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We know that if $f : X\to Y$ is a morphism between two 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 (but remains an algebraic set over $k$, i.e a finitely generated $k$-algebra) to the global statement?

Edit: to avoid ambiguity I am looking for a counterexample in case $X$ is not irreducible when the dimension of the fibers is measured globally, i.e. $n\geq 0$, the set of $x\in X$ such that $\dim(f^{-1}(f(x) ) ) \geq n$ is closed in $X$.

Edit2: in his comments @dorebell linked an answer here https://mathoverflow.net/a/184925/3333 where a counterexample to the upper semicontinuity of global dimension on the source is given with $X$ and $Y$ affine and irreducible (it works even if the counterexample is explained looking at the dimension of fibers from the target)

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    $\begingroup$ I know this is an old question, but I recently found it while reading about this problem. Be careful: in fact, the statement you give is not true even for $X$ irreducible when $X,Y$ are no longer supposed to be affine (and possibly when they are?) This answer gives a counterexample: mathoverflow.net/a/184925/56878 $\endgroup$ – dorebell Mar 19 '16 at 20:39
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    $\begingroup$ I'd love to see a proof/reference if you have one easily available! This would be a very convenient form of the result. $\endgroup$ – dorebell Mar 19 '16 at 21:14
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    $\begingroup$ @dorebell look at 14.8 in Eisenbud's Commutative Algebra for instance $\endgroup$ – brunoh Mar 19 '16 at 21:47
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    $\begingroup$ @dorebell You can also find detailed discussion in exercises 10.5 and 10.6 of Kemper's Course in Commutative Algebra $\endgroup$ – brunoh Mar 19 '16 at 23:02
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    $\begingroup$ @dorebell I just realize that even with the references I gave you the result is not completely obvious. I proved it the following way : using irreducibility of X and Y, you can apply the corollary 14.6 of Eisenbud to get it for $n=dim\ Y-dim\ X$ on an open set $U$ of $Y$. Outside of it, the irreducible components of $Y-U$ are of dimension < $dim\ Y$. A simple recurrence on $dim\ Y$ then works well. $\endgroup$ – brunoh Mar 20 '16 at 1:41
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Let $X = (\mathbb{A}^2 \setminus \{x = 0\}) \coprod \mathbb{A}^1$, let $Y = \mathbb{A}^1$, and let $f$ be projection onto the first coordinate on the first component and the identity on the second. Then every point of $X$ lives in a one-dimensional fiber except the origin of the second component.

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    $\begingroup$ @EvanJenkins Thank you for your answer, but there are things I do not understand. First of all, how can you make X into an algebraic set (i.e. the Spec of a finitely generated algebra) ? It seems to me you cannot because of the plane minus a point. Second, the fiber to which belongs the origin of the second component is $0\amalg (\mathbb{A}^1-0)$ which is one dimensional right ? $\endgroup$ – brunoh Jun 13 '13 at 8:33
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    $\begingroup$ @brunoh: It's not a plane minus a point, it's a plane minus a line, which answers both your questions. $\endgroup$ – Dan Petersen Jun 13 '13 at 14:16
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    $\begingroup$ @brunoh: Evan isn't removing just a point from $\mathbb A^2$ but the whole line $\{x=0\}$. I think this clarifies both of the points you raised. $\endgroup$ – Andreas Blass Jun 13 '13 at 14:19
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    $\begingroup$ No, the set of points of $X$ which live in a positive dimensional fiber is $\mathbb A^2 \setminus \{x=0\} \sqcup \mathbb A^1 \setminus \{0\}$, which is not closed in $\mathbb A^2 \setminus \{x=0\} \sqcup \mathbb A^1$. $\endgroup$ – Dan Petersen Jun 13 '13 at 14:52
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    $\begingroup$ @EvenJenkins Is there a counterexample in which X is irreducible? $\endgroup$ – YZhou Oct 4 '13 at 5:11
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For what its worth, I can give you a non-Noetherian example, even with both schemes affine, irreducible (and of finite Krull dimension).

Set $R = k[x,y,x/y, x/y^2, x/y^3, ...]$ and $S = k[y]$. We have the obvious map $S \hookrightarrow R$ which induces $$ X = \text{Spec }R \to Y = \text{Spec }S. $$ Now, away from the origin of $S$, $y$ is invertible and $R[y^{-1}] = k[x,y,y^{-1}]$ has all fibers with dimension $1$. On the other hand, once we set $y = 0$ in $R$, we notice that $x = (x/y) y$ is a multiple, as is $(x/y) = (x/y^2) y$, and so is $(x/y^n)$ for all $n$. This is already a maximal ideal, so the fiber over $y = 0$ is $0$-dimensional.

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  • $\begingroup$ @Karl First of all, thank you very much for taking the time to notice it was probably not an obvious question. I like very much your example also and upvoted it. Nevertheless I cannot consider it as an the counter example I was looking for, because of the non-noetherianness of $X$. Since the result on the upper semicontinuity of fiber dimension is based on the the Generic Freeness Lemma, a key hypothesis is finite generation of the algebra involved, right ? That is why I need a counterexample using algebraic sets (SpecMax of finitely generated algebra over $k$). $\endgroup$ – brunoh Jun 13 '13 at 15:27
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    $\begingroup$ I agree, it doesn't answer your question but I still thought it was amusing. $\endgroup$ – Karl Schwede Jun 13 '13 at 15:46
  • $\begingroup$ The origin in closed in $X$, is it not? $\endgroup$ – Karl Schwede Jun 13 '13 at 15:49
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Let $ X = \mathbb A^2 \cup pt$, let $Y = \mathbb P^1$. Let $f(\mathbb A^2) = \mathbb A^1$ by projection, and let $f(pt)=\infty$. Then the generic fiber dimension is $1$, but at one point the fiber dimension is $0$.

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  • $\begingroup$ Thank you for your your answer. But first of all, I wanted a counterexample using only affine varieties, not projective ones. Second, I fail to see why it is a counterexample, since the dimension function is still upper semicontinuous on $X$, right ? $\endgroup$ – brunoh Jun 12 '13 at 22:43
  • $\begingroup$ It is not upper semicontinuous because the fiber dimension at infinity is $0$, but on any neighborhood of $0$ all other fibers are $1$-dimensional. The dimension function is actually lower semicontinuous here. $\endgroup$ – Matt Jun 13 '13 at 0:11
  • $\begingroup$ Yes, good point. To solve this, connect the point to $\mathbb A^2$ using a line. $\endgroup$ – Will Sawin Jun 13 '13 at 0:13
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    $\begingroup$ Let $Y$ be irreducible of positive dimension, and $p$ a point in $Y$. Let $X = ((Y\setminus p) \times {\mathbb A}^1) \coprod \{q\}$, and $X\to Y$ be projection on the first component, and $q\mapsto p$ on the second. $\endgroup$ – Allen Knutson Jun 13 '13 at 0:22
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    $\begingroup$ @Allen_Knutson It seems to me the dimension function is still upper semicontinuous. I also found this counterexample but realized my mistake. I think to find a counterexample one has to find fibers in $X$ with irreducible components of different dimensions each one of them staying in different irreducible components of $X$. I failed to do that yet ... $\endgroup$ – brunoh Jun 13 '13 at 1:07

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