## 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

- (a) finitely generated;
- (b) finitely presented;
- (c) flat;
- (d) faithfully flat;
- (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 $\mathop{\mathrm{Coker}}(F \to M) \otimes_R S$ is zero and hence $\mathop{\mathrm{Coker}}(F \to M)$ 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}
```

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