Morphism of finite type

Remarks

Equivalently, $f$ is of finite type if and only if it is locally of finite type and quasi-compact. 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 it would be a finite extension of $k$, which it is not.