The Stacks Project


Tag 08XD

Chapter 34: Descent > Section 34.4: Descent for universally injective morphisms

Theorem 34.4.25. If $M \otimes_R S$ has one of the following properties as an $S$-module

  1. (a)    finitely generated;
  2. (b)    finitely presented;
  3. (c)    flat;
  4. (d)    faithfully flat;
  5. (e)    finite projective;

then so does $M$ as an $R$-module (and conversely).

Proof. To prove (a), choose a finite set $\{n_i\}$ of generators of $M \otimes_R S$ in $\text{Mod}_S$. Write each $n_i$ as $\sum_j m_{ij} \otimes s_{ij}$ with $m_{ij} \in M$ and $s_{ij} \in S$. Let $F$ be the finite free $R$-module with basis $e_{ij}$ and let $F \to M$ be the $R$-module map sending $e_{ij}$ to $m_{ij}$. Then $F \otimes_R S\to M \otimes_R S$ is surjective, so $\mathop{\rm Coker}(F \to M) \otimes_R S$ is zero and hence $\mathop{\rm Coker}(F \to M)$ is zero. This proves (a).

To see (b) assume $M \otimes_R S$ is finitely presented. Then $M$ is finitely generated by (a). Choose a surjection $R^{\oplus n} \to M$ with kernel $K$. Then $K \otimes_R S \to S^{\oplus r} \to M \otimes_R S \to 0$ is exact. By Algebra, Lemma 10.5.3 the kernel of $S^{\oplus r} \to M \otimes_R S$ is a finite $S$-module. Thus we can find finitely many elements $k_1, \ldots, k_t \in K$ such that the images of $k_i \otimes 1$ in $S^{\oplus r}$ generate the kernel of $S^{\oplus r} \to M \otimes_R S$. Let $K' \subset K$ be the submodule generated by $k_1, \ldots, k_t$. Then $M' = R^{\oplus r}/K'$ is a finitely presented $R$-module with a morphism $M' \to M$ such that $M' \otimes_R S \to M \otimes_R S$ is an isomorphism. Thus $M' \cong M$ as desired.

To prove (c), let $0 \to M' \to M'' \to M \to 0$ be a short exact sequence in $\text{Mod}_R$. Since $\bullet \otimes_R S$ is a right exact functor, $M'' \otimes_R S \to M \otimes_R S$ is surjective. So by Lemma 34.4.10 the map $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is injective. If $M \otimes_R S$ is flat, then Lemma 34.4.24 shows $C(M \otimes_R S)$ is an injective object of $\text{Mod}_S$, so the injection $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is split in $\text{Mod}_S$ and hence also in $\text{Mod}_R$. Since $C(M \otimes_R S) \to C(M)$ is a split surjection by Lemma 34.4.12, it follows that $C(M) \to C(M'')$ is a split injection in $\text{Mod}_R$. That is, the sequence $$ 0 \to C(M) \to C(M'') \to C(M') \to 0 $$ is split exact. For $N \in \text{Mod}_R$, by (34.4.11.1) we see that $$ 0 \to C(M \otimes_R N) \to C(M'' \otimes_R N) \to C(M' \otimes_R N) \to 0 $$ is split exact. By Lemma 34.4.10, $$ 0 \to M' \otimes_R N \to M'' \otimes_R N \to M \otimes_R N \to 0 $$ is exact. This implies $M$ is flat over $R$. Namely, taking $M'$ a free module surjecting onto $M$ we conclude that $\text{Tor}_1^R(M, N) = 0$ for all modules $N$ and we can use Algebra, Lemma 10.74.8. This proves (c).

To deduce (d) from (c), note that if $N \in \text{Mod}_R$ and $M \otimes_R N$ is zero, then $M \otimes_R S \otimes_S (N \otimes_R S) \cong (M \otimes_R N) \otimes_R S$ is zero, so $N \otimes_R S$ is zero and hence $N$ is zero.

To deduce (e) at this point, it suffices to recall that $M$ is finitely generated and projective if and only if it is finitely presented and flat. See Algebra, Lemma 10.77.2. $\square$

    The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 1347–1363 (see updates for more information).

    \begin{theorem}
    \label{theorem-descend-module-properties}
    If $M \otimes_R S$ has one of the following properties as an $S$-module
    \begin{enumerate}
    \item[(a)]
    finitely generated;
    \item[(b)]
    finitely presented;
    \item[(c)]
    flat;
    \item[(d)]
    faithfully flat;
    \item[(e)]
    finite projective;
    \end{enumerate}
    then so does $M$ as an $R$-module (and conversely).
    \end{theorem}
    
    \begin{proof}
    To prove (a), choose a finite set $\{n_i\}$ of generators of $M \otimes_R S$
    in $\text{Mod}_S$. Write each $n_i$ as $\sum_j m_{ij} \otimes s_{ij}$ with
    $m_{ij} \in M$ and $s_{ij} \in S$. Let $F$ be the finite free $R$-module with
    basis $e_{ij}$ and let $F \to M$ be the $R$-module map sending $e_{ij}$ to
    $m_{ij}$. Then $F \otimes_R S\to M \otimes_R S$ is surjective, so
    $\Coker(F \to M) \otimes_R S$ is zero and hence $\Coker(F \to M)$
    is zero. This proves (a).
    
    \medskip\noindent
    To see (b) assume $M \otimes_R S$ is finitely presented. Then $M$ is finitely
    generated by (a). Choose a surjection $R^{\oplus n} \to M$ with kernel $K$.
    Then $K \otimes_R S \to S^{\oplus r} \to M \otimes_R S \to 0$ is exact.
    By Algebra, Lemma \ref{algebra-lemma-extension}
    the kernel of $S^{\oplus r} \to M \otimes_R S$
    is a finite $S$-module. Thus we can find finitely many elements
    $k_1, \ldots, k_t \in K$ such that the images of $k_i \otimes 1$ in
    $S^{\oplus r}$ generate the kernel of $S^{\oplus r} \to M \otimes_R S$.
    Let $K' \subset K$ be the submodule generated by $k_1, \ldots, k_t$.
    Then $M' = R^{\oplus r}/K'$ is a finitely presented $R$-module
    with a morphism $M' \to M$ such that $M' \otimes_R S \to M \otimes_R S$
    is an isomorphism. Thus $M' \cong M$ as desired.
    
    \medskip\noindent
    To prove (c), let $0 \to M' \to M'' \to M \to 0$ be a short exact sequence in 
    $\text{Mod}_R$. Since $\bullet \otimes_R S$ is a right exact functor,
    $M'' \otimes_R S \to M \otimes_R S$ is surjective. So by
    Lemma \ref{lemma-C-is-faithful} the map
    $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is injective.
    If $M \otimes_R S$ is flat, then
    Lemma \ref{lemma-flat-to-injective} shows
    $C(M \otimes_R S)$ is an injective object of $\text{Mod}_S$, so the injection
    $C(M \otimes_R S) \to C(M'' \otimes_R S)$
    is split in $\text{Mod}_S$ and hence also in $\text{Mod}_R$.
    Since $C(M \otimes_R S) \to C(M)$ is a split surjection by 
    Lemma \ref{lemma-split-surjection}, it follows that 
    $C(M) \to C(M'')$ is a split injection in $\text{Mod}_R$. That is, the sequence
    $$
    0 \to C(M) \to C(M'') \to C(M') \to 0
    $$
    is split exact. 
    For $N \in \text{Mod}_R$, by (\ref{equation-adjunction}) we see that 
    $$
    0 \to C(M \otimes_R N) \to C(M'' \otimes_R N) \to C(M' \otimes_R N) \to 0
    $$
    is split exact. By Lemma \ref{lemma-C-is-faithful}, 
    $$
    0 \to M' \otimes_R N \to M'' \otimes_R N \to M \otimes_R N \to 0
    $$
    is exact. This implies $M$ is flat over $R$. Namely, taking
    $M'$ a free module surjecting onto $M$ we conclude that
    $\text{Tor}_1^R(M, N) = 0$ for all modules $N$ and we can use
    Algebra, Lemma \ref{algebra-lemma-characterize-flat}.
    This proves (c).
    
    \medskip\noindent
    To deduce (d) from (c), note that if $N \in \text{Mod}_R$ and $M \otimes_R N$ 
    is zero,
    then $M \otimes_R S \otimes_S (N \otimes_R S) \cong (M \otimes_R N) \otimes_R 
    S$ is zero,
    so $N \otimes_R S$ is zero and hence $N$ is zero.
    
    \medskip\noindent
    To deduce (e) at this point, it suffices to recall that $M$ is finitely 
    generated and projective if and only if it is finitely presented and flat.
    See Algebra, Lemma \ref{algebra-lemma-finite-projective}.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    There are also 4 comments on Section 34.4: Descent.

    Add a comment on tag 08XD

    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.




    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?