A question on base change 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?
 A: 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).
A: 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. 
