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 in $D(A)$ inducing an isomorphism of the pro-objects $\{ A/(f^ n)\} $ and $\{ (f^ n : A \to A)\} $ in $D(A)$.
\[ (A \xrightarrow {f^ n} A) \longrightarrow A/(f^ n), \quad \text{and}\quad A/(f^{n + c}) \longrightarrow (A \xrightarrow {f^ n} A) \]
Proof. The first displayed arrow is obvious. We can define the second arrow of the lemma by the diagram
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 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.
\[ K \mapsto K^\wedge = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)) \]
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 15.86.11. Conversely, we see from Lemma 15.87.11 that the condition is exactly that
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 15.86.14 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 where $K, L$ are $f$-adically complete modules whose $f$-torsion is zero. Namely, if there is a such a short exact sequence, then 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.
\[ 0 \to K \to L \to M \to 0 \]
\[ M \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A) = (K/f^ nK \to L/f^ nL) \]
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 by Lemma 15.91.18 and using Lemma 15.87.4 we obtain 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 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. Finally, for $K \in D(A)$ and $p \in \mathbf{Z}$ there is a short exact sequence 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); the next lemma gives more information.
\[ M^\wedge = R\mathop{\mathrm{lim}}\nolimits (M \xrightarrow {f^ n} M) \]
\[ H^{-1}(M^\wedge ) = \mathop{\mathrm{lim}}\nolimits M[f^ n] = T_ f(M) \]
\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits M[f^ n] \to H^0(M^\wedge ) \to \mathop{\mathrm{lim}}\nolimits M/f^ n M \to 0 \]
\[ 0 \to H^0(H^ p(K)^\wedge ) \to H^ p(K^\wedge ) \to T_ f(H^{p + 1}(K)) \to 0 \]
Lemma 15.93.6. Let $A$ be a ring and let $f \in A$. Let $K$ be an object of $D(A)$. Denote $K_ n = K \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)$. For all $p \in \mathbf{Z}$ there is a commutative diagram with exact rows and columns where $\widehat{H^ p(K)} = \mathop{\mathrm{lim}}\nolimits H^ p(K)/f^ nH^ p(K)$ is the usual $f$-adic completion. The left vertical short exact sequence and the middle horizontal short exact sequence are taken from Example 15.93.5 The middle vertical short exact sequence is the one from Lemma 15.87.4.
\[ \xymatrix{ & 0 & 0 \\ 0 \ar[r] & \widehat{H^ p(K)} \ar[r] \ar[u] & \mathop{\mathrm{lim}}\nolimits H^ p(K_ n) \ar[r] \ar[u] & T_ f(H^{p + 1}(K)) \ar[r] & 0 \\ 0 \ar[r] & H^0(H^ p(K)^\wedge ) \ar[r] \ar[u] & H^ p(K^\wedge ) \ar[r] \ar[u] & T_ f(H^{p + 1}(K)) \ar[r] \ar@{=}[u] & 0 \\ & R^1\mathop{\mathrm{lim}}\nolimits H^ p(K)[f^ n] \ar[u] \ar[r]^\cong & R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(K_ n) \ar[u] \\ & 0 \ar[u] & 0 \ar[u] } \]
Proof. To construct the top horizontal short exact sequence, observe that we have the following inverse system short exact sequences
coming from the construction of $K_ n$ as a shift of the cone on $f^ n : K \to K$. Taking the inverse limit of these we obtain the top horizontal short exact sequence, see Homology, Lemma 12.31.3.
Let us prove that we have a commutative diagram as in the lemma. We consider the map $L = \tau _{\leq p}K \to K$. Setting $L_ n = L \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)$ we obtain a map $(L_ n) \to (K_ n)$ of inverse systems which induces a map of short exact sequences
Since $H^ i(L) = 0$ for $i > p$ and $H^ p(L) = H^ p(K)$, a computation using the references in the statement of the lemma shows that $H^ p(L^\wedge ) = H^0(H^ p(K)^\wedge )$ and that $H^ p(L_ n) = H^ p(K)/f^ nH^ p(K)$. On the other hand, we have $H^{p - 1}(L_ n) = H^{p - 1}(K_ n)$ and hence we see that we get the isomorphism as indicated in the statement of the lemma since we already know the kernel of $H^0(H^ p(K)^\wedge ) \to \widehat{H^ p(K)}$ is equal to $R^1\mathop{\mathrm{lim}}\nolimits H^ p(K)[f^ n]$. We omit the verification that the rightmost square in the diagram commutes if we define the top row by the construction in the first paragraph of the proof. $\square$
Remark 15.93.7. With notation as in Lemma 15.93.6 we also see that the inverse system $H^ p(K_ n)$ has ML if and only if the inverse system $H^{p + 1}(K)[f^ n]$ has ML. This follows from the inverse system of short exact sequences $0 \to H^ p(K)/f^ nH^ p(K) \to H^ p(K_ n) \to H^{p + 1}(K)[f^ n] \to 0$ (see proof of the lemma) combined with Homology, Lemma 12.31.3 and Lemma 15.86.13.
Lemma 15.93.8 (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
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
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 15.93.9. 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 15.93.10. 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 15.93.9 and 15.11.2. $\square$
Lemma 15.93.11. 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 15.93.9. 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 15.93.12. 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 15.93.11 and Proposition 15.91.5. $\square$
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