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We know that if $f : X\to Y$ is a morphism between two irreducible affine varieties over an algebraically closed field $k$, then the function that assigns to each point of $X$ the dimension of the fiber it belongs to is upper semicontinuous on $X$.

Does anyone know of a simple counterexample when $X$ is not irreducible anymore (but remains an algebraic set over $k$, i.e a finitely generated $k$-algebra) ?

Edit : to avoid ambiguity about the definition of upper semicontinuity, it means here that for all $n\geq 0$, the set of $x\in X$ such as $\dim(f^{-1}(f(x) ) ) \geq n$ is closed in $X$.

It seems to me it is not so obvious to find a counterexample, since in fact the set of $x\in X$ such that the dimension of the irreducible component of $f^{-1}(f(x) )$ in $X$ that contains $x$ is $\geq n$ is always closed even when $X$ is not irreducible.

brunoh
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  • I got my answer on MO, here : http://mathoverflow.net/questions/133567/counter-example-of-upper-semicontinuity-of-fiber-dimension-in-classical-algebraic – brunoh Jun 13 '13 at 15:59
  • sorry for digging up that old question. could you provide a reference where I can find a proof of the claim from first paragraph. I wondering how the algebraical closedness of $k$ flows in the argumentation and if it is really neccessary to make this closedness of base field assumption. I hope that a careful reading of the proof might give an answer. – user267839 Dec 27 '21 at 00:18
  • @katalaveino look at 14.8 in Eisenbud's Commutative Algebra for instance. You can also find detailed discussion in exercises 10.5 and 10.6 of Kemper's Course in Commutative Algebra. See the discussion in the MO link. Reasoning is provided here. – brunoh Dec 27 '21 at 09:34
  • Ok, so as far as I see it correctly the statement from the first paragraph in this question is wrong, right? (https://mathoverflow.net/questions/193/when-is-fiber-dimension-upper-semi-continuous/184925#184925 is a conterexample) and therefore the statements in Kemper's Course on ComAlgebra in Exercise 10.6 (a) & (b) are still wrong even if we assume $X$ irreducible. – user267839 Dec 28 '21 at 00:11
  • On the other hand the proof that the statement on semicontinuity in Kemper's Ex 10.5 is an immerdiate consequence of the proof of 14.8(a) in Eisenbud's book. and here we indeed need the algebraic closedness of $k$ to apply Nullstellensatz to identify prime ideals 1-1 with irreducible subvarieties. – user267839 Dec 28 '21 at 00:11
  • but by the way: on page 143 Kemper wrote that the Ex. 10.5 can be concluded as a consequence of Corollary 10.6 (a certain estimation on dimension of fibers) and this Corollary seemingly uses somewhere that the base field $k$ is algebraically closed. do you see where in the proof of 10.6 it is used? The Nullstellensatz (which is based on alg alg closedness of $k$ gives a 1-1 correspondence between maximal primes and points. but going through the proof there it is only required that we can to any point $y \in Y$ associate a maximal ideal; that's trivial, here we not need Nullstellensatz – user267839 Dec 28 '21 at 01:01

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I got my answer on MO, here : mathoverflow.net/questions/133567/…

brunoh
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