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suppose $\pi:X\rightarrow \text{Spec}\ k$ is a quasifinite morphism. Show that $\pi$ is finite. (hint: deal first with the case where $X=\text{Spec}\ A,$ $A$ finitely generated over $k$. If $A$ contains an element $x$ not algebraic over $k$, we have a inclusion $k[x]\hookrightarrow A.$)

In fact, I finished the exercise but used the Nullstellensatz, Chinese remainder theorem and the fact that $A$ is a Jacobson ring with finitely many prime ideals to show:

1.$\text{Spec}\ A\rightarrow \text{Spec}\ k$ is finite.

This section is about Chevalley theorem, I want to know how to use this theorem directly to prove 1.

Thanks a lot.

Kenta S
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XiaYu
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1 Answers1

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Suppose $A$ contains an element $x$ that is not algebraic over $k$, i.e., we have an inclusion $k[x] \hookrightarrow A$, corresponding to a dominant morphism $\theta: X = \operatorname{Spec} A \to \mathbb{A}_k^1$ of finite type $k$-schemes. Note that $X$ has only finite points hence the generic point $p$ of $X$ is constructible. By Chevalley's theorem, the image of $p$ is not the generic point of $\mathbb{A}_k^1$. Thus $\theta$ is not dominant.

mathjiajia
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