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Tag 023N

Chapter 34: Descent > Section 34.3: Descent for modules

Effective descent for modules along faithfully flat ring maps.

Proposition 34.3.9. Let $R \to A$ be a faithfully flat ring map. Then

  1. any descent datum on modules with respect to $R \to A$ is effective,
  2. the functor $M \mapsto (A \otimes_R M, can)$ from $R$-modules to the category of descent data is an equivalence, and
  3. the inverse functor is given by $(N, \varphi) \mapsto H^0(s(N_\bullet))$.

Proof. We only prove (1) and omit the proofs of (2) and (3). As $R \to A$ is faithfully flat, there exists a faithfully flat base change $R \to R'$ such that $R' \to A' = R' \otimes_R A$ has a section (namely take $R' = A$ as in the proof of Lemma 34.3.6). Hence, using Lemma 34.3.8 we may assume that $R \to A$ as a section, say $\sigma : A \to R$. Let $(N, \varphi)$ be a descent datum relative to $R \to A$. Set $$ M = H^0(s(N_\bullet)) = \{n \in N \mid 1 \otimes n = \varphi(n \otimes 1)\} \subset N $$ By Lemma 34.3.7 it suffices to show that $A \otimes_R M \to N$ is an isomorphism.

Take an element $n \in N$. Write $\varphi(n \otimes 1) = \sum a_i \otimes x_i$ for certain $a_i \in A$ and $x_i \in N$. By Lemma 34.3.2 we have $n = \sum a_i x_i$ in $N$ (because $\sigma^0_0 \circ \delta^1_0 = \text{id}$ in any cosimplicial object). Next, write $\varphi(x_i \otimes 1) = \sum a_{ij} \otimes y_j$ for certain $a_{ij} \in A$ and $y_j \in N$. The cocycle condition means that $$ \sum a_i \otimes a_{ij} \otimes y_j = \sum a_i \otimes 1 \otimes x_i $$ in $A \otimes_R A \otimes_R N$. We conclude two things from this. First, by applying $\sigma$ to the first $A$ we conclude that $\sum \sigma(a_i) \varphi(x_i \otimes 1) = \sum \sigma(a_i) \otimes x_i$ which means that $\sum \sigma(a_i) x_i \in M$. Next, by applying $\sigma$ to the middle $A$ and multiplying out we conclude that $\sum_i a_i (\sum_j \sigma(a_{ij}) y_j) = \sum a_i x_i = n$. Hence by the first conclusion we see that $A \otimes_R M \to N$ is surjective. Finally, suppose that $m_i \in M$ and $\sum a_i m_i = 0$. Then we see by applying $\varphi$ to $\sum a_im_i \otimes 1$ that $\sum a_i \otimes m_i = 0$. In other words $A \otimes_R M \to N$ is injective and we win. $\square$

    The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 641–655 (see updates for more information).

    \begin{proposition}
    \label{proposition-descent-module}
    \begin{slogan}
    Effective descent for modules along faithfully flat ring maps.
    \end{slogan}
    Let $R \to A$ be a faithfully flat ring map.
    Then
    \begin{enumerate}
    \item any descent datum on modules with respect to $R \to A$
    is effective,
    \item the functor $M \mapsto (A \otimes_R M, can)$ from $R$-modules
    to the category of descent data is an equivalence, and
    \item the inverse functor is given by $(N, \varphi) \mapsto H^0(s(N_\bullet))$.
    \end{enumerate}
    \end{proposition}
    
    \begin{proof}
    We only prove (1) and omit the proofs of (2) and (3).
    As $R \to A$ is faithfully flat, there exists a faithfully flat
    base change $R \to R'$ such that $R' \to A' = R' \otimes_R A$ has
    a section (namely take $R' = A$ as in the proof of
    Lemma \ref{lemma-ff-exact}). Hence, using
    Lemma \ref{lemma-descent-descends}
    we may assume that $R \to A$ as a section, say $\sigma : A \to R$.
    Let $(N, \varphi)$ be a descent datum relative to $R \to A$.
    Set
    $$
    M = H^0(s(N_\bullet)) = \{n \in N \mid 1 \otimes n = \varphi(n \otimes 1)\}
    \subset
    N
    $$
    By Lemma \ref{lemma-recognize-effective} it suffices to show that
    $A \otimes_R M \to N$ is an isomorphism.
    
    \medskip\noindent
    Take an element $n \in N$. Write
    $\varphi(n \otimes 1) = \sum a_i \otimes x_i$ for certain
    $a_i \in A$ and $x_i \in N$. By Lemma \ref{lemma-descent-datum-cosimplicial}
    we have $n = \sum a_i x_i$ in $N$ (because
    $\sigma^0_0 \circ \delta^1_0 = \text{id}$ in any cosimplicial object).
    Next, write $\varphi(x_i \otimes 1) = \sum a_{ij} \otimes y_j$ for
    certain $a_{ij} \in A$ and $y_j \in N$.
    The cocycle condition means that
    $$
    \sum a_i \otimes a_{ij} \otimes y_j = \sum a_i \otimes 1 \otimes x_i
    $$
    in $A \otimes_R A \otimes_R N$. We conclude two things from this.
    First, by applying $\sigma$ to the first $A$ we conclude that
    $\sum \sigma(a_i) \varphi(x_i \otimes 1) = \sum \sigma(a_i) \otimes x_i$
    which means that $\sum \sigma(a_i) x_i \in M$. Next, by applying
    $\sigma$ to the middle $A$ and multiplying out we conclude that
    $\sum_i a_i (\sum_j \sigma(a_{ij}) y_j) = \sum a_i x_i = n$. Hence
    by the first conclusion we see that $A \otimes_R M \to N$ is
    surjective. Finally, suppose that $m_i \in M$ and
    $\sum a_i m_i = 0$. Then we see by applying $\varphi$ to
    $\sum a_im_i \otimes 1$ that $\sum a_i \otimes m_i = 0$.
    In other words $A \otimes_R M \to N$ is injective and we win.
    \end{proof}

    Comments (1)

    Comment #874 by Bhargav Bhatt on August 3, 2014 a 6:46 pm UTC

    Suggested slogan: One has effective descent for modules along faithfully flat ring maps.

    There are also 2 comments on Section 34.3: Descent.

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