits. Hence $\underline{I}$ is a summand of $J = \underline{I'}$ whence an injective object of the category $\mathcal{A}$. $\square$

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