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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 $f|_{SpecA_{ij}} : SpecA_{ij} \to SpecB_i$

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  • $\begingroup$ The two maps you write are already the same information because $\operatorname{Spec}$ is an equivalence of categories. Or did you mean something else? $\endgroup$
    – KReiser
    Nov 17 '19 at 4:14
  • $\begingroup$ 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? $\endgroup$
    – 11101
    Nov 17 '19 at 4:28
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Nov 17 '19 at 4:37
  • $\begingroup$ Thank you! This was not clear from the definition as stated by Hartshorne. I just wanted to make sure. $\endgroup$
    – 11101
    Nov 17 '19 at 4:41
  • $\begingroup$ Glad to help. I'll put the comments in an answer. $\endgroup$
    – KReiser
    Nov 17 '19 at 4:45
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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.)

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