1
$\begingroup$

With or without using Chevalley's theorem, show that if $X\to \text{Spec }k$ is a quasifinite morphism (according to Vakil, this means finite type morphism + finite fibers), then the morphism is actually finite.

This question was asked before (in FOAG exercise 7.4.D.), but I don't quite understand the solution:

First, even if solve the case where $X$ is affine, how do we show that the claim holds in general for any scheme $X$?

Moreover, the accepted solution suggests that $X=\text{Spec }A$ is integral ("consider the generic point of $X$"), that the morphism $\text{Spec }A \to\text{Spec }k[x]$ induced by the inclusion $k[x]\subset A$ is dominant, and that the condition on finite fibers implies that the generic point of $X$ is constructible, but none of these are particularly clear to me.

Do we need some sort of requirement that the fiber is discrete (this would make, for instance, the last claim easy to verify)?

$\endgroup$
  • 1
    $\begingroup$ I mean the way I would tackle this is to note the following, sorry if this is not directly answering your questions. Since this doesn't change if we base change to algebraic closure we can assume that $k$ is algebraically closed. If $X\to\mathrm{Spec}(k)$ is quasi-finite, then $X$ is certainly finite, and thus so is $X(k)$. But, let $U\subseteq X$ be any affine open. Then, certainly $U(k)$ is finite. But, if $\dim X>0$ then we can take $d=\dim U>0$. $\endgroup$ – Alex Youcis Aug 26 at 21:35
  • 1
    $\begingroup$ Moreover, by passing to the reduced subscheme of some irreducible component we can assume that $U$ is integral. So there is a surjective finite map $U\to \mathbb{A}^d_k$, which implies that $k^d$ is finite. This is a contradiction. Thus, $\dim X=0$. Since $X$ is Noetherian it has finitely many irreducible components, all of which have dimension $0$ so must be points. This implies that the reduced subscheme of those points are fields which, since they are finite type over $k$, are finite extensions by Zariski's lemma (which is just the Nullstellensatz). It then follows that the points $\endgroup$ – Alex Youcis Aug 26 at 21:40
  • 1
    $\begingroup$ themselves are finite Artinian $k$-algebras ans the conclusion follows. $\endgroup$ – Alex Youcis Aug 26 at 21:40
0
$\begingroup$

For the sake of completeness, I'll add a solution that's a little different than what Alex commented (i.e. using Chevalley's theorem, kind of):

First, we can write $X=\bigcup_{i=1}^n\text{Spec}A_i$, where the $A_i$ are finitely generated $k$-algebras. We can break up each affine guy into (finitely many because of compactness) irreducible components and assume that $A_i$ are actually irreducible and consider the reduced closed subscheme $\text{Spec}A_i/N(A_i)$, where $N(A_i)$ is the nilradical. By the Chevalley argument outlined (in the problem statement, I guess), it follows that $A_i/N(A_i)$ is actually a finite extension of $k$ (this also follows from the fact that integral + finite type is finite). This shows that $\text{Spec}A_i$ has support that is a point, from which it follows that $X$ is discrete and hence the Spec of the product of the $A_i$s.

To show that $X$ is actually a finite dimensional $k$-vector space, note that any finitely generated $k$-algebra is Artinian iff it is a finite dimensional $k$-vector space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.