## 15.97 Taking limits of complexes

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

1. 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

2. 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 15.97.1. 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

1. the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,

2. $K_1$ is pseudo-coherent, and

3. 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 15.64.1. By Lemma 15.75.4 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 15.87.1 and Remark 15.87.6. 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 15.97.2. 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

1. $A$ is $I$-adically complete,

2. $K_1$ is pseudo-coherent, 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 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 15.97.1. Finally, $K$ is derived complete by Lemma 15.91.14. $\square$

Lemma 15.97.3. 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

1. the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,

2. $K_1$ is a perfect object, and

3. 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 15.97.1. 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 15.77.3. Any element of $A$ which maps to a unit in $A_1$ is a unit in $A$ by Algebra, Lemma 10.32.4 and hence $\mathop{\mathrm{Ker}}(A \to A_1)$ is contained in the Jacobson radical of $A$ by Algebra, Lemma 10.19.1. Hence $A \to \kappa$ factors as $A \to A_1 \to \kappa$. Hence

$K \otimes _ A^\mathbf {L} \kappa = K \otimes _ A^\mathbf {L} A_1 \otimes _{A_1}^\mathbf {L} \kappa = K_1 \otimes _{A_1}^\mathbf {L} \kappa$

and we get what we want as $K_1$ has finite tor dimension by Lemma 15.74.2. $\square$

Lemma 15.97.4. 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

1. $A$ is $I$-adically complete,

2. $K_1$ is a perfect object, 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 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 15.97.3 and 15.97.2 (to get derived completeness). $\square$

We do not know if the following lemma holds for unbounded complexes.

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. $\square$

Here is a different type of result.

Lemma 15.97.6 (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

1. $H^ i(K)$ is a finite $A$-module, and

2. 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 15.94.2. By Lemma 15.94.4 we have $H^ i(K^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^ i(K)/I^ nH^ i(K)$. By Lemma 15.87.4 we get short exact sequences

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

The Mittag-Leffler condition guarantees that the left terms are zero (Lemma 15.87.1) and we conclude the lemma is true. $\square$

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