Suppose that $G$ is an algebraic group over a field $k$. Let $G^o$be the connected component of the identity. Since $G^0$ contains a $k$-rational point (the identity) therefore it is geometrically connected. I am wondering whether the remaining connected components of $G$ are also geometrically connected.
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1$\begingroup$ A good exercise: for any $k$-group scheme $G$ locally of finite type, the connected components of $G$ that are geometrically connected over $k$ are in bijective correspondence with the $k$-points of the etale quotient $G/G^0$. In particular, they're all geometrically connected if and only if $G/G^0$ is a constant $k$-scheme. (hint: the real content is descending from $k_s$ down to $k$) $\endgroup$– user74230Commented Dec 14, 2014 at 22:23
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No, take for example $k=\mathbb{Q}$ and $G=\mu_p$ for some prime $p>2$. Then $G$ has only two connected components, whereas $G_\overline{\mathbb{Q}}$ has $p$.
answered Dec 14, 2014 at 17:05