irreducible components of the fibre product Is it known if all the irreducible components of  the fibre product Xx_sY 
 are birationally equivalent? (Suppose X,Y,s are irreducible complex curves, for instance)
 A: No. If $E/F$ is a cubic field extension, separable but not normal, then $E\otimes_F E$ is the cartesian product of $E$ and the Galois closure of $E$. 
A: Certainly not: even in the case of $X=Y=S=\mathbb{P}^1$ and the two maps $X,Y \to S$ are the same and general of degree at least three.  In this case one component is the "diagonal" $\mathbb{P}^1$ and the remaining component is a curve of genus 1, I think.
A: In higher dimensions, the irreducible components don't even have to be of the same dimension. Consider a blow up of a smooth (closed) point $f:X\to S$ with exceptional divisor $E$. Then $X\times_S X$ has a component that is birational to $X$ and another one which is just $E\times E$ (here the $\times$ is over the point that is blown up). It is easy to see that this one has dimension $2\dim X-2$ which is larger than $\dim X$ as soon as the latter is at least $3$.
A: Outside of the curve case, this can even happen for birational maps.  One example is to take a non-rational surface $X$ over $\mathbb{C}$ and blow-up a smooth point, yielding $\pi : Y \to X$.  The product $Y \times_X Y$ has two components.  One is isomorphic to $Y$ (and thus birational to $X$) and the other is a rational surface (if I recall correctly, just $\mathbb{P}^1 \times_{\mathbb{C}} \mathbb{P}^1$) intersecting the $Y$ component along the exceptional divisor.  
