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Is there a field $k$ and a regular integral $k$-variety $X$ that is neither affine, projective, geometrically connected, geometrically reduced, nor geometrically regular?

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    $\begingroup$ Try something like $\mathbb{A}^2_L-\{(0,0)\}$ where $L=\mathbb{F}_{p^2}(T)(T^{\frac{1}{p}})$ over $\mathbb{F}_p(T)$. $\endgroup$ Commented Sep 7, 2015 at 4:51

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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.

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Thanks to Peter Xu for helpful discussion on this question.

Let $p$ be a prime, and let $k = \mathbb{F}_p(t)$. Let $L = \mathbb{F}_{p^2}(t^{1/p})$, which is neither separable nor purely inseparable over $k$. Let $X = \mathbb{A}^1 \times \mathbb{P}^1 \times \text{Spec}\,L$ considered as a $k$-variety.

The scheme $X = (\mathbb{A}^1 \times \mathbb{P}^1)_K$ is regular and integral, since these properties are absolute, not depending on whether the scheme is considered as $L$-variety or $k$-variety. We have $\Gamma(X, \mathcal{O}_X) = L[z]$, where $z$ is the coordinate on $\mathbb{A}^1$. Since $\dim X = 2 \neq 1 = \dim \text{Spec}\,L[z]$, the variety $X$ is not affine. Since $\dim_k L[z] = \infty$, we have that $X$ is not projective over $k$.

We have $L \cong \mathbb{F}_{p^2} \otimes_{\mathbb{F}_p} \mathbb{F}_p(t)[u]/(u^p - t)$, so$$L \otimes_k \overline{k} \cong \mathbb{F}_{p^2} \otimes_{\mathbb{F}_p} {{\overline{k}[u]}\over{(u^p - t)}} \cong {{\overline{k}[u]}\over{(u^p - t)}} \times {{\overline{k}[u]}\over{(u^p - t)}}.$$In particular, $\text{Spec}(L \otimes_k \overline{k})$ is disconnected, and above each connected component lies a nontrivial component of $X \times_k \overline{k} \cong (\mathbb{A}^1 \times \mathbb{P}^1)_{(L \otimes_k \overline{k})}$. Thus, $X$ is not geometrically connected. Also, $u - t^{1/p}$ is a nonzero nilpotent global section of the structure sheaf of $X \times_k \overline{k}$, so $X$ is not geometrically reduced. Finally, regular implies reduced, so geometrically regular implies geometrically reduced, so $X$ can not be geometrically regular.

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