# Definition of locally of finite type morphism

Hartshorne defines the locally finite type morphism as follows:

Definition: A morphism $$f : X \to Y$$ of schemes is locally of finite type if there exists a covering of $$Y$$ by affine open subsets $$V_i = Spec B_i$$, such that for each i, $$f^{-1}(V_i)$$ can be covered by open affine subsets $$U_{ij}=SpecA_{ij}$$, where each $$A_{ij}$$ is a finitely generated $$B_i$$-algebra.

My question is: Is there something missing in this definition?

I would expect a relation between the map $$B_i \to A_{ij}$$ that determines the algebra structure and the map $$f|_{SpecA_{ij}} : SpecA_{ij} \to SpecB_i$$

• The two maps you write are already the same information because $\operatorname{Spec}$ is an equivalence of categories. Or did you mean something else? Nov 17 '19 at 4:14
• What I mean is: Given $f : X \to Y$ we can get a map $B_i \to A_{ij}$. Do we actually require this map to turn $A_{ij}$ into a finitely generated $B_i$ algebra? Or is this completely unrelated with the definition of locally of finite type morphism as stated by Hartshorne? Nov 17 '19 at 4:28
• Yes, this map is the map which gives $A_{ij}$ the structure of a $B_i$ algebra. We require that the $B_i$-algebra structure on $A_{ij}$ given by this map make $A_{ij}$ a finitely generated $B_i$-algebra. Nov 17 '19 at 4:37
• Thank you! This was not clear from the definition as stated by Hartshorne. I just wanted to make sure. Nov 17 '19 at 4:41
From the comments: the $$B_i$$-algebra structure on $$A_{ij}$$ is induced via the map $$f|_{\operatorname{Spec} A_{ij}} \operatorname{Spec} A_{ij} \to \operatorname{Spec} B_{i}$$. This is the algebra structure Hartshorne refers to when he says that $$A_{ij}$$ is a finitely-generated $$B_i$$-algebra. (Compare to Stack's definition, where the connection between the map $$f$$ and the algebra structure is slightly more explicit.)