Lemma 35.4.24. If $M \in \text{Mod}_ R$ is flat, then $C(M)$ is an injective $R$-module.

### 35.4.23 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].

**Proof.**
Let $0 \to N \to P \to Q \to 0$ be an exact sequence in $\text{Mod}_ R$. Since $M$ is flat,

is exact. By Lemma 35.4.10,

is exact. By (35.4.11.1), this last sequence can be rewritten as

Hence $C(M)$ is an injective object of $\text{Mod}_ R$. $\square$

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

finitely generated;

finitely presented;

flat;

faithfully flat;

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

is split exact. For $N \in \text{Mod}_ R$, by (35.4.11.1) we see that

is split exact. By Lemma 35.4.10,

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. $\square$

There is a variant for $R$-algebras.

Theorem 35.4.26. If $A \otimes _ R S$ has one of the following properties as an $S$-algebra

of finite type;

of finite presentation;

formally unramified;

unramified;

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 $\mathop{\mathrm{Coker}}(F \to A) \otimes _ R S$ is zero and hence $\mathop{\mathrm{Coker}}(F \to A)$ 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 10.148.2 or [Proposition 17.2.1, EGA4]. Since $\Omega _{(A \otimes _ R S)/S} = \Omega _{A/R} \otimes _ R S$, the vanishing descends by Theorem 35.4.22.

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 10.151.2.

To prove (e), recall that by Algebra, Lemma 10.151.8 or [Théorème 17.6.1, EGA4] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

Remark 35.4.27. 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.

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