Lemma 46.3.3. Let $A$ be a ring. Let $F$ be an adequate functor on $\textit{Alg}_ A$. If $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of $A$-algebras, then $F(B) = \mathop{\mathrm{colim}}\nolimits F(B_ i)$.

**Proof.**
This holds because for any $A$-module $M$ we have $M \otimes _ A B = \mathop{\mathrm{colim}}\nolimits M \otimes _ A B_ i$ (see Algebra, Lemma 10.12.9) and because filtered colimits commute with exact sequences, see Algebra, Lemma 10.8.8.
$\square$

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