Lemma 46.3.19. Let $A = A_1 \times \ldots \times A_ n$ be a product of rings. An adequate functor over $A$ is the same thing as a sequence $F_1, \ldots , F_ n$ of adequate functors $F_ i$ over $A_ i$.

**Proof.**
This is true because an $A$-algebra $B$ is canonically a product $B_1 \times \ldots \times B_ n$ and the same thing holds for $A$-modules. Setting $F(B) = \coprod F_ i(B_ i)$ gives the correspondence. Details omitted.
$\square$

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