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Please help if you know a proof or a good reference for the following fact (exercise 3.5, Hartshorne's text).

Fact. A finite morphism of schemes $f: X \rightarrow Y$ is quasi-finite.

Here, the definition of quasi-finite is taken as $f^{-1}(q)$ is finite for all $q \in Y$.

mr.bigproblem
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  • I can't give a specific citation, but I'm sure you can find it here: http://stacks.math.columbia.edu/browse – Potato Jun 11 '13 at 04:34
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    Dear Potato, I have looked at that reference. They use another definition of quasi-finite and prove that the other definition implies the above definition and to understand the whole proof, one has to go through much machinery. I prefer a more direct approach. – mr.bigproblem Jun 11 '13 at 04:37
  • Wikipedia says "This follows from the fact that any finite k-algebra, for any field k is an Artinian ring," but I don't know enough about schemes to say if this is correct. Does that help at all? – Potato Jun 11 '13 at 04:54
  • I read that too and it sounds like the approach used in "stact project". The above fact is reduced to an algebraic statement as below.

    Suppose that $A \subset B$ are rings, $p \subset A$ is a prime ideal and $B$ is a finitely generated $A$-module then there are only finitely many prime ideals $q$ lying over $p$.

    I know a proof of the above fact with an extra assumption that $A$ is integrally closed. The approach in Stack Project seems to fix some ideal $q$ lying over $p$, do some trick with $B_q/pB_q$ and shows that the fibre of f, what ever it is, is finite.

     – mr.bigproblem Jun 11 '13 at 05:02

2 Answers2

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Let me show in an elementary way, as a complement to TTS's fine answer, that a finite dimensional algebra $A$ over a field $k$ has finite spectrum $\operatorname {Spec} A$, i.e. that $A$ has only finitely many prime ideals.

a) Every prime ideal $\mathfrak p\subset A$ is maximal
Indeed $A/\mathfrak p$ is an integral finite dimensional algebra over $k$, and is thus a field (here is a proof), so that $\mathfrak p$ is maximal.

b) There are only finitely many maximal ideals in $A$
If $\mathfrak m_1, \cdots, \mathfrak m_r\subset A $ are distinct maximal ideals, the Chinese remainder theorem (Atiyah-Macdonald, Proposition 1.10) implies that $A\to A/\mathfrak m_1\times \cdots \times A/\mathfrak m_r \;$ is surjective, so that $\operatorname { dim }\: A\geq \sum \operatorname {dim}(A/\mathfrak m_i)\geq r$.

Hence we deduce from b) that $A$ has at most $\operatorname { dim } A\: $ maximal ideals and, taking a) into account, at most $\operatorname { dim }\: A $ prime ideals.

NaNoS
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Georges Elencwajg
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  • I like this one, the same as my approach to this problem! It really supprises me to apply the Chinese remainder theorem! – Yuchen Liu Jul 18 '13 at 10:25
  • A little bit difference: I assume $A$ is reduced at first, then by the Chinese remainder theorem I prove that $A$ is isomorphic to $A/\mathfrak{m}_1\times\cdots\times A/\mathfrak{m}_r$ if $\cap_i\mathfrak{m}_i=(0)$, because any two distinct maximal ideals are coprime. – Yuchen Liu Jul 18 '13 at 10:28
  • Dear @Yuchen, you say that you are surprized that I use the Chinese remainder theorem. But if you will allow me a personal question: are you Chinese? :-) – Georges Elencwajg Jul 18 '13 at 11:12
  • Dear @Georges, of course I am Chinese. I am surprized because I haven't use the Chinese remainder theorem in abstract circumstance before, it usually occurs in concrete circumstance like number rings. So I know the proof but never tried to apply them in algebraic geometry, so this approach really surprises me! – Yuchen Liu Jul 18 '13 at 11:21
  • Dear @Yuchen, thanks for answering my rather indiscreet question. I quite agree that the use of the Chinese remainder theorem in an abstract context like the present one is unusual. I learned this too little known trick from Adrien Douady, a deceased brilliant French mathematician who was a member of Bourbaki. But actually my question was a benevolent little joke, originating in my immense admiration for Chinese mathematicians or mathematicians of Chinese origin in the USA and all over the world. – Georges Elencwajg Jul 18 '13 at 11:37
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Hartshorne indicates at some point that the fibre corresponds to $X \times_Y \operatorname{Spec} k(q)$. The morphism from this to $\operatorname{Spec} k(q)$ is still finite, so you just need to know that if $k$ is a field and $A$ a finite-dimensional $k$-algebra then $\operatorname{Spec} A$ is finite.

You could quote some theorem about Artinian rings, but the geometric language helps us give a proof. Some steps: (i) the points of $\operatorname{Spec} A$ are closed (ii) the irreducible components are just points (iii) $A$ is Noetherian.

TTS
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    Overall, this sounds right! According to Stack Project, the fibre of $f$ at $q$ is defined to be $X \times_Y \text{Spec}k(q)$ and there is a homeomorphism from this fibre to $f^{-1}(q)$ and the fact that $f$ is finite implies that the points in the fibre are isolated, so that they are exactly finitely many components of the fibre. The argument you provided sounds like that but you use the fact that the map $X \times_Y \text{Spec }k(q) \rightarrow \text{Spec }k(q)$ is finite. Is this true in general that if $X \rightarrow Y$ is finite then $X \times_Y Z \rightarrow Z$ is finite? – mr.bigproblem Jun 12 '13 at 07:29
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    I'm hoping to come back and make this a bit clearer, but I think this just boils down to the fact that if I have I have a finitely generated module $M$ over $A$ and an algebra $A \to B$ then the extension of scalars $M \otimes_A B$ is finitely generated over $B$: if I have generators $x_i$ for $M$ then I just need $x_i \otimes 1$. – TTS Jun 15 '13 at 04:33