TITLE: Not affine, projective, geometrically connected, geometrically reduced, nor geometrically regular...
QUESTION [6 upvotes]: Is there a field $k$ and a regular integral $k$-variety $X$ that is neither affine, projective, geometrically connected, geometrically reduced, nor geometrically regular?

REPLY [3 votes]: This is more of a comment on Kevin's answer, but it's too long. I think the exposition can be made clearer with less technicalities (I don't even follow the second isomorphism) and complications. The point is that affine/projective are easy to get rid of, so we focus on the latter three. If we want to make a regular thing non-reduced (hence non-regular) and non-connected after an extension of scalars, we work over an imperfect coefficient field like $k=\mathbb{F}_p(t)$, and take the spectrum associated to a one-variable polynomial which, on a suitable field extension, will split into relatively prime factors (making it disconnected) which also ramify (making it non-reduced, hence nonregular). So for example, $\text{Spec } k[x]/(x^{2p}-t)$, since upon extension to $k'=\mathbb{F}_p(t^{1/(2p)})$, we get that $x^{2p}-t=(x-t^{1/(2p)})^p(x+t^{1/(2p)})^p$. Here $p$ is any odd prime. 
Then you do the easy part, which is making it non-affine/projective. The easiest way is to take projective space over our variety (which preserves all the properties we want), then remove a point. Valuative criterion shows it's not even proper over the base field (hence not projective), and dimension-counting like Kevin did shows it's not affine.