finite morphism

Affine schemes

Let $X$ and $Y$ be affine schemes, so that $X=\mathrm{Spec}A$ and $Y=\mathrm{Spec}B$. Let $f:X\to Y$ be a morphism, so that it induces ahomomorphism of rings $g:B\to A$.

The homomorphism $g$ makes $A$ into a $B$-algebra. If $A$ isfinitely-generated as a $B$-algebra, then $f$ is said to be a morphismof finite type.

If $A$ is in fact finitely generated as a $B$-module, then $f$ is saidto be a finite morphism.

For example, if $k$ is a field, the scheme ${𝔸}^n(k)$ has anatural morphism to $\mathrm{Spec}k$ induced by the ring homomorphism $k\to k[X_1,\mathrm{\dots },X_n]$. This is a morphism of finite type, but if $n>0$then it is not a finite morphism.

On the other hand, if we take the affine scheme , it has a natural morphism to${𝔸}^1$ given by the ring homomorphism . Then this morphism is a finitemorphism. As a morphism of schemes, we see that every fiber is finite.

General schemes

Now, let $X$ and $Y$ be arbitrary schemes, and let $f:X\to Y$be a morphism. We say that $f$ is of finite type if there exist anopen cover of $Y$ by affine schemes $\{U_i\}$ and a finite open coverof each $U_i$ by affine schemes $\{V_{ij}\}$ such that ${f|}_{V_{ij}}$is a morphism of finite type for every $i$ and $j$. We say that $f$is finite if there exists an open cover of $Y$ by affineschemes $\{U_i\}$ such that each inverse image, $V_i=f^{-1}(U_i)$ isitself affine, and such that ${f|}_{V_i}$ is a finite morphism of affineschemes.

Example.

Let $X={\mathbb{P}}^1(k)$ and $Y=\mathrm{Spec}k$.We cover $X$ by two copies of ${𝔸}^1$ and consider the naturalmorphisms from each of these copies to $\mathrm{Spec}k$. Both of theseaffine morphisms are of finite type, but are not finite. The coveringmorphisms patch together to give a morphism from ${\mathbb{P}}^1$ to$\mathrm{Spec}k$. The overall morphism is of finite type, but again is notfinite.

References.

D. Eisenbud and J. Harris, The Geometry of Schemes, Springer.