Are birational morphisms stable under base change via a dominant morphism Let $f: X \to Y$ be a birational morphism of integral schemes and $g: Z \to Y$ a morphism of integral schemes which maps the generic point of $Z$ to the generic point of $Y$, i.e., the morphism $g$ is dominant.
Is then $X \times_Y Z \to Z$ birational?
Edit: My ideas: Denote the generic points of $X,Y,Z$ by $\eta_X, \eta_Y, \eta_Z$. Then $f$ induces an isomorphism $\eta_X = \eta_Y$. Denote the base change $X \times_Y Z \to Z$ by $f'$. Then $f'$ induces an isomorphism $g'^*(\eta_X) = \eta_Z$?
 A: 
The fiber product can be reducible

Yes, $X \times_Y Z \to Z$ will always be an isomorphism over an open set $W \subset Z$. However it's possible for $X \times_Y Z$ to be reducible.
Example
Let $k$ be a field and let  $Y = \mathbb{A}_k^3$; let $X = \mathrm{Bl}_p Y$ where $p \in Y$ is a closed point (e.g. the origin), and let $f: X \to Y$ be the projection. Let $E \subset X$ be the exceptional divisor (it's a copy of $\mathbb{P}^2$ over $p$), and let $\iota: E \to X$ be the inclusion. Take $Z = X$ and $f=g$ -- then $g$ is certainly dominant since it's also birational. 
In this situation the fiber product $X \times_Y X$ has 2 components: one is the image of the diagonal $X \xrightarrow{\Delta} X \times_Y X$ over $Y$ (so just a copy of $X$). The other component is the image of the map 
$$
E \times_{\mathrm{Spec} k(p)} E \xrightarrow{\iota \times \iota} X \times_Y X
$$
In particular, this component is 4 dimensional, whereas the other is 3 dimensional (being birational to $\mathbb{A}^3$)! Since $X \times_Y X$ isn't even equidimensional, it can't be irreducible. 
Requiring $Z$ to be flat might ensure the fiber product is integral. 
I think if we assume in addition that $g: Z \to Y$ is a flat morphism, we can guarantee that $X \times_Y Z$ is integral, but I'm having a hard time finding a reference for this. 
Note that flat implies dominant, since flat morphisms are open. Also note that flatness would rule out the example above ($f: X \to Y$ was not flat, for instance because its fiber dimension was not constant). 
A: Since $f: X \to Y$ is birational, we can find some open subsets $U \subset Y$ and $V \subset X$ so that $f$ restricts to an isomorphism $f: V \to U$. Then $g: W = g^{-1}(U) \to U$ is still dominant (really dominance of $g$ here just guarantees that $W$ is nonempty for any open $U$ we may need to restrict to). 
Then the pullback $V \times_U W \to W$ is an isomorphism. $V \times_U W$ is an open subset of $X \times_Y Z$ and the morphism is just the restriction of the pullback $X \times_Y Z \to Z$. Thus the pullback is birational since it induces an isomorphism on open subsets.
