Section 15.93 (0BKF): Derived completion for a principal ideal

15.93 Derived completion for a principal ideal

In this section we discuss what happens with derived completion when the ideal is generated by a single element.

Lemma 15.93.1. Let $A$ be a ring. Let $f \in A$. If there exists an integer $c \geq 1$ such that $A[f^ c] = A[f^{c + 1}] = A[f^{c + 2}] = \ldots $ (for example if $A$ is Noetherian), then for all $n \geq 1$ there exist maps

\[ (A \xrightarrow {f^ n} A) \longrightarrow A/(f^ n), \quad \text{and}\quad A/(f^{n + c}) \longrightarrow (A \xrightarrow {f^ n} A) \]

in $D(A)$ inducing an isomorphism of the pro-objects $\{ A/(f^ n)\} $ and $\{ (f^ n : A \to A)\} $ in $D(A)$.

Proof. The first displayed arrow is obvious. We can define the second arrow of the lemma by the diagram

\[ \xymatrix{ A/A[f^ c] \ar[r]_-{f^{n + c}} \ar[d]_{f^ c} & A \ar[d]^1 \\ A \ar[r]^{f^ n} & A } \]

Since the top horizontal arrow is injective the complex in the top row is quasi-isomorphic to $A/f^{n + c}A$. We omit the calculation of compositions needed to show the statement on pro objects. $\square$

Lemma 15.93.2. Let $A$ be a ring and $f \in A$. Set $I = (f)$. In this situation we have the naive derived completion $K \mapsto K' = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/f^ nA)$ and the derived completion

\[ K \mapsto K^\wedge = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)) \]

of Lemma 15.91.18. The natural transformation of functors $K^\wedge \to K'$ is an isomorphism if and only if the $f$-power torsion of $A$ is bounded.

Proof. If the $f$-power torsion is bounded, then the pro-objects $\{ (f^ n : A \to A)\} $ and $\{ A/f^ nA\} $ are isomorphic by Lemma 15.93.1. Hence the functors are isomorphic by Lemma 15.86.10. Conversely, we see from Lemma 15.87.11 that the condition is exactly that

\[ R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A[f^ n]) \]

is zero for all $K \in D(A)$. Here the maps of the system $(A[f^ n])$ are given by multiplication by $f$. Taking $K = A$ and $K = \bigoplus _{i \in \mathbf{N}} A$ we see from Lemma 15.86.13 this implies $(A[f^ n])$ is zero as a pro-object, i.e., $f^{n - 1}A[f^ n] = 0$ for some $n$, i.e., $A[f^{n - 1}] = A[f^ n]$, i.e., the $f$-power torsion is bounded. $\square$

Example 15.93.3. Let $A$ be a ring. Let $f \in A$ be a nonzerodivisor. An example to keep in mind is $A = \mathbf{Z}_ p$ and $f = p$. Let $M$ be an $A$-module. Claim: $M$ is derived complete with respect to $f$ if and only if there exists a short exact sequence

\[ 0 \to K \to L \to M \to 0 \]

where $K, L$ are $f$-adically complete modules whose $f$-torsion is zero. Namely, if there is a such a short exact sequence, then

\[ M \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A) = (K/f^ nK \to L/f^ nL) \]

because $f$ is a nonzerodivisor on $K$ and $L$ and we conclude that $R\mathop{\mathrm{lim}}\nolimits (M \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A))$ is quasi-isomorphic to $K \to L$, i.e., $M$. This shows that $M$ is derived complete by Lemma 15.91.17. Conversely, suppose that $M$ is derived complete. Choose a surjection $F \to M$ where $F$ is a free $A$-module. Since $f$ is a nonzerodivisor on $F$ the derived completion of $F$ is $L = \mathop{\mathrm{lim}}\nolimits F/f^ nF$. Note that $L$ is $f$-torsion free: if $(x_ n)$ with $x_ n \in F$ represents an element $\xi $ of $L$ and $f\xi = 0$, then $x_ n = x_{n + 1} + f^ nz_ n$ and $fx_ n = f^ ny_ n$ for some $z_ n, y_ n \in F$. Then $f^ n y_ n = fx_ n = fx_{n + 1} + f^{n + 1}z_ n = f^{n + 1}y_{n + 1} + f^{n + 1}z_ n$ and since $f$ is a nonzerodivisor on $F$ we see that $y_ n \in fF$ which implies that $x_ n \in f^ nF$, i.e., $\xi = 0$. Since $L$ is the derived completion, the universal property gives a map $L \to M$ factoring $F \to M$. Let $K = \mathop{\mathrm{Ker}}(L \to M)$ be the kernel. Again $K$ is $f$-torsion free, hence the derived completion of $K$ is $\mathop{\mathrm{lim}}\nolimits K/f^ nK$. On the other hand, both $M$ and $L$ are derived complete, hence $K$ is too by Lemma 15.91.6. It follows that $K = \mathop{\mathrm{lim}}\nolimits K/f^ nK$ and the claim is proved.

Example 15.93.4. Let $p$ be a prime number. Consider the map $\mathbf{Z}_ p[x] \to \mathbf{Z}_ p[y]$ of polynomial algebras sending $x$ to $py$. Consider the cokernel $M = \mathop{\mathrm{Coker}}(\mathbf{Z}_ p[x]^\wedge \to \mathbf{Z}_ p[y]^\wedge )$ of the induced map on (ordinary) $p$-adic completions. Then $M$ is a derived complete $\mathbf{Z}_ p$-module by Proposition 15.91.5 and Lemma 15.91.6; see also discussion in Example 15.93.3. However, $M$ is not $p$-adically complete as $1 + py + p^2 y^2 + \ldots $ maps to a nonzero element of $M$ which is contained in $\bigcap p^ nM$.

Example 15.93.5. Let $A$ be a ring and let $f \in A$. Denote $K \mapsto K^\wedge $ the derived completion with respect to $(f)$. Let $M$ be an $A$-module. Using that

\[ M^\wedge = R\mathop{\mathrm{lim}}\nolimits (M \xrightarrow {f^ n} M) \]

by Lemma 15.91.18 and using Lemma 15.87.4 we obtain

\[ H^{-1}(M^\wedge ) = \mathop{\mathrm{lim}}\nolimits M[f^ n] = T_ f(M) \]

the $f$-adic Tate module of $M$. Here the maps $M[f^ n] \to M[f^{n - 1}]$ are given by multiplication by $f$. Then there is a short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits M[f^ n] \to H^0(M^\wedge ) \to \mathop{\mathrm{lim}}\nolimits M/f^ n M \to 0 \]

describing $H^0(M^\wedge )$. We have $H^1(M^\wedge ) = R^1\mathop{\mathrm{lim}}\nolimits M/f^ nM = 0$ as the transition maps are surjective (Lemma 15.87.1). All the other cohomologies of $M^\wedge $ are zero for trivial reasons. We claim that for $K \in D(A)$ there are short exact sequences

\[ 0 \to H^0(H^ n(K)^\wedge ) \to H^ n(K^\wedge ) \to T_ f(H^{n + 1}(K)) \to 0 \]

Namely this follows from the spectral sequence of Example 15.91.22 because it degenerates at $E_2$ (as only $i = -1, 0$ give nonzero terms).

Lemma 15.93.6 (Bhatt). Let $I$ be a finitely generated ideal in a ring $A$. Let $M$ be a derived complete $A$-module. If $M$ is an $I$-power torsion module, then $I^ nM = 0$ for some $n$.

Proof. Say $I = (f_1, \ldots , f_ r)$. It suffices to show that for each $i$ there is an $n_ i$ such that $f_ i^{n_ i}M = 0$. Hence we may assume that $I = (f)$ is a principal ideal. Let $B = \mathbf{Z}[x] \to A$ be the ring map sending $x$ to $f$. By Lemma 15.91.23 we see that $M$ is derived complete as a $B$-module with respect to the ideal $(x)$. After replacing $A$ by $B$, we may assume that $f$ is a nonzerodivisor in $A$.

Assume $I = (f)$ with $f \in A$ a nonzerodivisor. According to Example 15.93.3 there exists a short exact sequence

\[ 0 \to K \xrightarrow {u} L \to M \to 0 \]

where $K$ and $L$ are $I$-adically complete $A$-modules whose $f$-torsion is zero ¹. Consider $K$ and $L$ as topological modules with the $I$-adic topology. Then $u$ is continuous. Let

\[ L_ n = \{ x \in L \mid f^ n x \in u(K)\} \]

Since $M$ is $f$-power torsion we see that $L = \bigcup L_ n$. Let $N_ n$ be the closure of $L_ n$ in $L$. By Lemma 15.36.4 we see that $N_ n$ is open in $L$ for some $n$. Fix such an $n$. Since $f^{n + m} : L \to L$ is a continuous open map, and since $f^{n + m} L_ n \subset u(f^ m K)$ we conclude that the closure of $u(f^ mK)$ is open for all $m \geq 1$. Thus by Lemma 15.36.5 we conclude that $u$ is open. Hence $f^ tL \subset \mathop{\mathrm{Im}}(u)$ for some $t$ and we conclude that $f^ t$ annihilates $M$ as desired. $\square$

Lemma 15.93.7. Let $f \in A$ be an element of a ring. Set $J = \bigcap f^ nA$. Let $M$ be an $A$-module derived complete with respect to $f$. Then $JM' = 0$ where $M' = \mathop{\mathrm{Ker}}(M \to \mathop{\mathrm{lim}}\nolimits M/f^ nM)$. In particular, if $A$ is derived complete then $J$ is an ideal of square zero.

Proof. Take $x \in M'$ and $g \in J$. For every $n \geq 1$ we may write $x = f^ n x_ n$. Since $g$ is in $f^ nA$ we see that the element $y_ n = gx_ n$ in $M'$ is independent of the choice of $x_ n$. In particular, we may take $x_ n = fx_{n + 1}$ and we find that $y_ n = fy_{n + 1}$. Thus we obtain a map $A_ f \to M$ sending $1/f^ n$ to $y_ n$. This map has to be zero as $M$ is derived complete (Lemma 15.91.1) and hence $y_ n = 0$ for all $n$. Since $gx = gfx_1 = fy_1$ this completes the proof. $\square$

Lemma 15.93.8. Let $A$ be a ring derived complete with respect to an ideal $I$. Then $(A, I)$ is a henselian pair.

Proof. Let $f \in I$. By Lemma 15.11.15 it suffices to show that $(A, fA)$ is a henselian pair. Observe that $A$ is derived complete with respect to $fA$ (follows immediately from Definition 15.91.4). By Lemma 15.91.3 the map from $A$ to the $f$-adic completion $A'$ of $A$ is surjective. By Lemma 15.11.4 the pair $(A', fA')$ is henselian. Thus it suffices to show that $(A, \bigcap f^ nA)$ is a henselian pair, see Lemma 15.11.9. This follows from Lemmas 15.93.7 and 15.11.2. $\square$

Lemma 15.93.9. Let $A$ be a ring derived complete with respect to an ideal $I$. Set $J = \bigcap I^ n$. If $I$ can be generated by $r$ elements then $J^ N = 0$ where $N = 2^ r$.

Proof. When $r = 1$ this is Lemma 15.93.7. Say $I = (f_1, \ldots , f_ r)$ with $r > 1$. By Lemma 15.91.6 the ring $A_ t = A/f_ r^ tA$ is derived complete with respect to $I$ and hence a fortiori derived complete with respect to $I_ t = (f_1, \ldots , f_{r - 1})A_ t$. Observe that $A \to A_ t$ sends $J$ into $J_ t = \bigcap I_ t^ n$. By induction $J_ t^{N/2} = 0$ with $N = 2^ r$. The ideal $\bigcap \mathop{\mathrm{Ker}}(A \to A_ t) = \bigcap f_ r^ t A$ has square zero by the case $r = 1$. This finishes the proof. $\square$

Lemma 15.93.10. Let $A$ be a reduced ring derived complete with respect to a finitely generated ideal $I$. Then $A$ is $I$-adically complete.

Proof. Follows from Lemma 15.93.9 and Proposition 15.91.5. $\square$

[1] For the proof it is enough to show that there exists a sequence $K \xrightarrow {u} L \to M \to 0$ where $K$ and $L$ are $I$-adically complete $A$-modules. This can be shown by choosing a presentation $F_1 \to F_0 \to M \to 0$ with $F_ i$ free and then setting $K$ and $L$ equal to the $f$-adic completions of $F_1$ and $F_0$. Namely, as $f$ is a nonzerodivisor these completions will be the derived completions and the sequence will remain exact.

Comments (2)

Comment #6425 by Arthur Ogus on July 19, 2021 at 13:55

If $I = (f)$ , then any derived $I$ -complete module $M$ with bounded $f$ -torsion is in fact classically $I$ -adically complete. By Lemma 15.90.3, the map from $M$ to its $I$ -adic completion is surjective, so it suffices to show injectivity, ie, that $\cap \{ I^n M : n \ge 0\} = 0$ . First suppose that $M$ is $f$ -torsion free. If $x = f^nx_n$ for every $n \ge 0$ , the torsion-freeness of $M$ implies that $fx_n = x_{n-1}$ for all $n$ , so the sequence $(x_n )$ defines an element of $(Hom A_{(f)}, M) = 0$ . Thus $x = 0$ . For the general case, choose $N$ such that the $f$ -torsion $M[f^\infty]$ is annihilated by $f^N$ , and note that $M[F^N] = M[f^\infty]$ is $I$ -adically separated and complete, hence also derived $I$ -complete. Then it follows from Lemma 15.90.6 that the quotient $M/M[f^\infty]$ is also derived $I$ -complete. Since it is $f$ -torsion free, it follows from the previous case that it is $I$ -adically separated. Now suppose that $x \in \cap \{ I^n M : n \ge 0\}$ , say $x = f^nx_n$ for all $n$ . Then the image of $x$ in $M/M[f^\infty]$ vanishes, so in fact $x \in M[f^\infty]$ . Since $x$ is an $f$ -torsion element, so is each $x_n$ . in particular, $x = f^Nx_N = 0$ .

Comment #6468 by Arthur Ogus on July 26, 2021 at 14:36

The fact that a derived $(f)$ -complete module with bounded $f$ -torsion is f-adically separated follows quite directly from the exact sequence $0 \to R^1\lim M[f^n] \to H^0(\hat M \ ) \to \lim M/f^n M \to 0$ of Example 0BKG. That exact sequence follows easily from the fact that the complex $\hat M \ = R\lim (M\xrightarrow {f^n} M)$ and the spectral sequence whose second page has $E_2^{p,q} = R^p\lim H^q (K_n) \Rightarrow R\lim K_n$

The Stacks project

15.93 Derived completion for a principal ideal

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