## Tag `06LD`

## 15.25. Completion and flatness

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

Lemma 15.25.1. 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 15.25.4 below.

Lemma 15.25.2. 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 15.25.1 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 10.89.5 (and Algebra, Lemma 10.89.4) we see that $\prod_{\alpha \in A} R$ is flat. $\square$Lemma 15.25.3. 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.

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 15.25.4. 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

- $M_n$ is a flat $A/I^n$-module,
- $M_{n + 1} \to M_n$ 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 15.25.3 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 15.25.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

- $I$ is finitely generated,
- $R/I$ is Noetherian,
- $M/IM$ is flat over $R/I$,
- $\text{Tor}_1^R(M, R/I) = 0$.
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 15.25.4 we conclude that $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^nM$ is flat as an $R^\wedge$-module. $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 5321–5564 (see updates for more information).

```
\section{Completion and flatness}
\label{section-completion-flat}
\noindent
In this section we discuss when the completion of a ``big'' flat module
is flat.
\begin{lemma}
\label{lemma-ui-completion-direct-sum-into-product}
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.
\end{lemma}
\begin{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.
\medskip\noindent
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 \ref{algebra-theorem-universally-exact-criteria}.
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 \ref{algebra-lemma-Artin-Rees})
there exists a constant $c$ such that
$\Im(R^{\oplus k} \to R^{\oplus m}) \cap (I^N)^{\oplus m}
\subset \Im((I^{N - c})^{\oplus k} \to R^{\oplus m})$.
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.
\end{proof}
\noindent
The following lemma can also be deduced from
Lemma \ref{lemma-limit-flat} below.
\begin{lemma}
\label{lemma-completed-direct-sum-flat}
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.
\end{lemma}
\begin{proof}
Denote $R^\wedge$ the completion of $R$ with respect to $I$. As
$R \to R^\wedge$ is flat by
Algebra, Lemma \ref{algebra-lemma-completion-flat}
it suffices to prove that
$(\bigoplus\nolimits_{\alpha \in A} R)^\wedge$ is a flat
$R^\wedge$-module (use
Algebra, Lemma \ref{algebra-lemma-composition-flat}).
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 \ref{algebra-lemma-completion-complete}).
In this case
Lemma \ref{lemma-ui-completion-direct-sum-into-product}
tells us the map
$(\bigoplus\nolimits_{\alpha \in A} R)^\wedge \to \prod_{\alpha \in A} R$
is universally injective. Thus, by
Algebra, Lemma \ref{algebra-lemma-ui-flat-domain}
it suffices to show that $\prod_{\alpha \in A} R$ is flat. By
Algebra, Proposition \ref{algebra-proposition-characterize-coherent}
(and
Algebra, Lemma \ref{algebra-lemma-Noetherian-coherent})
we see that $\prod_{\alpha \in A} R$ is flat.
\end{proof}
\begin{lemma}
\label{lemma-tor-strictly-pro-zero}
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.
\end{lemma}
\begin{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 \ref{algebra-lemma-Artin-Rees})
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.
\end{proof}
\begin{lemma}
\label{lemma-limit-flat}
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
\begin{enumerate}
\item $M_n$ is a flat $A/I^n$-module,
\item $M_{n + 1} \to M_n$ is surjective.
\end{enumerate}
Then $M = \lim M_n$ is a flat $A$-module and
$Q \otimes_A M = \lim Q \otimes_A M_n$ for every finite $A$-module $Q$.
\end{lemma}
\begin{proof}
We first show that $Q \otimes_A M = \lim Q \otimes_A M_n$ 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 \ref{lemma-tor-strictly-pro-zero}
we see that this system is essentially constant (with value $0$).
It follows from Homology, Lemma \ref{homology-lemma-apply-Mittag-Leffler-again}
that $\lim Q \otimes_A A/I^n =
\Coker(\lim F_1 \otimes_A M_n \to \lim F_0 \otimes_A M_n)$.
Since $F_i$ is finite free this equals
$\Coker(F_1 \otimes_A M \to F_0 \otimes_A M) = Q \otimes_A M$.
\medskip\noindent
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 \ref{algebra-lemma-flat}). By the above we see
$$
\Ker(Q \otimes_A M \to Q' \otimes_A M) =
\Ker(\lim Q \otimes_A M_n \to \lim Q' \otimes_A M_n).
$$
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 $Q'' = \Coker(Q \to Q')$. Above we have seen that the
inverse systems of Tor's are essentially constant with value $0$.
It follows from
Homology, Lemma \ref{homology-lemma-apply-Mittag-Leffler-again}
that the inverse limit of the right most maps is injective.
\end{proof}
\begin{lemma}
\label{lemma-flat-after-completion}
Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be
an $R$-module. Assume
\begin{enumerate}
\item $I$ is finitely generated,
\item $R/I$ is Noetherian,
\item $M/IM$ is flat over $R/I$,
\item $\text{Tor}_1^R(M, R/I) = 0$.
\end{enumerate}
Then the $I$-adic completion $R^\wedge$
is a Noetherian ring and $M^\wedge$ is flat over $R^\wedge$.
\end{lemma}
\begin{proof}
By Algebra, Lemma \ref{algebra-lemma-what-does-it-mean}
the modules $M/I^nM$ are flat over $R/I^n$ for all $n$.
By Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated} 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 \ref{algebra-lemma-completion-Noetherian}
the ring $R^\wedge$ is Noetherian.
Applying Lemma \ref{lemma-limit-flat} we conclude that
$M^\wedge = \lim M/I^nM$ is flat as an $R^\wedge$-module.
\end{proof}
```

## Comments (0)

## Add a comment on tag `06LD`

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.