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

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

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

**Proof.**
Setting $M = N/K$ we find that part (1) follows from part (2). Let $0 \to K \to N \to M \to 0$ be as in (2). For each $n$ we get the short exact sequence

\[ 0 \to K/(I^ nN \cap K) \to N/I^ nN \to M/I^ nM \to 0. \]

By Lemma 10.86.1 we obtain the exact sequence

\[ 0 \to \mathop{\mathrm{lim}}\nolimits K/(I^ nN \cap K) \to N^\wedge \to M^\wedge \to 0. \]

By the Artin-Rees Lemma 10.50.2 we may choose $c$ such that $I^ nK \subset I^ n N \cap K \subset I^{n-c} K$ for $n \geq c$. Hence $K^\wedge = \mathop{\mathrm{lim}}\nolimits K/I^ nK = \mathop{\mathrm{lim}}\nolimits K/(I^ nN \cap K)$ and we conclude that (2) is true.

Let $M$ be as in (3) and let $0 \to K \to R^{\oplus t} \to M \to 0$ be a presentation of $M$. We get a commutative diagram

\[ \xymatrix{ & K \otimes _ R R^\wedge \ar[r] \ar[d] & R^{\oplus t} \otimes _ R R^\wedge \ar[r] \ar[d] & M \otimes _ R R^\wedge \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K^\wedge \ar[r] & (R^{\oplus t})^\wedge \ar[r] & M^\wedge \ar[r] & 0 } \]

The top row is exact, see Section 10.38. The bottom row is exact by part (2). By Lemma 10.95.1 the vertical arrows are surjective. The middle vertical arrow is an isomorphism. We conclude (3) holds by the Snake Lemma 10.4.1.
$\square$

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 $J \otimes _ R R^\wedge \to R \otimes _ R R^\wedge = R^\wedge $ where $J$ is an arbitrary ideal of $R$. According to Lemma 10.96.1 this is identified with $J^\wedge \to R^\wedge $ and $J^\wedge \to R^\wedge $ is injective. Part (1) follows from Lemma 10.38.5. Part (2) is a reformulation of 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 $J$.
$\square$

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