The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.95 Completion

Suppose that $R$ is a ring and $I$ is an ideal. We define the completion of $R$ with respect to $I$ to be the limit

\[ R^\wedge = \mathop{\mathrm{lim}}\nolimits _ n R/I^ n. \]

An element of $R^\wedge $ is given by a sequence of elements $f_ n \in R/I^ n$ such that $f_ n \equiv f_{n + 1} \bmod I^ n$ for all $n$. We will view $R^\wedge $ as an $R$-algebra. Similarly, if $M$ is an $R$-module then we define the completion of $M$ with respect to $I$ to be the limit

\[ M^\wedge = \mathop{\mathrm{lim}}\nolimits _ n M/I^ nM. \]

An element of $M^\wedge $ is given by a sequence of elements $m_ n \in M/I^ nM$ such that $m_ n \equiv m_{n + 1} \bmod I^ nM$ for all $n$. We will view $M^\wedge $ as an $R^\wedge $-module. From this description it is clear that there are always canonical maps

\[ M \longrightarrow M^\wedge \quad \text{and}\quad M \otimes _ R R^\wedge \longrightarrow M^\wedge . \]

Moreover, given a map $\varphi : M \to N$ of modules we get an induced map $\varphi ^\wedge : M^\wedge \to N^\wedge $ on completions making the diagram

\[ \xymatrix{ M \ar[r] \ar[d] & N \ar[d] \\ M^\wedge \ar[r] & N^\wedge } \]

commute. In general completion is not an exact functor, see Examples, Section 102.8. Here are some initial positive results.

Lemma 10.95.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\varphi : M \to N$ be a map of $R$-modules.

  1. If $M/IM \to N/IN$ is surjective, then $M^\wedge \to N^\wedge $ is surjective.

  2. If $M \to N$ is surjective, then $M^\wedge \to N^\wedge $ is surjective.

  3. If $0 \to K \to M \to N \to 0$ is a short exact sequence of $R$-modules and $N$ is flat, then $0 \to K^\wedge \to M^\wedge \to N^\wedge \to 0$ is a short exact sequence.

  4. The map $M \otimes _ R R^\wedge \to M^\wedge $ is surjective for any finite $R$-module $M$.

Proof. Assume $M/IM \to N/IN$ is surjective. Then the map $M/I^ nM \to N/I^ nN$ is surjective for each $n \geq 1$ by Nakayama's lemma. More precisely, apply Lemma 10.19.1 part (11) to the map $M/I^ nM \to N/I^ nN$ over the ring $R/I^ n$ and the nilpotent ideal $I/I^ n$ to see this. Set $K_ n = \{ x \in M \mid \varphi (x) \in I^ nN\} $. Thus we get short exact sequences

\[ 0 \to K_ n/I^ nM \to M/I^ nM \to N/I^ nN \to 0 \]

We claim that the canonical map $K_{n + 1}/I^{n + 1}M \to K_ n/I^ nM$ is surjective. Namely, if $x \in K_ n$ write $\varphi (x) = \sum z_ j n_ j$ with $z_ j \in I^ n$, $n_ j \in N$. By assumption we can write $n_ j = \varphi (m_ j) + \sum z_{jk}n_{jk}$ with $m_ j \in M$, $z_{jk} \in I$ and $n_{jk} \in N$. Hence

\[ \varphi (x - \sum z_ j m_ j) = \sum z_ jz_{jk} n_{jk}. \]

This means that $x' = x - \sum z_ j m_ j \in K_{n + 1}$ maps to $x$ which proves the claim. Now we may apply Lemma 10.86.1 to the inverse system of short exact sequences above to see (1). Part (2) is a special case of (1). If the assumptions of (3) hold, then for each $n$ the sequence

\[ 0 \to K/I^ nK \to M/I^ nM \to N/I^ nN \to 0 \]

is short exact by Lemma 10.38.12. Hence we can directly apply Lemma 10.86.1 to conclude (3) is true. To see (4) choose generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $R^{\oplus n} \to M$, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Hence by (2) we see $(R^\wedge )^{\oplus n} \to M^\wedge $, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Assertion (4) follows from this. $\square$

Definition 10.95.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. We say $M$ is $I$-adically complete if the map

\[ M \longrightarrow M^\wedge = \mathop{\mathrm{lim}}\nolimits _ n M/I^ nM \]

is an isomorphism1. We say $R$ is $I$-adically complete if $R$ is $I$-adically complete as an $R$-module.

It is not true that the completion of an $R$-module $M$ with respect to $I$ is $I$-adically complete. For an example see Examples, Section 102.6. If the ideal is finitely generated, then the completion is complete.

reference

Lemma 10.95.3. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $M$ be an $R$-module. Then

  1. the completion $M^\wedge $ is $I$-adically complete, and

  2. $I^ nM^\wedge = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM) = (I^ nM)^\wedge $ for all $n \geq 1$.

In particular $R^\wedge $ is $I$-adically complete, $I^ nR^\wedge = (I^ n)^\wedge $, and $R^\wedge /I^ nR^\wedge = R/I^ n$.

Proof. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Applying Lemma 10.95.1 part (2) to the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ yields a surjection

\[ (M^\wedge )^{\oplus r} \xrightarrow {(f_1, \ldots , f_ r)} (I^ n M)^\wedge = \mathop{\mathrm{lim}}\nolimits _{m \geq n} I^ n M/I^ m M = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ n M). \]

On the other hand, the image of $(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \to M^\wedge $ is $I^ n M^\wedge $. Thus $M^\wedge / I^ n M^\wedge \simeq M/I^ n M$. Taking inverse limits yields $(M^\wedge )^\wedge \simeq M^\wedge $; that is, $M^\wedge $ is $I$-adically complete. $\square$

Lemma 10.95.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $0 \to M \to N \to Q \to 0$ be an exact sequence of $R$-modules such that $Q$ is annihilated by a power of $I$. Then completion produces an exact sequence $0 \to M^\wedge \to N^\wedge \to Q \to 0$.

Proof. Say $I^ c Q = 0$. Then $Q/I^ nQ = Q$ for $n \geq c$. On the other hand, it is clear that $I^ nM \subset M \cap I^ nN \subset I^{n - c}M$ for $n \geq c$. Thus $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/(M \cap I^ n N)$. Apply Lemma 10.86.1 to the system of exact sequences

\[ 0 \to M/(M \cap I^ n N) \to N/I^ n N \to Q \to 0 \]

for $n \geq c$ to conclude. $\square$

reference

Lemma 10.95.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Denote $K_ n = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM)$. Then $M^\wedge $ is $I$-adically complete if and only if $K_ n$ is equal to $I^ nM^\wedge $ for all $n \geq 1$.

Proof. The module $I^ n M^\wedge $ is contained in $K_ n$. Thus for each $n \geq 1$ there is a canonical exact sequence

\[ 0 \to K_ n/I^ nM^\wedge \to M^\wedge /I^ nM^\wedge \to M/I^ nM \to 0. \]

As $I^ nM^\wedge $ maps onto $I^ nM/I^{n + 1}M$ we see that $K_{n + 1} + I^ n M^\wedge = K_ n$. Thus the inverse system $\{ K_ n/I^ n M^\wedge \} _{n \geq 1}$ has surjective transition maps. By Lemma 10.86.1 we see that there is a short exact sequence

\[ 0 \to \mathop{\mathrm{lim}}\nolimits _ n K_ n/I^ n M^\wedge \to (M^\wedge )^\wedge \to M^\wedge \to 0 \]

Hence $M^\wedge $ is complete if and only if $K_ n/I^ n M^\wedge = 0$ for all $n \geq 1$. $\square$

Lemma 10.95.6. Let $R$ be a ring, let $I \subset R$ be an ideal, and let $R^\wedge = \mathop{\mathrm{lim}}\nolimits R/I^ n$.

  1. any element of $R^\wedge $ which maps to a unit of $R/I$ is a unit,

  2. any element of $1 + I$ maps to an invertible element of $R^\wedge $,

  3. any element of $1 + IR^\wedge $ is invertible in $R^\wedge $, and

  4. the ideals $IR^\wedge $ and $\mathop{\mathrm{Ker}}(R^\wedge \to R/I)$ are contained in the Jacobson radical of $R^\wedge $.

Proof. Let $x \in R^\wedge $ map to a unit $x_1$ in $R/I$. Then $x$ maps to a unit $x_ n$ in $R/I^ n$ for every $n$ by Lemma 10.31.4. Hence $y = (x_ n^{-1}) \in \mathop{\mathrm{lim}}\nolimits R/I^ n = R^\wedge $ is an inverse to $x$. Parts (2) and (3) follow immediately from (1). Part (4) follows from (1) and Lemma 10.18.1. $\square$

Lemma 10.95.7. Let $A$ be a ring. Let $I = (f_1, \ldots , f_ r)$ be a finitely generated ideal. If $M \to \mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$ is surjective for each $i$, then $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is surjective.

Proof. Note that $\mathop{\mathrm{lim}}\nolimits M/I^ nM = \mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ as $I^ n \supset (f_1^ n, \ldots , f_ r^ n) \supset I^{rn}$. An element $\xi $ of $\mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ can be symbolically written as

\[ \xi = \sum \nolimits _{n \geq 0} \sum \nolimits _ i f_ i^ n x_{n, i} \]

with $x_{n, i} \in M$. If $M \to \mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$ is surjective, then there is an $x_ i \in M$ mapping to $\sum x_{n, i} f_ i^ n$ in $\mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$. Then $x = \sum x_ i$ maps to $\xi $ in $\mathop{\mathrm{lim}}\nolimits M/I^ nM$. $\square$

Lemma 10.95.8. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. If $M$ is $J$-adically complete and $I$ is finitely generated, then $M$ is $I$-adically complete.

Proof. Assume $M$ is $J$-adically complete and $I$ is finitely generated. We have $\bigcap I^ nM = 0$ because $\bigcap J^ nM = 0$. By Lemma 10.95.7 it suffices to prove the surjectivity of $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ in case $I$ is generated by a single element. Say $I = (f)$. Let $x_ n \in M$ with $x_{n + 1} - x_ n \in f^ nM$. We have to show there exists an $x \in M$ such that $x_ n - x \in f^ nM$ for all $n$. As $x_{n + 1} - x_ n \in J^ nM$ and as $M$ is $J$-adically complete, there exists an element $x \in M$ such that $x_ n - x \in J^ nM$. Replacing $x_ n$ by $x_ n - x$ we may assume that $x_ n \in J^ nM$. To finish the proof we will show that this implies $x_ n \in I^ nM$. Namely, write $x_ n - x_{n + 1} = f^ nz_ n$. Then

\[ x_ n = f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots ) \]

The sum $z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots $ converges in $M$ as $f^ c \in J^ c$. The sum $f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots )$ converges in $M$ to $x_ n$ because the partial sums equal $x_ n - x_{n + c}$ and $x_{n + c} \in J^{n + c}M$. $\square$

Lemma 10.95.9. Let $R$ be a ring. Let $I$, $J$ be ideals of $R$. Assume there exist integers $c, d > 0$ such that $I^ c \subset J$ and $J^ d \subset I$. Then completion with respect to $I$ agrees with completion with respect to $J$ for any $R$-module. In particular an $R$-module $M$ is $I$-adically complete if and only if it is $J$-adically complete.

Proof. Consider the system of maps $M/I^ nM \to M/J^{\lfloor n/d \rfloor }M$ and the system of maps $M/J^ mM \to M/I^{\lfloor m/c \rfloor }M$ to get mutually inverse maps between the completions. $\square$

Lemma 10.95.10. Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $I$-adically complete $R$-module, and let $K \subset M$ be an $R$-submodule. The following are equivalent

  1. $K = \bigcap (K + I^ nM)$ and

  2. $M/K$ is $I$-adically complete.

Proof. Set $N = M/K$. By Lemma 10.95.1 the map $M = M^\wedge \to N^\wedge $ is surjective. Hence $N \to N^\wedge $ is surjective. It is easy to see that the kernel of $N \to N^\wedge $ is the module $\bigcap (K + I^ nM) / K$. $\square$

Lemma 10.95.11. Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $R$-module. If (a) $R$ is $I$-adically complete, (b) $M$ is a finite $R$-module, and (c) $\bigcap I^ nM = (0)$, then $M$ is $I$-adically complete.

Proof. By Lemma 10.95.1 the map $M = M \otimes _ R R = M \otimes _ R R^\wedge \to M^\wedge $ is surjective. The kernel of this map is $\bigcap I^ nM$ hence zero by assumption. Hence $M \cong M^\wedge $ and $M$ is complete. $\square$

Lemma 10.95.12. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

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

  2. $\bigcap _{n \geq 1} I^ nM = (0)$, and

  3. $M/IM$ is a finite $R/I$-module.

Then $M$ is a finite $R$-module.

Proof. Let $x_1, \ldots , x_ n \in M$ be elements whose images in $M/IM$ generate $M/IM$ as a $R/I$-module. Denote $M' \subset M$ the $R$-submodule generated by $x_1, \ldots , x_ n$. By Lemma 10.95.1 the map $(M')^\wedge \to M^\wedge $ is surjective. Since $\bigcap I^ nM = 0$ we see in particular that $\bigcap I^ nM' = (0)$. Hence by Lemma 10.95.11 we see that $M'$ is complete, and we conclude that $M' \to M^\wedge $ is surjective. Finally, the kernel of $M \to M^\wedge $ is zero since it is equal to $\bigcap I^ nM = (0)$. Hence we conclude that $M \cong M' \cong M^\wedge $ is finitely generated. $\square$

[1] This includes the condition that $\bigcap I^ nM = (0)$.

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