A Faithfully flat morphisms

In this appendix, we document some important facts about flat and faithfully flat morphisms.

If $X$ is quasi-compact and quasi-separated, then $p^*$ being exact implies $p$ being flat.

We only have to show that the pullback of flat maps is flat. Since flatness is local, it suffices to show that if $f: A \to B$ is a flat map of rings and $g:A \to C$ is any map of rings, then $C \to B \otimes _A C$ is flat. But if $M_\bullet$ is a exact sequence of chain complexes, then

$$B \otimes _A C \otimes _C M_\bullet = B \otimes _A M_\bullet ,$$

where we think of $M_\bullet$ as an $A$-module via $g$. So this is exact.

• (1) $\Rightarrow$ (2): Take $h = \mathrm{id}: \mathcal{F} \to \mathcal{F}$.

• (2) $\Rightarrow$ (3): Apply (2) to the homology groups of $\mathcal{F}_\bullet$.

• (3) $\Rightarrow$ (1): $h = 0$ iff $\mathcal{F} \overset {h}{\to } \mathcal{F}' \overset {1}{\to } \mathcal{F}'$ is exact.

• (4) $\Rightarrow$ (2): Take $\mathcal{F} \not= 0$. We may assume $\mathcal{F}$ is in fact coherent, for $\mathcal{F}$ contains a coherent subsheaf $\mathcal{G}$ and $p^*$ preserves subsheaves by flatness. So if $p^* \mathcal{G} \not= 0$, then $p^* \mathcal{F} \not= 0$.

  Pick $x \in X$ such that $\mathcal{F}_x \not= 0$. By surjectivity, there is a field $k$ and a map $\tilde{x}: \operatorname{Spec}k \to X$ that sends the unique point to $x$ and has a lift to $Y$ (e.g. by first picking a map to $Y$ that hits a preimage of $x$). This means the pullback $Y \times _{\operatorname{Spec}k} X$ is non-empty. Moreover, $\tilde{x}^* \mathcal{F} \not= 0$ by Nakayama, and is free since $\operatorname{Spec}k$ is a field. So the pullback of $\mathcal{F}$ to $Y \times _{\operatorname{Spec}k} X$ is non-zero. Hence $p^* \mathcal{F} \not= 0$.

• (2) $\Rightarrow$ (4): Let $\mathfrak {p} \in X$, and $\operatorname{Spec}A \subseteq X$ an affine open containing $\mathfrak {p}$. Set $\mathcal{F}|_{\operatorname{Spec}A} = \widetilde{A_\mathfrak {p}/\mathfrak {p} A_\mathfrak {p}}$ and extend by zero. Then $p^* \mathcal{F} \not= 0$ implies there is some affine open $\operatorname{Spec}B \subseteq Y$ such that $B \otimes _A \frac{A_\mathfrak {p}}{\mathfrak {p}A_\mathfrak {p}} =B_\mathfrak {p}/\mathfrak {p}B_\mathfrak {p} \not= 0$. Then a prime of $B_\mathfrak {p}/\mathfrak {p}B_\mathfrak {p}$ is a prime of $B$ that gets mapped to $\mathfrak {p}$ under $p$.

Using (4), it is clear that

We say a property P of morphisms satisfies fpqc descent if whenever we have a pullback diagram

with $f$ faithfully flat, then $p$ has property P iff $p'$ does.

Suppose $p'$ is flat. If we have a sequence $\mathcal{F}_\bullet$ of quasi-coherent sheaves on $X$, then $p'^* f^* \mathcal{F}_\bullet$ is exact since $f$ and $p'$ are flat, and since $p'^* f^* = f'^* p^*$, we know $p^* \mathcal{F}_\bullet$ is exact by faithfulness.

[Proof sketch] The pullback of a finite morphism is clearly finite. For the other direction, We will prove the affine version. The gluing step part (which is the hard step) is annoying and will be omitted.

Suppose that $f: R \to S$ is faithfully flat and $M$ is an $R$-module. We want to show that $M \otimes _R S$ being finitely-generated implies $M$ is finitely generated.

Suppose $y_1, \ldots y_m$ generate $M \otimes _R S$, and $y_j = \sum x_{i, j} \otimes f_{i, j}$. Then the $x_{i, j}$ generate $M$, since they generate $M \otimes _R S$ as an $S$-module and $R \to S$ is faithfully flat, hence reflects surjectivity.