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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 $\mathop{\rm Im}(R^{\oplus k} \to R^{\oplus m}) \cap (I^N)^{\oplus m} \subset \mathop{\rm Im}((I^{N - c})^{\oplus k} \to R^{\oplus m})$. 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

  1. $M_n$ is a flat $A/I^n$-module,
  2. $M_{n + 1} \to M_n$ is surjective.

Then $M = \mathop{\rm lim}\nolimits M_n$ is a flat $A$-module and $Q \otimes_A M = \mathop{\rm lim}\nolimits Q \otimes_A M_n$ for every finite $A$-module $Q$.

Proof. We first show that $Q \otimes_A M = \mathop{\rm lim}\nolimits Q \otimes_A M_n$ 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 $\mathop{\rm lim}\nolimits Q \otimes_A A/I^n = \mathop{\rm Coker}(\mathop{\rm lim}\nolimits F_1 \otimes_A M_n \to \mathop{\rm lim}\nolimits F_0 \otimes_A M_n)$. Since $F_i$ is finite free this equals $\mathop{\rm Coker}(F_1 \otimes_A M \to F_0 \otimes_A M) = Q \otimes_A M$.

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 $$ \mathop{\rm Ker}(Q \otimes_A M \to Q' \otimes_A M) = \mathop{\rm Ker}(\mathop{\rm lim}\nolimits Q \otimes_A M_n \to \mathop{\rm lim}\nolimits Q' \otimes_A M_n). $$ 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 $Q'' = \mathop{\rm Coker}(Q \to Q')$. 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

  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 15.25.4 we conclude that $M^\wedge = \mathop{\rm lim}\nolimits M/I^nM$ 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 5218–5461 (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}

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