# Dominant morphism, equal dimensions: always finite?

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 $f:X\to X$.

• 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. Commented May 20, 2013 at 14:39
• 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. Commented Jun 18, 2013 at 18:01

No, $f$ might be an inclusion of an open set; for example, the morphism corresponding to the (not module finite) extension $\mathbb{C}[x] \subseteq \mathbb{C}[x,x^{-1}]$.
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.
• Thank you, Steve. Sorry to move the goal post, but how about if I add the condition $X=Y$ to the question?
• @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). Commented May 20, 2013 at 14:40