Dominant map is dominant in generic fiber

Question

Suppose we have a dominant morphism of varieties $f : X \rightarrow Y$ over $Z$ (i.e we have two fibrations $X \rightarrow Z$ and $Y \rightarrow Z$ on which we can add all the properties needed as flat, smooths, etc., and $X \rightarrow Y$ is a morphism of $Z$-scheme). We also suppose that $f$ is dominant. Is it true that $f_z : X_z \rightarrow Y_z$ is dominant above a general $z \in Z$ (maybe in an dense open subset of $Z$)) ?

Answer 1

We assume the following:

1. $Z$ is Noetherian
2. $f$ and $y: Y \rightarrow Z$ are of finite type
3. $f(X)$ is dense in $Y$

I claim that there is an everywhere dense open subset $U \subset Z$ such that for every $z \in U$, $f(X_z)$ is dense in $Y_z$.

Clearly, we can assume that $Z$ is affine irreducible with generic point $\eta$ (then $X,Y$ are quasi-compact) and that $Y_{\eta} \neq \emptyset$ (otherwise the statement is trivial).

By Chevalley, $f(X)$ is a dense constructible subset of $Y$, so contains an everywhere dense open subset $W \subset Y$.

First, let’s show that $W_{\eta}$ is dense in $Y_{\eta}$. Let $v \in Y_{\eta}$ be generic, then $y$ is the specialization of some generic $u \in Y$. But then $y(v)=\eta$ is a specialization of $y(u)$, so $u \in Y_{\eta}$, thus $u=v$ is generic in $Y$ so is in $W \cap Y_{\eta}=W_{\eta}$.

So we use the “complement” of Stacks, 37.24.3 to conclude.

If you don’t want to track the dependencies, here’s how the argument goes (to show from there that the fibers of $W$ are dense in the fibers of $Y$, when restricted to some open subset of $Z$).

First, we can assume that $Z$ is affine and integral (by changing base by the reduction of some affine open subscheme). We write $T$ for the reduced closed subscheme of $Y$ the underlying set of which is $Y \backslash W$.

Let $V \subset Z$ be a nonempty open subset such that $Y_V, T_V$ are flat over $V$ (it exists, it’s standard, but not trivial).

Assume that $z \in V$ is such that there is a generic point $u \in T_z$ of $Y_z$. Then $O_{Z,z} \rightarrow O_{T,u}$ is local and flat, so is faithfully flat, and thus there is a point $v \in T_{\eta}$ specializing to $u$.

Flatness implies that $\dim_u (T_z)=\dim_v (T_{\eta})$, and same for $Y$ (“the dimension of fibers of a flat morphism is locally constant”). But $\dim_u(T_z)=\dim_u(Y_z)$, therefore $\dim_v(T_{\eta})=\dim_v(Y_{\eta})$.

So there is an irreducible component $v \in S \subset T_{\eta}$ whose dimension is at least that any irreducible component of $Y_{\eta}$ going through $v$: but $S$ must be contained in an irreducible component $S’$ of $Y_{\eta}$, thus by dimension $S=S’$ and $T_{\eta}$ contains one of the generic points of $Y_{\eta}$, a contradiction. QED.