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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)

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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 $E$.

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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.

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Interestingly enough when the degree is $3$ this is a special case of Tom's example. –  Torsten Ekedahl Oct 18 '10 at 19:03
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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.

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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 $3$.

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