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This is a statement made in Gathmann's Commutative Algebra Notes, Chapter 9.

"The following corollary is essentially a restatement of finite fiber property: it says that in integral ring extensions only maximal ideals can contract to maximal ideals, i.e. that points are only subvarieties that can map to a single point in the target space.

Corollary 9.21 Let $R\subset R'$ be an integral ring extension.

(a) If $R$ and $R'$ are integral domains, then $R$ is a field iff $R'$ is a field.

(b) A prime ideal $P'\subset R'$ is maximal iff $P'\cap R$ is maximal."

Where is finiteness coming from? Or how should I understand the finiteness of fiber here? Is the inverse image of induced morphism of each point of target variety, finite points of source variety? Should I expect each maximal ideal of $R$ corresponds to finite number of $R'$ maximal ideals? This is true if $R'$ is artinian.

It essentially relates to Determine whether these extensions of $\mathbb{C}[x]$ are integral's finite fiber property and I do not see where finite fiber comes from as I did the problem in a purely algebraic way.

user26857
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user45765
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  • Have you read this answer? – user26857 Jul 04 '17 at 22:42
  • @user26857 Yep. That is the reason I am asking. For the specific case, I can see there is finite fiber explicitly from geometry but not general case. Can I conclude that any integral extension should imply that for each smaller ring's maximal ideal, there is finite number of larger ring's maximal ideals? – user45765 Jul 04 '17 at 22:56
  • Yes, you can; see here. – user26857 Jul 04 '17 at 23:04
  • @user26857 Is there a reason to expect the larger ring being f.g. algebra? If I require f.g. algebra, it is clear to me that there is finitely many primes by inducting on the number of generators. I mean in general, $R\subset R'$ as integral extension, is it clear that $R'$ is always f.g. algebra given that integral extension? – user45765 Jul 04 '17 at 23:07
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    Oh, you asked in general, for any integral ring extension. No, then this is wrong; see here. – user26857 Jul 04 '17 at 23:09
  • @user26857 Thanks. So in gathmann's notes context, the finite morphism is presumed. – user45765 Jul 04 '17 at 23:31

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By Corollary 9.21-(b), we know that points are the only subvarieties that can be mapped to a single point in the target space (as Gathmann pointed).

Now given morphism $f:X\to Y$. Since to any $y\in Y$, $\{y\}$ is closed, by continuity, we know $f^{-1}(y)$ is an algebraic variety. By $k^n$ is an Noetherian topological space and any closed subset of a Noetherian topological space can be expressed as the union of finitely many irreducible closed subsets, we know $f^{-1}(y)$ has finitely many irreducible closed components, i.e. finitely many points.

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