Let $X$ be a regular integral projective scheme of dimension 1 over a field $k$ (not algebraically closed). Further, assume $X$ satisfies $dim_kH^0(X,\mathscr{O}_X)=1$. Let $\bar{X}$ denote the fibered product $X\times_k\bar{k}$. Then is it true that $\bar{X}$ is integral?
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$\begingroup$ Oops! Thanks Angelo, that was quite simple. Let me modify the question. Let $X\to Y$ be a flat morphism of smooth projective varieties defined over an algebraically closed field. Suppose that for every closed point $y\in Y$, the fiber $X_y$ is integral. Then is it true that $\bar{X_{\eta}}$ is integral? $\endgroup$– RexDec 16, 2010 at 13:40
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$\begingroup$ Sorry, $\eta$ in the above is the generic point of $Y$. $\endgroup$– RexDec 16, 2010 at 13:41
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2 Answers
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This is an answer to the updated question, and it is positive in far more general situations. By EGA, IV.9.7.7, for any morphism of finite presentation $X\to Y$, the set $E$ of $y\in Y$ such that $X_y$ is geometrically integral is locally constructible. In your situation, $Y$ is noetherian and the set of the closed points is dense in $Y$. As $E$ contains the closed points of $Y$, it is equal to $Y$. So every fiber of $X\to Y$ (including the generic fiber when $Y$ is irreducible) is geometrically integral.
You don't need flatness neither smoothness neither properness hypothesis.
EDIT: The fact that $E=Y$ is only true when $Y$ is an algebraic variety because the set of the closed point is very dense. In general, $E$ contains a dense open subset, so the generic fiber is geometrically integral.
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No, there are many counterexamples. Suppose that $a \in k$ is not a square; then the conic in $\mathbb P^2_k$ with equation $x^2 - ay^2$ is integral, while after base changing to $\overline k$ it splits into two components (these are switched by the Galois group).
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$\begingroup$ Optimally nice and simple, Angelo, bravo! Mediocre nit-picking: in characteristic two there is no Galois group; the conic remains irreducible but becomes non-reduced. Anyway, it ceases to be integral in that case too, which is what Rex wanted to know. $\endgroup$ Dec 16, 2010 at 13:18
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$\begingroup$ Thanks, Georges; of course any seasoned algebraic geometer would have answered as quickly, I just happened to be in front of the computer at the right time. And of course you are right about the characteristic 2 case. $\endgroup$– AngeloDec 16, 2010 at 13:27