The Stacks project

17.20 Flat morphisms of ringed spaces

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

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 $x \in X$.

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

\[ \textit{Mod}(f^{-1}\mathcal{O}_ Y) \to \textit{Mod}(\mathcal{O}_ X), \quad \mathcal{F} \longmapsto \mathcal{F} \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{O}_ X. \]

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.

  1. 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.

  2. 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

\[ \textit{Mod}(\mathcal{O}_ Y) \to \textit{Mod}(\mathcal{O}_ X),\quad \mathcal{G} \longmapsto f^*\mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \]

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02N2. Beware of the difference between the letter 'O' and the digit '0'.