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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