
## 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 10.81.3. 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 10.50.2) 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 10.96.2 it suffices to prove that $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge$ is a flat $R^\wedge$-module (use Algebra, Lemma 10.38.4). 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 10.96.4). 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 10.81.7 it suffices to show that $\prod _{\alpha \in A} R$ is flat. By Algebra, Proposition 10.89.6 (and Algebra, Lemma 10.89.5) 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 $\text{Tor}_ p^ A(M, A/I^{n + c}) \to \text{Tor}_ p^ A(M, A/I^ n)$ is zero for all $n \geq 1$.

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 10.50.2) there is a constant $c \geq 0$ such that $K \cap (I^{n + c})^{\oplus t} \subset I^ nK$. 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 12.28.7 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 10.38.5). 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 12.28.7 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 10.98.8 the modules $M/I^ nM$ are flat over $R/I^ n$ for all $n$. By Algebra, Lemma 10.95.3 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 10.96.5 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. $\square$

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