# 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.

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 $p$.

Using (4), it is clear that

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.

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.

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