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'finite morphism'
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| Title of object: |
finite morphism |
| Canonical Name: |
FiniteMorphism |
| Type: |
Definition |
| Created on: |
2002-07-24 05:38:57 |
| Modified on: |
2004-04-10 01:58:19 |
| Classification: |
msc:14-00, msc:14A10, msc:14A15 |
| Defines: |
affine morphism, finite type, locally of finite type |
Revision comment (for changes between this and next version):
| Changes for correction #6705 ('reversals of A and B'). |
Preamble:
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\DeclareMathOperator{\Spec}{Spec} |
Content:
\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 $B$ is finitely-generated as an $A$-algebra, then $f$ is said to be a morphism of \emph{finite type}.
If $B$ is in fact finitely generated as an $A$-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}
Let now $X$ and $Y$ be arbitrary schemes, and let $f \colon X\to Y$ be a morphism.
Let $\{U_i\}$ be an open cover of $Y$ by affine schemes, and for each $i$ let $\{V_{ij}\}$ be an open cover of $f^{-1}(U_i)$ by affine schemes. Then if $f|_{V_{ij}}$ is a morphism of finite type for every $i$ and $j$, then $f$ is said to be \emph{locally of finite type}.
Suppose $Y$ has a covering by open affine schemes $U_i$ such that for each $i$, $f^{-1}(U_i) = V_i$ is an open affine subscheme of $X$. Then $f$ is said to be an \emph{affine morphism}.
Suppose that $f$ is an affine morphism, and that $\{U_i\}$ are a cover such that each $f^{-1}(U_i)$ is affine. If $f|_{U_i}$ is a morphism of finite type for every $i$, then $f$ is said to be a morphism of \emph{finite type}. If $f|_{U_i}$ is a finite morphism for every $i$, then $f$ is said to be a finite morphism.
For example, if we take $X$ to be $\mathbb{P}^1(k)$ and $Y$ to be $\Spec k$. Then there is a covering of $X$ by two copies of $\mathbb{A}^1$, and there is a natural morphism from each of these copies to $\Spec k$; these patch together to give a morphism from $\mathbb{P}^1$ to $\Spec k$. This morphism is locally of finite type but not of finite type, since it is not affine. |
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