Lemma 15.97.5. Let $A$ be a ring and $I \subset A$ an ideal. Suppose given $K_ n \in D(A/I^ n)$ and maps $K_{n + 1} \to K_ n$ in $D(A/I^{n + 1})$. If

1. $A$ is Noetherian,

2. $K_1$ is bounded above, and

3. the maps induce isomorphisms $K_{n + 1} \otimes _{A/I^{n + 1}}^\mathbf {L} A/I^ n \to K_ n$,

then $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is a derived complete object of $D^-(A)$ and $K \otimes _ A^\mathbf {L} A/I^ n \to K_ n$ is an isomorphism for all $n$.

Proof. The object $K$ of $D(A)$ is derived complete by Lemma 15.91.14.

Suppose that $H^ i(K_1) = 0$ for $i > b$. Then we can find a complex of free $A/I$-modules $P_1^\bullet$ representing $K_1$ with $P_1^ i = 0$ for $i > b$. By Lemma 15.75.3 we can, by induction on $n > 1$, find complexes $P_ n^\bullet$ of free $A/I^ n$-modules representing $K_ n$ and maps $P_ n^\bullet \to P_{n - 1}^\bullet$ representing the maps $K_ n \to K_{n - 1}$ inducing isomorphisms (!) of complexes $P_ n^\bullet /I^{n - 1}P_ n^\bullet \to P_{n - 1}^\bullet$.

Thus we have arrived at the situation where $R\mathop{\mathrm{lim}}\nolimits K_ n$ is represented by $P^\bullet = \mathop{\mathrm{lim}}\nolimits P_ n^\bullet$, see Lemma 15.87.1 and Remark 15.87.6. The complexes $P_ n^\bullet$ are uniformly bounded above complexes of flat $A/I^ n$-modules and the transition maps are termwise surjective. Then $P^\bullet$ is a bounded above complex of flat $A$-modules by Lemma 15.27.4. It follows that $K \otimes _ A^\mathbf {L} A/I^ t$ is represented by $P^\bullet \otimes _ A A/I^ t$. We have $P^\bullet \otimes _ A A/I^ t = \mathop{\mathrm{lim}}\nolimits P_ n^\bullet \otimes _ A A/I^ t$ termwise by Lemma 15.27.4. The transition maps $P_{n + 1}^\bullet \otimes _ A A/I^ t \to P_ n^\bullet \otimes _ A A/I^ t$ are isomorphisms for $n \geq t$ by our choice of $P_ n^\bullet$, hence we have $\mathop{\mathrm{lim}}\nolimits P_ n^\bullet \otimes _ A A/I^ t = P_ t^\bullet \otimes _ A A/I^ t = P_ t^\bullet$. Since $P_ t^\bullet$ represents $K_ t$, we see that $K \otimes _ A^\mathbf {L} A/I^ t \to K_ t$ is an isomorphism. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).