## 15.27 Completion and flatness

In this section we discuss when the completion of a “big” flat module is flat.

Lemma 15.27.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $A$ be a set. Assume $R$ is Noetherian and complete with respect to $I$. There is a canonical map

$\left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \longrightarrow \prod \nolimits _{\alpha \in A} R$

from the $I$-adic completion of the direct sum into the product which is universally injective.

Proof. By definition an element $x$ of the left hand side is $x = (x_ n)$ where $x_ n = (x_{n, \alpha }) \in \bigoplus \nolimits _{\alpha \in A} R/I^ n$ such that $x_{n, \alpha } = x_{n + 1, \alpha } \bmod I^ n$. As $R = R^\wedge$ we see that for any $\alpha$ there exists a $y_\alpha \in R$ such that $x_{n, \alpha } = y_\alpha \bmod I^ n$. Note that for each $n$ there are only finitely many $\alpha$ such that the elements $x_{n, \alpha }$ are nonzero. Conversely, given $(y_\alpha ) \in \prod _\alpha R$ such that for each $n$ there are only finitely many $\alpha$ such that $y_{\alpha } \bmod I^ n$ is nonzero, then this defines an element of the left hand side. Hence we can think of an element of the left hand side as infinite “convergent sums” $\sum _\alpha y_\alpha$ with $y_\alpha \in R$ such that for each $n$ there are only finitely many $y_\alpha$ which are nonzero modulo $I^ n$. The displayed map maps this element to the element to $(y_\alpha )$ in the product. In particular the map is injective.

Let $Q$ be a finite $R$-module. We have to show that the map

$Q \otimes _ R \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \longrightarrow Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right)$

is injective, see Algebra, Theorem 10.82.3. Choose a presentation $R^{\oplus k} \to R^{\oplus m} \to Q \to 0$ and denote $q_1, \ldots , q_ m \in Q$ the corresponding generators for $Q$. By Artin-Rees (Algebra, Lemma 10.51.2) there exists a constant $c$ such that $\mathop{\mathrm{Im}}(R^{\oplus k} \to R^{\oplus m}) \cap (I^ N)^{\oplus m} \subset \mathop{\mathrm{Im}}((I^{N - c})^{\oplus k} \to R^{\oplus m})$. Let us contemplate the diagram

$\xymatrix{ \bigoplus _{l = 1}^ k \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & \bigoplus _{j = 1}^ m \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & Q \otimes _ R \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & 0 \\ \bigoplus _{l = 1}^ k \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & \bigoplus _{j = 1}^ m \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & 0 }$

with exact rows. Pick an element $\sum _ j \sum _\alpha y_{j, \alpha }$ of $\bigoplus _{j = 1, \ldots , m} \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge$. If this element maps to zero in the module $Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right)$, then we see in particular that $\sum _ j q_ j \otimes y_{j, \alpha } = 0$ in $Q$ for each $\alpha$. Thus we can find an element $(z_{1, \alpha }, \ldots , z_{k, \alpha }) \in \bigoplus _{l = 1, \ldots , k} R$ which maps to $(y_{1, \alpha }, \ldots , y_{m, \alpha }) \in \bigoplus _{j = 1, \ldots , m} R$. Moreover, if $y_{j, \alpha } \in I^{N_\alpha }$ for $j = 1, \ldots , m$, then we may assume that $z_{l, \alpha } \in I^{N_\alpha - c}$ for $l = 1, \ldots , k$. Hence the sum $\sum _ l \sum _\alpha z_{l, \alpha }$ is “convergent” and defines an element of $\bigoplus _{l = 1, \ldots , k} \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge$ which maps to the element $\sum _ j \sum _\alpha y_{j, \alpha }$ we started out with. Thus the right vertical arrow is injective and we win. $\square$

The following lemma can also be deduced from Lemma 15.27.4 below.

Lemma 15.27.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $A$ be a set. Assume $R$ is Noetherian. The completion $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge$ is a flat $R$-module.

Proof. Denote $R^\wedge$ the completion of $R$ with respect to $I$. As $R \to R^\wedge$ is flat by Algebra, Lemma 10.97.2 it suffices to prove that $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge$ is a flat $R^\wedge$-module (use Algebra, Lemma 10.39.4). Since

$(\bigoplus \nolimits _{\alpha \in A} R)^\wedge = (\bigoplus \nolimits _{\alpha \in A} R^\wedge )^\wedge$

we may replace $R$ by $R^\wedge$ and assume that $R$ is complete with respect to $I$ (see Algebra, Lemma 10.97.4). In this case Lemma 15.27.1 tells us the map $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge \to \prod _{\alpha \in A} R$ is universally injective. Thus, by Algebra, Lemma 10.82.7 it suffices to show that $\prod _{\alpha \in A} R$ is flat. By Algebra, Proposition 10.90.6 (and Algebra, Lemma 10.90.5) we see that $\prod _{\alpha \in A} R$ is flat. $\square$

Lemma 15.27.3. Let $A$ be a Noetherian ring. Let $I$ be an ideal of $A$. Let $M$ be a finite $A$-module. For every $p > 0$ there exists a $c > 0$ such that $\text{Tor}_ p^ A(M, A/I^ n) \to \text{Tor}_ p^ A(M, A/I^{n - c})$ is zero for all $n \geq c$.

Proof. Proof for $p = 1$. Choose a short exact sequence $0 \to K \to R^{\oplus t} \to M \to 0$. Then $\text{Tor}_1^ A(M, A/I^ n) = K \cap (I^ n)^{\oplus t}/I^ nK$. By Artin-Rees (Algebra, Lemma 10.51.2) there is a constant $c \geq 0$ such that $K \cap (I^ n)^{\oplus t} \subset I^{n - c}K$ for $n \geq c$. Thus the result for $p = 1$. For $p > 1$ we have $\text{Tor}_ p^ A(M, A/I^ n) = \text{Tor}^ A_{p - 1}(K, A/I^ n)$. Thus the lemma follows by induction. $\square$

Lemma 15.27.4. Let $A$ be a Noetherian ring. Let $I$ be an ideal of $A$. Let $(M_ n)$ be an inverse system of $A$-modules such that

1. $M_ n$ is a flat $A/I^ n$-module,

2. $M_{n + 1} \to M_ n$ is surjective.

Then $M = \mathop{\mathrm{lim}}\nolimits M_ n$ is a flat $A$-module and $Q \otimes _ A M = \mathop{\mathrm{lim}}\nolimits Q \otimes _ A M_ n$ for every finite $A$-module $Q$.

Proof. We first show that $Q \otimes _ A M = \mathop{\mathrm{lim}}\nolimits Q \otimes _ A M_ n$ for every finite $A$-module $Q$. Choose a resolution $F_2 \to F_1 \to F_0 \to Q \to 0$ by finite free $A$-modules $F_ i$. Then

$F_2 \otimes _ A M_ n \to F_1 \otimes _ A M_ n \to F_0 \otimes _ A M_ n$

is a chain complex whose homology in degree $0$ is $Q \otimes _ A M_ n$ and whose homology in degree $1$ is

$\text{Tor}_1^ A(Q, M_ n) = \text{Tor}_1^ A(Q, A/I^ n) \otimes _{A/I^ n} M_ n$

as $M_ n$ is flat over $A/I^ n$. By Lemma 15.27.3 we see that this system is essentially constant (with value $0$). It follows from Homology, Lemma 12.31.7 that $\mathop{\mathrm{lim}}\nolimits Q \otimes _ A A/I^ n = \mathop{\mathrm{Coker}}(\mathop{\mathrm{lim}}\nolimits F_1 \otimes _ A M_ n \to \mathop{\mathrm{lim}}\nolimits F_0 \otimes _ A M_ n)$. Since $F_ i$ is finite free this equals $\mathop{\mathrm{Coker}}(F_1 \otimes _ A M \to F_0 \otimes _ A M) = Q \otimes _ A M$.

Next, let $Q \to Q'$ be an injective map of finite $A$-modules. We have to show that $Q \otimes _ A M \to Q' \otimes _ A M$ is injective (Algebra, Lemma 10.39.5). By the above we see

$\mathop{\mathrm{Ker}}(Q \otimes _ A M \to Q' \otimes _ A M) = \mathop{\mathrm{Ker}}(\mathop{\mathrm{lim}}\nolimits Q \otimes _ A M_ n \to \mathop{\mathrm{lim}}\nolimits Q' \otimes _ A M_ n).$

For each $n$ we have an exact sequence

$\text{Tor}_1^ A(Q', M_ n) \to \text{Tor}_1^ A(Q'', M_ n) \to Q \otimes _ A M_ n \to Q' \otimes _ A M_ n$

where $Q'' = \mathop{\mathrm{Coker}}(Q \to Q')$. Above we have seen that the inverse systems of Tor's are essentially constant with value $0$. It follows from Homology, Lemma 12.31.7 that the inverse limit of the right most maps is injective. $\square$

Lemma 15.27.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

1. $I$ is finitely generated,

2. $R/I$ is Noetherian,

3. $M/IM$ is flat over $R/I$,

4. $\text{Tor}_1^ R(M, R/I) = 0$.

Then the $I$-adic completion $R^\wedge$ is a Noetherian ring and $M^\wedge$ is flat over $R^\wedge$.

Proof. By Algebra, Lemma 10.99.8 the modules $M/I^ nM$ are flat over $R/I^ n$ for all $n$. By Algebra, Lemma 10.96.3 we have (a) $R^\wedge$ and $M^\wedge$ are $I$-adically complete and (b) $R/I^ n = R^\wedge /I^ nR^\wedge$ for all $n$. By Algebra, Lemma 10.97.5 the ring $R^\wedge$ is Noetherian. Applying Lemma 15.27.4 we conclude that $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is flat as an $R^\wedge$-module. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

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 06LD. Beware of the difference between the letter 'O' and the digit '0'.