# The Stacks Project

## Tag 08XB

#### 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 $\mathop{\rm Coker}(F \to M) \otimes_R S$ 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. $\square$

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 $\mathop{\rm Coker}(F \to A) \otimes_R S$ is zero and hence $\mathop{\rm Coker}(F \to A)$ 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}

There are no comments yet for this tag.

There are also 4 comments on Section 34.4: Descent.

## Add a comment on tag 08XB

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).