# finite morphism

## Affine schemes

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

For example, if $k$ is a field, the scheme $\mathbb{A}^{n}(k)$ has a natural morphism to $\operatorname{Spec}k$ induced by the ring homomorphism $k\to k[X_{1},\ldots,X_{n}]$. 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 $\operatorname{Spec}k[X,Y]/\left$, it has a natural morphism to $\mathbb{A}^{1}$ given by the ring homomorphism $k[X]\to k[X,Y]/\left$. Then this morphism is a finite morphism. As a morphism of schemes, we see that every fiber is finite.

## General schemes

Now, let $X$ and $Y$ be arbitrary schemes, and let $f\colon X\to Y$ be a morphism. We say that $f$ is of finite type if there exist an open cover of $Y$ by affine schemes $\{U_{i}\}$ and a finite open cover of 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 affine schemes $\{U_{i}\}$ such that each inverse image, $V_{i}=f^{-1}(U_{i})$ is itself affine, and such that $f|_{V_{i}}$ is a finite morphism of affine schemes.

### Example.

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

## References.

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

Title finite morphism FiniteMorphism 2013-03-22 12:51:47 2013-03-22 12:51:47 rmilson (146) rmilson (146) 9 rmilson (146) Definition msc 14A10 msc 14-00 msc 14A15 affine morphism finite type