---
tag: T09DXZAT6R
kind: definition
name: Morphism of finite type
msc: [14A15]
defines: AHTE5DVX61
uses: [JXRKH6P1Z4, B3V3XJ2W8G, AC5TFFQT9M, 41009J4M5D, YBJ2VQHFDQ]
sources:
  - url: https://en.wikipedia.org/wiki/Morphism_of_finite_type
---

## Definition

Let $f \colon X \to Y$ be a [[morphism of schemes|JXRKH6P1Z4]].
It is *locally of finite type* if there is a cover of $Y$ by open [[affine|B3V3XJ2W8G]] subschemes $V_i = \operatorname{Spec} B_i$ with the following property.
For each $i$, the open set $f^{-1}(V_i)$ is covered by open affine subschemes $U_{ij} = \operatorname{Spec} A_{ij}$, where each $A_{ij}$ is a [[finitely generated|AC5TFFQT9M]] $B_i$-algebra.
It is *of finite type* if moreover each $f^{-1}(V_i)$ can be covered by finitely many such $U_{ij}$.

## Remarks

Equivalently, $f$ is of finite type if and only if it is locally of finite type and [[quasi-compact|41009J4M5D]].
A morphism $\operatorname{Spec} A \to \operatorname{Spec} B$ is of finite type if and only if $A$ is a finitely generated $B$-algebra.
Example: for a field $k$, the morphism $\mathbb{A}^n_k \to \operatorname{Spec} k$ is of finite type, since $k[x_1, \ldots, x_n]$ is generated by the $x_i$.
Projective space $\mathbb{P}^n_k \to \operatorname{Spec} k$ is of finite type too, being covered by $n + 1$ such affine pieces.
Non-example: $\operatorname{Spec} k(x) \to \operatorname{Spec} k$ is not of finite type.
If the field $k(x)$ were a finitely generated $k$-algebra, then by [[Zariski's lemma|YBJ2VQHFDQ]] it would be a finite extension of $k$, which it is not.