Let $f:X\to Y$ be a dominant morphism of varieties (integral separated schemes of finite type over an algebraically closed field) such that dim $X$ = dim $Y$.

Question: must f be finite?

It seems that f must have finite fibers by looking at dimensions. So perhaps it is the same question to ask if $f$ must be proper, because finite fibers + proper = finite. I am not familiar enough with standard non-examples of properness to have a good intuition on this.

Finally, if it makes any difference to assume X and Y are quasi-projective, please do so.

Edit: After Steve's answer to the original question, I would like to ask the same question for a self-morphism $f:X\to X$.

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    $\begingroup$ Your comment about finite fibers is not quite right---the generic fiber will be finite by looking at dimensions, but take e.g. the blowup of the plane at the origin. This has an exceptional fiber over $0$, isomorphic to the projective line, but of course the generic fiber is just a point. $\endgroup$
    – Stephen
    Commented May 20, 2013 at 14:39
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    $\begingroup$ It is an exercise in Hartshorne to show that a dominant, generically finite (finite generic fiber), finite type morphism of integral schemes restricts to a finite morphism of dense open subschemes. Therefore, although Steve has shown that $f$ need not be finite, it is finite on a dense subset. $\endgroup$ Commented Jun 18, 2013 at 18:01

1 Answer 1


No, $f$ might be an inclusion of an open set; for example, the morphism corresponding to the (not module finite) extension $\mathbb{C}[x] \subseteq \mathbb{C}[x,x^{-1}]$.

Edited in light of edit to question:

Here's another type of thing that can happen: look at the map $(x,y) \mapsto (x,xy)$ from $\mathbb{C}^2$ to itself. This is evidently dominant, but it does not have finite fibers (since the fiber over the origin is a line). I bet one can find an example with finite fibers which is not finite, but I can't think of one at the moment.

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    $\begingroup$ PS--- I like that you give that definition of "variety". It reminds me of the old joke about the graduate student in the back of Grothendieck's seminar who, confused when Serre mentions "variety" during a talk, is happy when Serre, er, clarifies "integral separated scheme of finite type over a field". $\endgroup$
    – Stephen
    Commented May 20, 2013 at 3:56
  • $\begingroup$ Thank you, Steve. Sorry to move the goal post, but how about if I add the condition $X=Y$ to the question? $\endgroup$
    – POJ
    Commented May 20, 2013 at 4:59
  • $\begingroup$ @POJ, I don't know off the top of my head, but suspect the answer is still no (that is, there exists a variety $X$ and a dominant map $f:X \rightarrow X$ with finite fibers that is not finite). $\endgroup$
    – Stephen
    Commented May 20, 2013 at 14:40

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