Lemma 15.91.20. Let $A$ be a ring and let $I \subset A$ be an ideal which can be generated by $r$ elements. Then derived completion has finite cohomological dimension:

1. Let $K \to L$ be a morphism in $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$.

2. Let $K \to L$ be a morphism of $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \leq -1$ and injective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \leq -r - 1$ and injective for $i = -r$.

Proof. Say $I$ is generated by $f_1, \ldots , f_ r$. For any $K \in D(A)$ by Lemma 15.91.18 we have $K^\wedge = R\mathop{\mathrm{lim}}\nolimits K \otimes _ A^\mathbf {L} K_ n$ where $K_ n$ is the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ and hence we obtain a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(K \otimes _ A^\mathbf {L} K_ n) \to H^ i(K^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^ i(K \otimes _ A^\mathbf {L} K_ n) \to 0$

by Lemma 15.87.4.

Proof of (1). Pick a distinguished triangle $K \to L \to C \to K$. Then $H^ i(C) = 0$ for $i \geq 0$. Since $K_ n$ is sitting in degrees $\leq 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i \geq 0$ and that $H^{-1}(C \otimes _ A^\mathbf {L} K_ n) = H^{-1}(C) \otimes _ A A/(f_1^ n, \ldots , f_ r^ n)$ is a system with surjective transition maps. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i \geq 0$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge $ we get (1).

Proof of (2). Pick a distinguished triangle $K \to L \to C \to K$. Then $H^ i(C) = 0$ for $i < 0$. Since $K_ n$ is sitting in degrees $-r, \ldots , 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i < -r$. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i < r$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge $ we get (2). $\square$

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