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Suppose $X$ is an irreducible nonsingular projective variety over a field $k$ (not necessarily algebraically closed) Let $K$ be a field extension of $k$ ( If $K/k$ is not algebraic, we can assume that the field $k$ is infinite. I don't know if this assumption is useful or not. But if necessary one can assume this). Now can we prove that the base extension $X'$= $X$ x Spec $K$ over Spec $k$ is irreducible?

If $k$ is algebraically closed, then I know that this is true. Otherwise I don't know how to prove?

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  • $\begingroup$ I'm sure this has been asked before, albeit without the hypothesis of projectiveness. Anyway, the field extension itself will often give the counterexample, as pointed out below. $\endgroup$ – Zhen Lin Jul 25 '14 at 14:28
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If $k$ is separably algebraically closed, then irreducible is the same as geometrically irreducible; otherwise it isn't. Take $X$ to be the spectrum of a finite separable extension of $k^\prime$ of $k$. This is both projective and étale (hence smooth) over $k$, but if we take $K=k^\prime$, then $X_K=\mathrm{Spec}(k^\prime\otimes_kk^\prime)$ is not irreducible.

In general, a $k$-scheme $X$ is geometrically irreducible over $k$ if and only if for each generic point $\eta$ of an irreducible component of $X$, $k$ is separably closed in $k(\eta)$. Note this uses no hypotheses (no smoothness, no projectivity, no finiteness, etc.).

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