FiniteMorphism 2002-07-24 05:38:57 2006-06-08 15:01:30 Definition affine morphism finite type % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \DeclareMathOperator{\Spec}{Spec} \section*{Affine schemes} Let $X$ and $Y$ be affine schemes, so that $X=\Spec A$ and $Y=\Spec B$. Let $f\colon X\to Y$ be a morphism, so that it induces a homomorphism of rings $g\colon B\to A$. 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 \emph{finite type}. If $A$ is in fact finitely generated as a $B$-module, then $f$ is said to be a \emph{finite} morphism. For example, if $k$ is a field, the scheme $\mathbb{A}^n(k)$ has a natural morphism to $\Spec k$ 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 $\Spec k[X,Y]/\left<Y^2-X^3-X\right>$, it has a natural morphism to $\mathbb{A}^1$ given by the ring homomorphism $k[X]\to k[X,Y]/\left<Y^2-X^3-X\right>$. Then this morphism is a finite morphism. As a morphism of schemes, we see that every fiber is finite. \section*{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 \emph{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 \emph{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. \paragraph{Example.} Let $X=\mathbb{P}^1(k)$ and $Y=\Spec k$. We cover $X$ by two copies of $\mathbb{A}^1$ and consider the natural morphisms from each of these copies to $\Spec k$. 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 $\Spec k$. The overall morphism is of finite type, but again is not finite. \section*{References.} D. Eisenbud and J. Harris, \textit{The Geometry of Schemes}, Springer.