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Tag 08XB

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

34.4.23. Descent for properties of modules

Throughout this subsection, fix a universally injective ring map $f : R \to S$, an object $M \in \text{Mod}_R$, and a ring map $R \to A$. We now investigate the question of which properties of $M$ or $A$ can be checked after base extension along $f$. We start with some results from [mesablishvili2].

Lemma 34.4.24. If $M \in \text{Mod}_R$ is flat, then $C(M)$ is an injective $R$-module.

Proof. Let $0 \to N \to P \to Q \to 0$ be an exact sequence in $\text{Mod}_R$. Since $M$ is flat, $$ 0 \to N \otimes_R M \to P \otimes_R M \to Q \otimes_R M \to 0 $$ is exact. By Lemma 34.4.10, $$ 0 \to C(Q \otimes_R M) \to C(P \otimes_R M) \to C(N \otimes_R M) \to 0 $$ is exact. By (34.4.11.1), this last sequence can be rewritten as $$ 0 \to \mathop{\rm Hom}\nolimits_R(Q, C(M)) \to \mathop{\rm Hom}\nolimits_R(P, C(M)) \to \mathop{\rm Hom}\nolimits_R(N, C(M)) \to 0. $$ Hence $C(M)$ is an injective object of $\text{Mod}_R$. $\square$

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 $\text{Coker}(F \to M) \otimes_R S$ is zero and hence $\text{Coker}(F \to M)$ 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. $\square$

There is a variant for $R$-algebras.

Theorem 34.4.26. If $A \otimes_R S$ has one of the following properties as an $S$-algebra

  1. (a)    of finite type;
  2. (b)    of finite presentation;
  3. (c)    formally unramified;
  4. (d)    unramified;
  5. (e)    étale;

then so does $A$ as an $R$-algebra (and of course conversely).

Proof. To prove (a), choose a finite set $\{x_i\}$ of generators of $A \otimes_R S$ over $S$. Write each $x_i$ as $\sum_j y_{ij} \otimes s_{ij}$ with $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending $e_{ij}$ to $y_{ij}$. Then $F \otimes_R S\to A \otimes_R S$ is surjective, so $\text{Coker}(F \to A) \otimes_R S$ is zero and hence $\text{Coker}(F \to A)$ is zero. This proves (a).

To see (b) assume $A \otimes_R S$ is a finitely presented $S$-algebra. Then $A$ is finite type over $R$ by (a). Choose a surjection $R[x_1, \ldots, x_n] \to A$ with kernel $I$. Then $I \otimes_R S \to S[x_1, \ldots, x_n] \to A \otimes_R S \to 0$ is exact. By Algebra, Lemma 10.6.3 the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$ is a finitely generated ideal. Thus we can find finitely many elements $y_1, \ldots, y_t \in I$ such that the images of $y_i \otimes 1$ in $S[x_1, \ldots, x_n]$ generate the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$. Let $I' \subset I$ be the ideal generated by $y_1, \ldots, y_t$. Then $A' = R[x_1, \ldots, x_n]/I'$ is a finitely presented $R$-algebra with a morphism $A' \to A$ such that $A' \otimes_R S \to A \otimes_R S$ is an isomorphism. Thus $A' \cong A$ as desired.

To prove (c), recall that $A$ is formally unramified over $R$ if and only if the module of relative differentials $\Omega_{A/R}$ vanishes, see Algebra, Lemma 10.144.2 or [EGA4, Proposition 17.2.1]. Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$, the vanishing descends by Theorem 34.4.22.

To deduce (d) from the previous cases, recall that $A$ is unramified over $R$ if and only if $A$ is formally unramified and of finite type over $R$, see Algebra, Lemma 10.147.2.

To prove (e), recall that by Algebra, Lemma 10.147.8 or [EGA4, Théorème 17.6.1] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

Remark 34.4.27. It would make things easier to have a faithfully flat ring homomorphism $g: R \to T$ for which $T \to S \otimes_R T$ has some extra structure. For instance, if one could ensure that $T \to S \otimes_R T$ is split in $\textit{Rings}$, then it would follow that every property of a module or algebra which is stable under base extension and which descends along faithfully flat morphisms also descends along universally injective morphisms. An obvious guess would be to find $g$ for which $T$ is not only faithfully flat but also injective in $\text{Mod}_R$, but even for $R = \mathbf{Z}$ no such homomorphism can exist.

    The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 1313–1531 (see updates for more information).

    \subsection{Descent for properties of modules}
    \label{subsection-descent-properties-modules}
    
    \noindent
    Throughout this subsection, fix a universally injective ring map $f : R \to S$,
    an object $M \in \text{Mod}_R$, and a ring map $R \to A$. We now investigate
    the question of which properties of $M$ or $A$ can be checked after base
    extension along $f$. We start with some results from 
    \cite{mesablishvili2}.
    
    \begin{lemma}
    \label{lemma-flat-to-injective}
    If $M \in \text{Mod}_R$ is flat, then $C(M)$ is an injective $R$-module.
    \end{lemma}
    
    \begin{proof}
    Let $0 \to N \to P \to Q \to 0$ be an exact sequence in $\text{Mod}_R$. Since 
    $M$ is flat,
    $$
    0 \to N \otimes_R M \to P \otimes_R M \to Q \otimes_R M \to 0
    $$
    is exact.
    By Lemma \ref{lemma-C-is-faithful},
    $$
    0 \to C(Q \otimes_R M) \to C(P \otimes_R M) \to C(N \otimes_R M) \to 0
    $$
    is exact. By (\ref{equation-adjunction}), this last sequence can be rewritten
    as
    $$
    0 \to \Hom_R(Q, C(M)) \to \Hom_R(P, C(M)) \to \Hom_R(N, C(M)) \to 0.
    $$
    Hence $C(M)$ is an injective object of $\text{Mod}_R$.
    \end{proof}
    
    \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}
    
    \noindent
    There is a variant for $R$-algebras.
    
    \begin{theorem}
    \label{theorem-descend-algebra-properties}
    If $A \otimes_R S$ has one of the following properties as an $S$-algebra
    \begin{enumerate}
    \item[(a)]
    of finite type;
    \item[(b)]
    of finite presentation;
    \item[(c)]
    formally unramified;
    \item[(d)]
    unramified;
    \item[(e)]
    \'etale;
    \end{enumerate}
    then so does $A$ as an $R$-algebra (and of course conversely).
    \end{theorem}
    
    \begin{proof}
    To prove (a), choose a finite set $\{x_i\}$ of generators of $A \otimes_R S$
    over $S$. Write each $x_i$ as $\sum_j y_{ij} \otimes s_{ij}$ with
    $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra
    on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending
    $e_{ij}$ to $y_{ij}$. Then $F \otimes_R S\to A \otimes_R S$ is surjective, so
    $\Coker(F \to A) \otimes_R S$ is zero and hence $\Coker(F \to A)$
    is zero. This proves (a).
    
    \medskip\noindent
    To see (b) assume $A \otimes_R S$ is a finitely presented $S$-algebra.
    Then $A$ is finite type over $R$ by (a). Choose a surjection
    $R[x_1, \ldots, x_n] \to A$ with kernel $I$.
    Then $I \otimes_R S \to S[x_1, \ldots, x_n] \to A \otimes_R S \to 0$ is exact.
    By Algebra, Lemma \ref{algebra-lemma-finite-presentation-independent}
    the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$
    is a finitely generated ideal. Thus we can find finitely many elements
    $y_1, \ldots, y_t \in I$ such that the images of $y_i \otimes 1$ in
    $S[x_1, \ldots, x_n]$ generate the kernel of
    $S[x_1, \ldots, x_n] \to A \otimes_R S$.
    Let $I' \subset I$ be the ideal generated by $y_1, \ldots, y_t$.
    Then $A' = R[x_1, \ldots, x_n]/I'$ is a finitely presented $R$-algebra
    with a morphism $A' \to A$ such that $A' \otimes_R S \to A \otimes_R S$
    is an isomorphism. Thus $A' \cong A$ as desired.
    
    \medskip\noindent
    To prove (c), recall that $A$ is formally unramified over $R$ if and only
    if the module of relative differentials $\Omega_{A/R}$ vanishes, see
    Algebra, Lemma \ref{algebra-lemma-characterize-formally-unramified} or
    \cite[Proposition~17.2.1]{EGA4}.
    Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$,
    the vanishing descends by Theorem \ref{theorem-descent}.
    
    \medskip\noindent
    To deduce (d) from the previous cases, recall that $A$ is unramified
    over $R$ if and only if $A$ is formally unramified and of finite type
    over $R$, see
    Algebra, Lemma \ref{algebra-lemma-formally-unramified-unramified}.
    
    \medskip\noindent
    To prove (e), recall that by
    Algebra, Lemma \ref{algebra-lemma-etale-flat-unramified-finite-presentation}
    or \cite[Th\'eor\`eme~17.6.1]{EGA4} the algebra
    $A$ is \'etale over $R$ if and only if
    $A$ is flat, unramified, and of finite presentation over $R$.
    \end{proof}
    
    \begin{remark}
    \label{remark-when-locally-split}
    It would make things easier to have a faithfully
    flat ring homomorphism $g: R \to T$ for which $T \to S \otimes_R T$ has some 
    extra structure.
    For instance, if one could ensure that $T \to S \otimes_R T$ is split in 
    $\textit{Rings}$,
    then it would follow that every property of a module or algebra which is stable 
    under base extension
    and which descends along faithfully flat morphisms also descends along 
    universally injective morphisms.
    An obvious guess would be to find $g$ for which $T$ is not only faithfully flat 
    but also injective in $\text{Mod}_R$,
    but even for $R = \mathbf{Z}$ no such homomorphism can exist.
    \end{remark}

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