Example 33.12.7. Let $k =\mathbf{F}_ p(t)$. It is quite easy to give an example of a regular variety $V$ over $k$ which is not geometrically reduced. For example we can take $\mathop{\mathrm{Spec}}(k[x]/(x^ p - t))$. In fact, there exists an example of a regular variety $V$ which is geometrically reduced, but not even geometrically normal. Namely, take for $p > 2$ the scheme $V = \mathop{\mathrm{Spec}}(k[x, y]/(y^2 - x^ p + t))$. This is a variety as the polynomial $y^2 - x^ p + t \in k[x, y]$ is irreducible. The morphism $V \to \mathop{\mathrm{Spec}}(k)$ is smooth at all points except at the point $v_0 \in V$ corresponding to the maximal ideal $(y, x^ p - t)$ (because $2y$ is invertible). In particular we see that $V$ is (geometrically) regular at all points, except possibly $v_0$. The local ring

$\mathcal{O}_{V, v_0} = \left(k[x, y]/(y^2 - x^ p + t)\right)_{(y, x^ p - t)}$

is a domain of dimension $1$. Its maximal ideal is generated by $1$ element, namely $y$. Hence it is a discrete valuation ring and regular. Let $k' = k[t^{1/p}]$. Denote $t' = t^{1/p} \in k'$, $V' = V_{k'}$, $v'_0 \in V'$ the unique point lying over $v_0$. Over $k'$ we can write $x^ p - t = (x - t')^ p$, but the polynomial $y^2 - (x - t')^ p$ is still irreducible and $V'$ is still a variety. But the element

$\frac{y}{x - t'} \in (\text{fraction field of }\mathcal{O}_{V', v'_0})$

is integral over $\mathcal{O}_{V', v'_0}$ (just compute its square) and not contained in it, so $V'$ is not normal at $v'_0$. This concludes the example.

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