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Let $X,Y$ be noetherian schemes and $f: X \rightarrow Y$ a morphism. What does (besides of the definition) the property finite imply. Can one make any conclusions like: dominant, birational, isomorphic on open subsets,...?

Thanks and greetings from Torrance

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Finite morphisms are proper, so, e.g. universally closed. Closed immersions are finite, so finite morphisms definitely don't have to be dominant. –  Keenan Kidwell Nov 13 '12 at 18:52
    
Finite is equivalent to "proper and affine". For example, a variety is at the same time proper and affine if and only if it has dimension zero. –  Brenin Nov 13 '12 at 19:16
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2 Answers 2

Given a field $k$ , consider the closed immersion $j:Spec(k)\to \mathbb A^1_k$ of a point into the affine line, dual to the ring morphism $k[T]\to k: f(t)\mapsto f(0)$.
This morphism $j$ between noetherian schemes is finite and gives a negative answer to all your questions: it is not dominant, not birational and not isomorphic on open sets.

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Dear @George, That's quite an efficient example! –  Keenan Kidwell Nov 13 '12 at 18:54
    
Dear @Keenan, after having written my answer I had the pleasant surprise to see your comment, which is based on exactly the same idea but is much more general! –  Georges Elencwajg Nov 13 '12 at 18:57
    
The example puts the light on the usefulness of considering even trivial examples when looking for information, as in the OP! –  Mariano Suárez-Alvarez Nov 13 '12 at 19:03
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$f$ is affine, i.e. $f^{-1}(V)$ is an affine open subset of $X$ for every affine open subet $V$ of $Y$(Hartshorne, Ex. 3.4, Ch. II).

$f$ is closed, i.e. $f(F)$ is closed for every closed subset $F$ of $X$(Hartshorne, Ex. 3.5, Ch. II).

$f^{-1}(y)$ is finite for every $y \in Y$(Hartshorne, Ex. 3.5, Ch. II).

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