The Stacks project

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 zero1. 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

If , then any derived -complete module with bounded -torsion is in fact classically -adically complete. By Lemma 15.90.3, the map from to its -adic completion is surjective, so it suffices to show injectivity, ie, that . First suppose that is -torsion free. If for every , the torsion-freeness of implies that for all , so the sequence defines an element of . Thus . For the general case, choose such that the -torsion is annihilated by , and note that is -adically separated and complete, hence also derived -complete. Then it follows from Lemma 15.90.6 that the quotient is also derived -complete. Since it is -torsion free, it follows from the previous case that it is -adically separated. Now suppose that , say for all . Then the image of in vanishes, so in fact . Since is an -torsion element, so is each . in particular, .

Comment #6468 by Arthur Ogus on

The fact that a derived -complete module with bounded -torsion is f-adically separated follows quite directly from the exact sequence of Example 0BKG. That exact sequence follows easily from the fact that the complex and the spectral sequence whose second page has

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