Lemma 29.15.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. The morphism $f$ is locally of finite type.

2. For all 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 of finite type.

3. There exist 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 locally of finite type.

4. There exist 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 the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is of finite type, for all $j\in J, i\in I_ j$.

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

Proof. This follows from Lemma 29.14.3 if we show that the property “$R \to A$ is of finite type” is local. We check conditions (a), (b) and (c) of Definition 29.14.1. By Algebra, Lemma 10.14.2 being of finite type is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 10.6.2 being of finite type is stable under composition and trivially for any ring $R$ the ring map $R \to R_ f$ is of finite type. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 10.23.3. $\square$

Comment #2783 by David Hansen on

Very pedantic comment: Part (2) of this Lemma is ungrammatical; it could be fixed by writing either "For all affine opens..." or "For every pair of affine opens...".

There are also:

• 2 comment(s) on Section 29.15: Morphisms of finite type

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).