Lemma 29.25.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. The morphism $f$ is flat.

2. For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is flat.

3. There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is flat.

4. There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is flat, for all $j\in J, i\in I_ j$.

Moreover, if $f$ is flat then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is flat.

Proof. This is a special case of Lemma 29.25.2 above. $\square$

## Comments (0)

There are also:

• 4 comment(s) on Section 29.25: Flat morphisms

## 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Your email address will not be published.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 01U5. Beware of the difference between the letter 'O' and the digit '0'.