The Stacks project

Theorem 35.4.25. If $M \otimes _ R S$ has one of the following properties as an $S$-module

  1. finitely generated;

  2. finitely presented;

  3. flat;

  4. faithfully flat;

  5. 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 35.4.10 the map $C(M \otimes _ R S) \to C(M'' \otimes _ R S)$ is injective. If $M \otimes _ R S$ is flat, then Lemma 35.4.24 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 35.4.12, 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 ( 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 35.4.10,

\[ 0 \to M' \otimes _ R N \to M'' \otimes _ R N \to M \otimes _ R N \to 0 \]

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 10.75.8. 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 10.78.2. $\square$

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