In this section we discuss what happens when we have a “formal deformation” of a complex and we take its limit. We will consider two cases
we have a limit $A = \mathop{\mathrm{lim}}\nolimits A_ n$ of an inverse system of rings whose transition maps are surjective with locally nilpotent kernels and objects $K_ n \in D(A_ n)$ which fit together in the sense that $K_ n = K_{n + 1} \otimes _{A_{n + 1}}^\mathbf {L} A_ n$, or
we have a ring $A$, an ideal $I$, and objects $K_ n \in D(A/I^ n)$ which fit together in the sense that $K_ n = K_{n + 1} \otimes _{A/I^{n + 1}}^\mathbf {L} A/I^ n$.
Under additional hypotheses we can show that $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ reproduces the system in the sense that $K_ n = K \otimes _ A^\mathbf {L} A_ n$ or $K_ n = K \otimes _ A^\mathbf {L} A/I^ n$.
Lemma 15.91.1. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $K_ n \in D(A_ n)$ and maps $K_{n + 1} \to K_ n$ in $D(A_{n + 1})$. Assume
the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,
$K_1$ is pseudo-coherent, and
the maps induce isomorphisms $K_{n + 1} \otimes _{A_{n + 1}}^\mathbf {L} A_ n \to K_ n$.
Then $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is a pseudo-coherent object of $D(A)$ and $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$.
Proof. By assumption we can find a bounded above complex of finite free $A_1$-modules $P_1^\bullet $ representing $K_1$, see Definition 15.63.1. By Lemma 15.72.4 we can, by induction on $n > 1$, find complexes $P_ n^\bullet $ of finite free $A_ n$-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 \otimes _{A_ n} A_{n - 1} \to P_{n - 1}^\bullet $. Thus $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is represented by $P^\bullet = \mathop{\mathrm{lim}}\nolimits P_ n^\bullet $, see Lemma 15.82.1 and Remark 15.82.6. Since $P_ n^ i$ is a finite free $A_ n$-module for each $n$ and $A = \mathop{\mathrm{lim}}\nolimits A_ n$ we see that $P^ i$ is finite free of the same rank as $P_1^ i$ for each $i$. This means that $K$ is pseudo-coherent. It also follows that $K \otimes _ A^\mathbf {L} A_ n$ is represented by $P^\bullet \otimes _ A A_ n = P_ n^\bullet $ which proves the final assertion. $\square$
Lemma 15.91.2. 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 $D(A/I^{n + 1})$. Assume
$A$ is $I$-adically complete,
$K_1$ is pseudo-coherent, and
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 pseudo-coherent, 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. We already know that $K$ is pseudo-coherent and that $K \otimes _ A^\mathbf {L} A/I^ n \to K_ n$ is an isomorphism for all $n$, see Lemma 15.91.1. To finish the proof it suffices to show that $K$ is derived complete. This follows from Lemma 15.86.8. $\square$
Lemma 15.91.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $K_ n \in D(A_ n)$ and maps $K_{n + 1} \to K_ n$ in $D(A_{n + 1})$. Assume
the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,
$K_1$ is a perfect object, and
the maps induce isomorphisms $K_{n + 1} \otimes _{A_{n + 1}}^\mathbf {L} A_ n \to K_ n$.
Then $K = R\mathop{\mathrm{lim}}\nolimits K_ n$ is a perfect object of $D(A)$ and $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$.
Proof. We already know that $K$ is pseudo-coherent and that $K \otimes _ A^\mathbf {L} A_ n \to K_ n$ is an isomorphism for all $n$ by Lemma 15.91.1. Thus it suffices to show that $H^ i(K \otimes _ A^\mathbf {L} \kappa ) = 0$ for $i \ll 0$ and every surjective map $A \to \kappa $ whose kernel is a maximal ideal $\mathfrak m$, see Lemma 15.73.7. Any element of $A$ which maps to a unit in $A_1$ is a unit in $A$ by Algebra, Lemma 10.31.4 and hence $\mathop{\mathrm{Ker}}(A \to A_1)$ is contained in the Jacobson radical of $A$ by Algebra, Lemma 10.18.1. Hence $A \to \kappa $ factors as $A \to A_1 \to \kappa $. Hence
and we get what we want as $K_1$ has finite tor dimension by Lemma 15.71.2. $\square$
Lemma 15.91.4. 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 $D(A/I^{n + 1})$. Assume
$A$ is $I$-adically complete,
$K_1$ is a perfect object, and
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 perfect, 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. Combine Lemmas 15.91.3 and 15.91.2 (to get derived completeness). $\square$
The following lemma is more of a curiosity. It is rather subtle and esoteric and should be used with care. We do not know if it holds for unbounded complexes.
Lemma 15.91.5. 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
$A$ is Noetherian,
$K_1$ is bounded above, and
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. 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 15.72.3 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 15.82.1 and Remark 15.82.6. 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 $n \geq t$. Hence we have $\mathop{\mathrm{lim}}\nolimits P_ n^\bullet \otimes _ A A/I^ t = R\mathop{\mathrm{lim}}\nolimits P_ n^\bullet \otimes _ A A/I^ t$. By assumption and our choice of $P_ n^\bullet $ the complex $P_ n^\bullet \otimes _ A A/I^ t = P_ n^\bullet \otimes _{A/I^ n} A/I^ t$ represents $K_ n \otimes _{A/I^ n}^\mathbf {L} A/I^ t = K_ t$ for all $n \geq t$. We conclude
In other words, we have $K \otimes _ A^\mathbf {L} A/I^ t = K_ t$. This proves the lemma as it follows that $K$ is derived complete by Proposition 15.88.2. $\square$
Here is a different type of result.
Lemma 15.91.6 (Kollár-Kovács). Let $I$ be an ideal of a Noetherian ring $A$. Let $K \in D(A)$. Set $K_ n = K \otimes _ A^\mathbf {L} A/I^ n$. Assume for all $i \in \mathbf{Z}$ we have
$H^ i(K)$ is a finite $A$-module, and
the system $H^ i(K_ n)$ satisfies Mittag-Leffler.
Then $\mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$ is equal to $\mathop{\mathrm{lim}}\nolimits H^ i(K_ n)$ for all $i \in \mathbf{Z}$.
Proof. Recall that $K^\wedge = R\mathop{\mathrm{lim}}\nolimits K_ n$ is the derived completion of $K$, see Proposition 15.88.2. By Lemma 15.88.4 we have $H^ i(K^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$. By Lemma 15.82.4 we get short exact sequences
The Mittag-Leffler condition guarantees that the left terms are zero (Lemma 15.82.1) and we conclude the lemma is true. $\square$
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