# A Faithfully flat morphisms

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

Let $f: A \to B$ be a ring homomorphism. We say $f$ is flat if the functor

$-\otimes _A B: A\text{-Mod} \to B\text{-Mod}$is exact.

A morphism $p: Y \to X$ is flat if for all $y \in Y$ and $x = f(y)$, the map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is flat. This in particular implies the pullback functor

$p^*: \operatorname{QCoh}(Y) \to \operatorname{QCoh}(X)$is exact.

Compositions and pullbacks of flat maps are flat.

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

Let $p: Y \to X$ be flat. Then the following are equivalent:

$p^*$ is faithful, i.e. if $h: \mathcal{F} \to \mathcal{F}'$ is a morphism of quasi-coherent sheaves over $Y$, and $p^*h = 0$, then $h = 0$.

If $p^* \mathcal{F} = 0$, then $\mathcal{F} = 0$.

$p^*$ reflects exactness, i.e. if a sequence $\mathcal{F}_\bullet$ is such that $p^* \mathcal{F}_\bullet$ is exact, then so is $\mathcal{F}_\bullet$.

$p$ is surjective.

When these hold, we say $p$ is *faithfully flat*.

(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 $p$.

Compositions and pullbacks of faithfully flat maps are faithfully flat.

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.

Flat morphisms satisfy fpqc descent.

Finite morphisms satisfy fpqc descent.

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.