Let $K$ be a field of characteristic zero and $X$ be a projective variety over $\mbox{Spec} K$. Denote by $\bar{K}$ the algebraic closure of $K$. Let $\bar{X}$ be the fiber product $X \times_K \bar{K}$. Suppose that $\bar{X}$ is a smooth Fano variety. Is it true that $X$ is Fano?
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1$\begingroup$ Yes. $X$ is smooth because $X_{\bar K}$ is smooth (apply the definition). Moreover, the anti-canonical bundle of $\omega_{X/K}$ is ample as the anti-canonical bundle of $X_{\bar K}$ is ample. (There are many ways to see the latter.) $\endgroup$– Ariyan JavanpeykarCommented May 30, 2014 at 19:15
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Let $f:X\rightarrow Y$ be a morphism of schemes over $S$. If $g :T\rightarrow S$ is faithfully flat and quasi-compact, and $F:X^{'}\rightarrow Y^{'}$ is the base changes by $g$, you have the follwing: an invertible sheaf $\mathcal{L}$ is $f$-ample if and only if its pull-back $L^{'}$ is ample.
In you case this yields that $X$ is Fano if and only if $\overline{X}$ is Fano.
