## 10.96 Completion for Noetherian rings

In this section we discuss completion with respect to ideals in Noetherian rings.

Lemma 10.96.1. Let $I$ be an ideal of a Noetherian ring $R$. Denote ${}^\wedge $ completion with respect to $I$.

If $N \to M$ is an injective map of finite $R$-modules, then the map on completions $N^\wedge \to M^\wedge $ is injective.

If $M$ is a finite $R$-module, then $M^\wedge = M \otimes _ R R^\wedge $.

**Proof.**
For the first statement, by the Artin-Rees Lemma 10.50.2, we have a constant $c$ such that $I^ nM \cap N$ equals $I^{n-c}(I^ cM \cap N) \subset I^{n-c}N$. Thus if $(n_ i) \in N^\wedge $ maps to zero in $M^\wedge $, then each $n_ i$ maps to zero in $N/I^{i-c}N$. And hence $n_{i-c} = 0$. Thus $N^\wedge \to M^\wedge $ is injective.

For the second statement let $0 \to K \to R^ t \to M \to 0$ be the presentation of $M$ corresponding to the generators $x_1, \ldots , x_ t$ of $M$. By Lemma 10.95.1 $(R^ t)^\wedge \to M^\wedge $ is surjective, and for any finitely generated $R$-module the canonical map $M \otimes _ R R^\wedge \to M^\wedge $ is surjective. Hence to prove the second statement it suffices to prove the kernel of $(R^ t)^\wedge \to M^\wedge $ is exactly $K^\wedge $.

Let $(x_ n) \in (R^ t)^\wedge $ be in the kernel. Note that each $x_ n$ is in the image of the map $K/I^ nK \to (R/I^ n)^ t$. Choose $c$ such that $(I^ n)^ t \cap K \subset I^{n-c} K$, which is possible by Artin-Rees (Lemma 10.50.2). For each $n \geq 0$ choose $y_ n \in K/I^{n + c}K$ mapping to $x_{n + c}$, and set $z_ n = y_ n \bmod I^ nK$. The elements $z_ n$ satisfy $z_{n + 1} - z_ n \bmod I^ nK = y_{n + 1} - y_ n \bmod I^ nK$, and $y_{n + 1} - y_ n \in I^{n + c}R^ t$ by construction. Hence $z_{n + 1} = z_ n \bmod I^ nK$ by the choice of $c$ above. In other words $(z_ n) \in K^\wedge $ maps to $(x_ n)$ as desired.
$\square$

Lemma 10.96.2. Let $I$ be a ideal of a Noetherian ring $R$. Denote ${}^\wedge $ completion with respect to $I$.

The ring map $R \to R^\wedge $ is flat.

The functor $M \mapsto M^\wedge $ is exact on the category of finitely generated $R$-modules.

**Proof.**
Consider $I \otimes _ R R^\wedge \to R \otimes _ R R^\wedge = R^\wedge $. According to Lemma 10.96.1 this is identified with $I^\wedge \to R^\wedge $ and $I^\wedge \to R^\wedge $ is injective. Part (1) follows from Lemma 10.38.5. Part (2) follows from part (1) and Lemma 10.96.1 part (2).
$\square$

Lemma 10.96.3. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $I \subset \mathfrak m$ be an ideal. Denote $R^\wedge $ the completion of $R$ with respect to $I$. The ring map $R \to R^\wedge $ is faithfully flat. In particular the completion with respect to $\mathfrak m$, namely $\mathop{\mathrm{lim}}\nolimits _ n R/\mathfrak m^ n$ is faithfully flat.

**Proof.**
By Lemma 10.96.2 it is flat. The composition $R \to R^\wedge \to R/\mathfrak m$ where the last map is the projection map $R^\wedge \to R/I$ combined with $R/I \to R/\mathfrak m$ shows that $\mathfrak m$ is in the image of $\mathop{\mathrm{Spec}}(R^\wedge ) \to \mathop{\mathrm{Spec}}(R)$. Hence the map is faithfully flat by Lemma 10.38.15.
$\square$

Lemma 10.96.4. Let $R$ be a Noetherian ring. Let $I$ be an ideal of $R$. Let $M$ be an $R$-module. Then the completion $M^\wedge $ of $M$ with respect to $I$ is $I$-adically complete, $I^ n M^\wedge = (I^ nM)^\wedge $, and $M^\wedge /I^ nM^\wedge = M/I^ nM$.

**Proof.**
This is a special case of Lemma 10.95.3 because $I$ is a finitely generated ideal.
$\square$

Lemma 10.96.5. Let $I$ be an ideal of a ring $R$. Assume

$R/I$ is a Noetherian ring,

$I$ is finitely generated.

Then the completion $R^\wedge $ of $R$ with respect to $I$ is a Noetherian ring complete with respect to $IR^\wedge $.

**Proof.**
By Lemma 10.95.3 we see that $R^\wedge $ is $I$-adically complete. Hence it is also $IR^\wedge $-adically complete. Since $R^\wedge /IR^\wedge = R/I$ is Noetherian we see that after replacing $R$ by $R^\wedge $ we may in addition to assumptions (1) and (2) assume that also $R$ is $I$-adically complete.

Let $f_1, \ldots , f_ t$ be generators of $I$. Then there is a surjection of rings $R/I[T_1, \ldots , T_ t] \to \bigoplus I^ n/I^{n + 1}$ mapping $T_ i$ to the element $\overline{f}_ i \in I/I^2$. Hence $\bigoplus I^ n/I^{n + 1}$ is a Noetherian ring. Let $J \subset R$ be an ideal. Consider the ideal

\[ \bigoplus J \cap I^ n/J \cap I^{n + 1} \subset \bigoplus I^ n/I^{n + 1}. \]

Let $\overline{g}_1, \ldots , \overline{g}_ m$ be generators of this ideal. We may choose $\overline{g}_ j$ to be a homogeneous element of degree $d_ j$ and we may pick $g_ j \in J \cap I^{d_ j}$ mapping to $\overline{g}_ j \in J \cap I^{d_ j}/J \cap I^{d_ j + 1}$. We claim that $g_1, \ldots , g_ m$ generate $J$.

Let $x \in J \cap I^ n$. There exist $a_ j \in I^{\max (0, n - d_ j)}$ such that $x - \sum a_ j g_ j \in J \cap I^{n + 1}$. The reason is that $J \cap I^ n/J \cap I^{n + 1}$ is equal to $\sum \overline{g}_ j I^{n - d_ j}/I^{n - d_ j + 1}$ by our choice of $g_1, \ldots , g_ m$. Hence starting with $x \in J$ we can find a sequence of vectors $(a_{1, n}, \ldots , a_{m, n})_{n \geq 0}$ with $a_{j, n} \in I^{\max (0, n - d_ j)}$ such that

\[ x = \sum \nolimits _{n = 0, \ldots , N} \sum \nolimits _{j = 1, \ldots , m} a_{j, n} g_ j \bmod I^{N + 1} \]

Setting $A_ j = \sum _{n \geq 0} a_{j, n}$ we see that $x = \sum A_ j g_ j$ as $R$ is complete. Hence $J$ is finitely generated and we win.
$\square$

Lemma 10.96.6. Let $R$ be a Noetherian ring. Let $I$ be an ideal of $R$. The completion $R^\wedge $ of $R$ with respect to $I$ is Noetherian.

**Proof.**
This is a consequence of Lemma 10.96.5. It can also be seen directly as follows. Choose generators $f_1, \ldots , f_ n$ of $I$. Consider the map

\[ R[[x_1, \ldots , x_ n]] \longrightarrow R^\wedge , \quad x_ i \longmapsto f_ i. \]

This is a well defined and surjective ring map (details omitted). Since $R[[x_1, \ldots , x_ n]]$ is Noetherian (see Lemma 10.30.2) we win.
$\square$

Suppose $R \to S$ is a local homomorphism of local rings $(R, \mathfrak m)$ and $(S, \mathfrak n)$. Let $S^\wedge $ be the completion of $S$ with respect to $\mathfrak n$. In general $S^\wedge $ is not the $\mathfrak m$-adic completion of $S$. If $\mathfrak n^ t \subset \mathfrak mS$ for some $t \geq 1$ then we do have $S^\wedge = \mathop{\mathrm{lim}}\nolimits S/\mathfrak m^ nS$ by Lemma 10.95.9. In some cases this even implies that $S^\wedge $ is finite over $R^\wedge $.

Lemma 10.96.7. Let $R \to S$ be a local homomorphism of local rings $(R, \mathfrak m)$ and $(S, \mathfrak n)$. Let $R^\wedge $, resp. $S^\wedge $ be the completion of $R$, resp. $S$ with respect to $\mathfrak m$, resp. $\mathfrak n$. If $\mathfrak m$ and $\mathfrak n$ are finitely generated and $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $, then

$S^\wedge $ is equal to the $\mathfrak m$-adic completion of $S$, and

$S^\wedge $ is a finite $R^\wedge $-module.

**Proof.**
We have $\mathfrak mS \subset \mathfrak n$ because $R \to S$ is a local ring map. The assumption $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $ implies that $S/\mathfrak mS$ is an Artinian ring, see Lemma 10.52.2. Hence has dimension $0$, see Lemma 10.59.4, hence $\mathfrak n = \sqrt{\mathfrak mS}$. This and the fact that $\mathfrak n$ is finitely generated implies that $\mathfrak n^ t \subset \mathfrak mS$ for some $t \geq 1$. By Lemma 10.95.9 we see that $S^\wedge $ can be identified with the $\mathfrak m$-adic completion of $S$. As $\mathfrak m$ is finitely generated we see from Lemma 10.95.3 that $S^\wedge $ and $R^\wedge $ are $\mathfrak m$-adically complete. At this point we may apply Lemma 10.95.12 to $S^\wedge $ as an $R^\wedge $-module to conclude.
$\square$

Lemma 10.96.8. Let $R$ be a Noetherian ring. Let $R \to S$ be a finite ring map. Let $\mathfrak p \subset R$ be a prime and let $\mathfrak q_1, \ldots , \mathfrak q_ m$ be the primes of $S$ lying over $\mathfrak p$ (Lemma 10.35.21). Then

\[ R_\mathfrak p^\wedge \otimes _ R S = (S_\mathfrak p)^\wedge = S_{\mathfrak q_1}^\wedge \times \ldots \times S_{\mathfrak q_ m}^\wedge \]

where the $(S_\mathfrak p)^\wedge $ is the completion with respect to $\mathfrak p$ and the local rings $R_\mathfrak p$ and $S_{\mathfrak q_ i}$ are completed with respect to their maximal ideals.

**Proof.**
The first equality follows from Lemma 10.96.1. We may replace $R$ by the localization $R_\mathfrak p$ and $S$ by $S_\mathfrak p = S \otimes _ R R_\mathfrak p$. Hence we may assume that $R$ is a local Noetherian ring and that $\mathfrak p = \mathfrak m$ is its maximal ideal. The $\mathfrak q_ iS_{\mathfrak q_ i}$-adic completion $S_{\mathfrak q_ i}^\wedge $ is equal to the $\mathfrak m$-adic completion by Lemma 10.96.7. For every $n \geq 1$ prime ideals of $S/\mathfrak m^ nS$ are in 1-to-1 correspondence with the maximal ideals $\mathfrak q_1, \ldots , \mathfrak q_ m$ of $S$ (by going up for $S$ over $R$, see Lemma 10.35.22). Hence $S/\mathfrak m^ nS = \prod S_{\mathfrak q_ i}/\mathfrak m^ nS_{\mathfrak q_ i}$ by Lemma 10.52.6 (using for example Proposition 10.59.6 to see that $S/\mathfrak m^ nS$ is Artinian). Hence the $\mathfrak m$-adic completion $S^\wedge $ of $S$ is equal to $\prod S_{\mathfrak q_ i}^\wedge $. Finally, we have $R^\wedge \otimes _ R S = S^\wedge $ by Lemma 10.96.1.
$\square$

Lemma 10.96.9. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $0 \to K \to P \to M \to 0$ be a short exact sequence of $R$-modules. If $M$ is flat over $R$ and $M/IM$ is a projective $R/I$-module, then the sequence of $I$-adic completions

\[ 0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0 \]

is a split exact sequence.

**Proof.**
As $M$ is flat, each of the sequences

\[ 0 \to K/I^ nK \to P/I^ nP \to M/I^ nM \to 0 \]

is short exact, see Lemma 10.38.12 and the sequence $0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0$ is a short exact sequence, see Lemma 10.95.1. It suffices to show that we can find splittings $s_ n : M/I^ nM \to P/I^ nP$ such that $s_{n + 1} \bmod I^ n = s_ n$. We will construct these $s_ n$ by induction on $n$. Pick any splitting $s_1$, which exists as $M/IM$ is a projective $R/I$-module. Assume given $s_ n$ for some $n > 0$. Set $P_{n + 1} = \{ x \in P \mid x \bmod I^ nP \in \mathop{\mathrm{Im}}(s_ n)\} $. The map $\pi : P_{n + 1}/I^{n + 1}P_{n + 1} \to M/I^{n + 1}M$ is surjective (details omitted). As $M/I^{n + 1}M$ is projective as a $R/I^{n + 1}$-module by Lemma 10.76.6 we may choose a section $t : M/I^{n + 1}M \to P_{n + 1}/I^{n + 1}P_{n + 1}$ of $\pi $. Setting $s_{n + 1}$ equal to the composition of $t$ with the canonical map $P_{n + 1}/I^{n + 1}P_{n + 1} \to P/I^{n + 1}P$ works.
$\square$

Lemma 10.96.10. Let $A$ be a Noetherian ring. Let $I, J \subset A$ be ideals. If $A$ is $I$-adically complete and $A/I$ is $J$-adically complete, then $A$ is $J$-adically complete.

**Proof.**
Let $B$ be the $(I + J)$-adic completion of $A$. By Lemma 10.96.2 $B/IB$ is the $J$-adic completion of $A/I$ hence isomorphic to $A/I$ by assumption. Moreover $B$ is $I$-adically complete by Lemma 10.95.8. Hence $B$ is a finite $A$-module by Lemma 10.95.12. By Nakayama's lemma (Lemma 10.19.1 using $I$ is in the radical of $A$ by Lemma 10.95.6) we find that $A \to B$ is surjective. The map $A \to B$ is flat by Lemma 10.96.2. The image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ contains $V(I)$ and as $I$ is contained in the radical of $A$ we find $A \to B$ is faitfully flat (Lemma 10.38.16). Thus $A \to B$ is injective. Thus $A$ is complete with respect to $I + J$, hence a fortiori complete with respect to $J$.
$\square$

## Comments (2)

Comment #3389 by Vignesh on

Comment #3390 by Vignesh on