Definition 17.20.1. Let $f : X \to Y$ be a morphism of ringed spaces. Let $x \in X$. We say $f$ is *flat at $x$* if the map of rings $\mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ is flat. We say $f$ is *flat* if $f$ is flat at every $x \in X$.

## 17.20 Flat morphisms of ringed spaces

The pointwise definition is motivated by Lemma 17.17.2 and Definition 17.17.3 above.

Consider the map of sheaves of rings $f^\sharp : f^{-1}\mathcal{O}_ Y \to \mathcal{O}_ X$. We see that the stalk at $x$ is the ring map $f^\sharp _ x : \mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$. Hence $f$ is flat at $x$ if and only if $\mathcal{O}_ X$ is flat at $x$ as an $f^{-1}\mathcal{O}_ Y$-module. And $f$ is flat if and only if $\mathcal{O}_ X$ is flat as an $f^{-1}\mathcal{O}_ Y$-module. A very special case of a flat morphism is an open immersion.

Lemma 17.20.2. Let $f : X \to Y$ be a flat morphism of ringed spaces. Then the pullback functor $f^* : \textit{Mod}(\mathcal{O}_ Y) \to \textit{Mod}(\mathcal{O}_ X)$ is exact.

**Proof.**
The functor $f^*$ is the composition of the exact functor $f^{-1} : \textit{Mod}(\mathcal{O}_ Y) \to \textit{Mod}(f^{-1}\mathcal{O}_ Y)$ and the change of rings functor

Thus the result follows from the discussion following Definition 17.20.1. $\square$

Definition 17.20.3. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules.

We say that $\mathcal{F}$ is

*flat over $Y$ at a point $x \in X$*if the stalk $\mathcal{F}_ x$ is a flat $\mathcal{O}_{Y, f(x)}$-module.We say that $\mathcal{F}$ is

*flat over $Y$*if $\mathcal{F}$ is flat over $Y$ at every point $x$ of $X$.

With this definition we see that $\mathcal{F}$ is flat over $Y$ at $x$ if and only if $\mathcal{F}$ is flat at $x$ as an $f^{-1}\mathcal{O}_ Y$-module because $(f^{-1}\mathcal{O}_ Y)_ x = \mathcal{O}_{Y, f(x)}$ by Sheaves, Lemma 6.21.5.

Lemma 17.20.4. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module flat over $Y$. Then the functor

is exact.

**Proof.**
This is true because $f^*\mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} = f^{-1}\mathcal{G} \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{F}$, the functor $f^{-1}$ is exact, and $\mathcal{F}$ is a flat $f^{-1}\mathcal{O}_ Y$-module.
$\square$

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