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Let $X,Y,Z$ are irrreducible varieties. $f:X\to Y$ is prpoer surjective and $g:Z \to Y$ is dominant.

Then, $X\times_Y Z$ is irreducible?

Moreover, it will be very helpful for me if there are other conditions of morphisms $f,g$ that makes $X\times_Z Y$ irreducible.

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    $\begingroup$ There are easy counterexamples where $X$, $Y$, and $Z$ are fields and the morphisms are finite Galois extensions. For example, $\mathbb C \otimes_{\mathbb R} \mathbb C \cong \mathbb C \times \mathbb C$. You can make more geometric examples by taking finite coverings between curves. $\endgroup$ – R. van Dobben de Bruyn Apr 15 '17 at 5:01
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    $\begingroup$ There are much stranger counterexamples where the morphisms even have geometrically integral fibres. For example, consider a non-small blow-up like $\operatorname{Bl}_p(\mathbb P^3) \to \mathbb P^3$. Take the product of this morphism $X \to Y$ with itself, and note that its fibre above $y \in Y$ is $X_y \times X_y$. Above $p$, we get $\mathbb P^2 \times \mathbb P^2$, and every other fibre is an isomorphism. Thus, $X \times_Y X$ has components of different dimensions! $\endgroup$ – R. van Dobben de Bruyn Apr 15 '17 at 5:16
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A good criterion for the irreducibility of a fiber product is when one of the morphisms, say $f$, is flat with irreducible fibers. Flatness implies that $X\times_YZ\to Z$ is also flat. Hence every irreducible component maps dominantly to $Z$. The second condition implies that there is only one component of that type.

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    $\begingroup$ At the very least, you'll want to assume that the fibres are geometrically irreducible, to prevent examples like field extensions. But it seems that under those assumptions, it should indeed be ok. $\endgroup$ – R. van Dobben de Bruyn Apr 15 '17 at 17:52

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