Lemma 46.3.18. Let $A \to A'$ be a ring map. If $F'$ is an adequate functor on $\textit{Alg}_{A'}$, then the module-valued functor $F : B \mapsto F'(A' \otimes _ A B)$ on $\textit{Alg}_ A$ is adequate too.

Proof. Choose an exact sequence $0 \to F' \to \underline{M'} \to \underline{N'}$. Then

\begin{align*} F(B) & = F'(A' \otimes _ A B) \\ & = \mathop{\mathrm{Ker}}(M' \otimes _{A'} ( A' \otimes _ A B) \to N' \otimes _{A'} (A' \otimes _ A B)) \\ & = \mathop{\mathrm{Ker}}(M' \otimes _ A B \to N' \otimes _ A B) \end{align*}

Thus $F$ is the kernel of $\underline{M} \to \underline{N}$ where $M = M'$ and $N = N'$ viewed as $A$-modules. $\square$

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