## 10.96 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 110.9. Here are some initial positive results.

Lemma 10.96.1. 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 10.20.1 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 $x \bmod I^ nM$ which proves the claim. Now we may apply Lemma 10.87.1 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 10.39.12. Hence we can directly apply Lemma 10.87.1 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 10.96.2. 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 110.7. If the ideal is finitely generated, then the completion is complete.

Lemma 10.96.3. 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 10.96.1 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 10.96.4. 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 10.87.1 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. $\square$

Lemma 10.96.5. 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 10.87.1 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 10.96.6. 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 10.32.4. 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 10.19.1. $\square$

Lemma 10.96.7. 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 10.96.8. 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 10.96.7 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 10.96.9. 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 10.96.10. 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 10.96.1 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 10.96.11. 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 10.96.1 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 10.96.12. 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 10.96.1 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 10.96.11 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 $\bigcap I^ nM = 0$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00M9. Beware of the difference between the letter 'O' and the digit '0'.