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I am following Vakil's FOAG, exercise 7.3.H: Let $X\to $Spec $K$ be a finite morphism, prove that $X$ is a finite union of points with the discrete topology.

I am following the guidance there. If we write $X=$Spec $A$ then $A$ is a finite dimensional vector space over $K$. If $A$ was a domain then it is easy to show that it is a field, and so we get that all primes of $A$ are maximal, and hence $X$ consists only of closed points.

The next part should be to prove that $X$ is discrete, and then finiteness would follow from quasicompactness.

My question is why is $X$ discrete? I will be glad for anything you can say about the general problem, but I am looking to understand how it is possible to show discreteness now, before finiteness, say.

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  • $\begingroup$ If $X$ is finite (as a set) and all points of $X$ are closed it is basic topology to show $X$ must be a discrete topological space (just from the axioms of the topological space). Note: $X$ is obviously finite as a set since your map has finite fibres. $\endgroup$
    – PVAL-inactive
    Aug 19, 2013 at 17:01
  • $\begingroup$ Thank you! but I cant use this result yet (before fibres have been defined) $\endgroup$
    – edo arad
    Aug 19, 2013 at 17:07
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    $\begingroup$ Dear edo, You know that $A$ is finite dim'l over $K$. Surely you can prove that all of its prime ideals are maximal, and that there are only finitely many of them. (You have already implicitly proved the first statement in what you've written in your post.) Regards, $\endgroup$
    – Matt E
    Aug 19, 2013 at 17:12
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    $\begingroup$ By fibre I just meant the preimage of a point (as a set). I did not mean the scheme-theoretic fibre. $\endgroup$
    – PVAL-inactive
    Aug 19, 2013 at 17:15

2 Answers 2

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(An answer based on an exchange of comments with the OP.)

As the OP observes, a finite-dimensional $K$-algebra that is a domain is necessarily a field, and so all prime ideals in $A$ are maximal.

Now, by CRT, if $\mathfrak m_1, \ldots,\mathfrak m_k$ are distinct maximal ideals, then $$A/(\mathfrak m_1 \cap \cdots \cap \mathfrak m_k) \cong A/\mathfrak m_1 \times \cdots \times A/\mathfrak m_k, $$ and so $k \leq \dim_K A.$ In particular, $A$ admits no more than $\dim_K A$ maximal ideals, and so Spec $A$ is a finite set of closed points.

Unfortunately, I don't see how to directly follow the hint (i.e. to first prove discreteness) in a natural way.

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    $\begingroup$ Dear MattE, if we know that $A$ is a Noetherian ring that is also Artinian, isn't it a result (IIRC) that $\operatorname{Spec} A$ is finite and discrete? Perhaps that will help us here. $\endgroup$
    – user38268
    Aug 19, 2013 at 23:28
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    $\begingroup$ Dear user, If $A$ is Artinian, it is automatically Noetherian. Furthermore, basic structure theory then implies that Spec $A$ is finite and discrete. So certainly the claimed result follows from the basic theory of Artinian rings. But (a) the above answer was based on hints to the OP, who clearly wasn't familiar with the general theory, and who was solving a problem that probably didn't presume the general theory; (b) the problem hint suggests that discreteness can be proved separately from finiteness, but I don't see how to do that in a natural way. Regards, $\endgroup$
    – Matt E
    Aug 20, 2013 at 1:48
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    $\begingroup$ Dear MattE, we know all prime ideals are maximal, in particular all minimal primes are maximal, then each irreducible component is a closed point. Then the topology is discrete. $\endgroup$
    – Linax Dio
    Jul 1, 2018 at 15:28
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Applying the definition goes a long way; assuming that $K$ is a field, $\operatorname{Spec}K$ is a point. For a morphism $f:\ X\ \longrightarrow\ \operatorname{Spec} K$ to be finite means precisely that $f^{-1}(\operatorname{Spec}K)=X=\operatorname{Spec} V$ where $V$ is a $K$-algebra that is a finitely generated as a $K$-module, i.e. it is a finite dimensional $K$-vector space with a ring map $K\ \longrightarrow V$.

EDIT: As Georges Elencwajg points out in the comments below, I was a bit hasty in my conclusions. I won't say too much in attempt to avoid saying more silly things.

Note that $V$ is Artinian, hence its spectrum is finite and all prime ideals are maximal.

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  • $\begingroup$ thank you, this is really the direction I had thought about. Maybe I should have included it in my question, but anyway this is exactly the point I got stuck on (I understand why every prime is maximal but not how to impose some discreteness or finiteness on the set of primes) $\endgroup$
    – edo arad
    Aug 19, 2013 at 17:20
  • $\begingroup$ Dear User, vector spaces don't have ideals. And algebras don't have ideals of each dimension. $\endgroup$
    – Georges Elencwajg
    Aug 19, 2013 at 17:41
  • $\begingroup$ My phrasing was sloppy and my conclusion incorrect. Hopefully fixed now. $\endgroup$
    – Servaes
    Aug 19, 2013 at 17:52
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    $\begingroup$ @edoarad: Dear edo, If you know that all prime ideals are maximal (and hence give closed points), why don't you try to prove that there are only finitely many by using the Chinese Remainder Theorem. Regards, $\endgroup$
    – Matt E
    Aug 19, 2013 at 18:56
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    $\begingroup$ @MattE great! so if $M_1,M_2,\ldots$ are maximal ideals, and $I_k=M_1\cap\cdots\cap M_k$ then $A/I_k\equiv A/M_1\oplus\cdots\oplus A/M_k$ where the dimension on the left is no more than $[A:K]$ but the dimension on the right is no less than $k$, which show that this series cant be infinite. $\endgroup$
    – edo arad
    Aug 19, 2013 at 19:10

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