Let $f:X\to S$ and $g:Y\to T$ be dominant morphisms. Then under what conditions does it hold that $f\times g$ is dominant? Is it enough to suppose that $X,Y$ are irreducible? I need it only in fairly restrictive conditions (e.g. smooth projective over an algebraically closed field), but still I wonder in what generality it holds, and can't seem to find any references for it.
Under what conditions is a product of dominant morphisms dominant
algebraic-geometryschemes
Related Solutions
[This answer has been edited to discuss the general case.]
I will assume that variety means irreducible (otherwise you could work on individual irreducible components). Then $f:X \to Y$ is dominant by assumption, and has closed image since its source is projective, thus it is surjective. Thus $f^{-1}(Z) \to Z$ is also surjective.
Now, as described in e.g. this MO answer (or in a Hartshorne exercise, maybe in Section 4 of Chapter II), for the proper map $f$, the function $y \mapsto \dim f^{-1}(y)$ is upper semicontinuous, so if $z \in Z$, then the dimensions of $f^{-1}(z)$ are at least $\dim X - \dim Y$. This implies that $f^{-1}(Z)$ contains components of codimension $1$. (The intuition is just that we can add the dimension of $Z$ and of a typical fibre. One way to make this precise is to note that if every component of $f^{-1}(Z)$ were of codimension at least $2$, then since it dominates $Z$, we would see that a generic fibre would be of dimension $\dim X - \dim Y -1$, whereas we already noted that every fibre has dimension at least $\dim X - \dim Y$.)
In general you can't do better than this, because there are morphisms of $3$-folds (just to take an example) which are birational, but in which the preimage of some particular point $y \in Y$ is a curve. Then if you take $Z$ to be a generic codim'n one subvariety passing through $y$, its preimage will be the union of a codim'n subvariety of $X$ (the proper transform of $Z$) and the curve $f^{-1}(P)$. So in general you can't expect $f^{-1}(Z)$ to be equidimensional.
Note also that $f^{-1}(Z)$ can also contain multiple components of codimension $1$. (E.g. let $X \to Y$ be the blow up of a surface at a point, and let $Z$ be a curve that passes through the blown up point.)
Rereading the question, I see that the point of the question might be the equidimensionality, and so you might be interested in the counterexample involving $3$-folds. This MO question and answers gives one such example.
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.
Best Answer
you need some flatness condition. just look at the affine case: for integral affine schemes dominance is equivalent to the condition that $A\to B$ is injective. now if $A\to B$ is injective and $C\to D$ is injective there is no reason for $A\otimes C\to B\otimes D$ to be injective.
but if you are working over an algebraically closed field then all is fine: $f:X\to Y$ is dominant between integral schemes $X,Y$ if and only if there are some open affine subschemes of $X,Y$ like $U=Spec\, A,V=Spec\, B,f(U)\subset V$ such that the induced map between $A\to B$ is injective. by this and the fact that everything is flat over a field(and product of two integral scheme over an algebraically closed field remain Integral) you can deduce the dominance of product.
the problem is topological so you can deduce the same for irreducible schemes over a field form the integral case.