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 $f:X\to X$.
Best Answer
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.