A regular point that is not smooth: a purely inseparable thickening

The field and its characteristic. By [F1] the field $k={\mathbb{F}}_p(s)=\mathrm{Frac}({\mathbb{F}}_p[s])$ consists of the fractions $f\mathrm{/}g$ with $f,g\in {\mathbb{F}}_p[s]$, $g\mathrm{\ne }0$, and contains an embedded copy of ${\mathbb{F}}_p=\mathbb{Z}\mathrm{/}p$. In ${\mathbb{F}}_p$ one has $p\cdot 1=0$ while $n\cdot 1\mathrm{\ne }0$ for every integer $n$ with $0<n<p$, since such an $n$ is not a multiple of $p$; a ring embedding preserves natural multiples of the identity, so in $k$ also $p\cdot 1_k=0$ and $n\cdot 1_k\mathrm{\ne }0$ for $0<n<p$. By the definition of characteristic in [F1] this is exactly $\mathrm{char}k=p$, so $k$ is a field of characteristic $p>0$.

The element $s$ is not a $p$th power. Suppose $s=(f\mathrm{/}g{)}^p$ with $f,g\in {\mathbb{F}}_p[s]$ and $g\mathrm{\ne }0$; then $s{g}^p=f^p$ in the polynomial ring ${\mathbb{F}}_p[s]$, which is a UFD by [F2]. Evaluation at $0$ is a surjective ring homomorphism ${\mathbb{F}}_p[s]\to {\mathbb{F}}_p$ with kernel $(s)$, so ${\mathbb{F}}_p[s]\mathrm{/}(s)\cong {\mathbb{F}}_p$ is a field and $(s)$ is a maximal, hence prime, ideal by [F3]; therefore $s$ is a prime element, and the $s$-adic order ${\mathrm{ord}}_s$ on nonzero polynomials, which records the largest power of $s$ dividing an element, is additive over products. Comparing orders in $s{g}^p=f^p$ gives $1+p{\mathrm{ord}}_s(g)=p{\mathrm{ord}}_s(f)$, which is impossible because $p\ge 2$ does not divide $1$. Hence no element of $k$ has $p$th power $s$, that is, $s\mathrm{\notin }{k}^p$.

The general theorem applies to this pair. The field $k$ has characteristic $p>0$ by step 1.1, and $s\mathrm{\notin }{k}^p$ by step 1.2, so [F4] applies with $F=k$ and $a=s$: the ring $L=k[t]\mathrm{/}(t^p-s)$ is a field, the structural map $k\to L$ is injective, ${\alpha }^p=s$ and $L=k[\alpha ]$; the affine $k$-scheme $X=\mathrm{Spec}L$ is of finite type over $k$ and regular; for every field extension $K\mathrm{/}k$ and every $\beta \in K$ with ${\beta }^p=s$ there is a $K$-algebra isomorphism $L{\otimes }_{k}K\cong K[u]\mathrm{/}(u^p)$, where $K[u]\mathrm{/}(u^p)$ is a Noetherian local ring with unique prime $(u)$, Krull dimension $0$ and embedding dimension $1$, and is not regular; and consequently $X\to \mathrm{Spec}k$ is not smooth, although $X$ is regular. The Axiom of Choice is used only here, through [F4], as declared in [F8].

Adjoining the $p$th root: the base change at $K=L$. By step 2.1 the element $\alpha \in L$ satisfies ${\alpha }^p=s$, so the witness clause of [F4] applies with $K=L$ and $\beta =\alpha$ and gives the $L$-algebra isomorphism $L{\otimes }_{k}L\cong L[u]\mathrm{/}(u^p)$; this is the coordinate ring of the base change $X_L$ of $X$ along $\mathrm{Spec}L\to \mathrm{Spec}k$. The ring $L[u]\mathrm{/}(u^p)$ is a Noetherian local ring with unique prime and maximal ideal $(u)$, Krull dimension $0$, embedding dimension $1$, and it is not regular; every element outside $(u)$ is a unit, so its localisation at $(u)$ is the ring itself, and by [F9] that localisation is the local ring of the unique point of $X_L$. Hence that point is nonregular, although its image under $X_L\to X$ is the regular point of $X$.

The extension is purely inseparable. Every element of $L=k[\alpha ]$ is a $k$-linear combination $z={\sum }_{i=0}^{p-1}{c}_i{\alpha }^i$ with $c_i\in k$: each power ${\alpha }^m$ with $m\ge p$ is reduced by ${\alpha }^m={\alpha }^{m-p}{\alpha }^p=s{\alpha }^{m-p}$. By [F7] the $p$th power map of the field $L$ is additive and multiplicative, so $z^p={\sum }_{i=0}^{p-1}{c}_i^p{\alpha }^{ip}={\sum }_{i=0}^{p-1}{c}_i^p{s}^i\in k$, because ${\alpha }^{ip}=({\alpha }^p{)}^i=s^i$ for $i\ge 1$, and $c_i^p\in k$ and $s^i\in k$ for every $i$. Hence every $z\in L$ is a root of the nonzero polynomial $x^p-z^p$ over $k$, so every element of $L$ is algebraic over $k$ by [F6] and $L\mathrm{/}k$ is an algebraic extension; with $\mathrm{char}k=p>0$ from step 1.1 and $z^p\in k$ for every $z$, the definition [F5] shows that $k\subseteq L$ is purely inseparable.

Boundaries and conclusion. The argument includes $p=2$, where $L[u]\mathrm{/}(u^2)$ is the dual-numbers ring over $L$, local with Krull dimension $0$ and embedding dimension $1$ and not regular. The hypothesis $s\mathrm{\notin }{k}^p$ is genuinely used: it holds in ${\mathbb{F}}_p(s)$ by step 1.2, but over $K=L$ the same element satisfies $s={\alpha }^p$, and correspondingly $t^p-s=(t-\alpha {)}^p$ over $K$, so the base change acquires the class $u=t-\alpha$ with $u^p=0$; this is why the regularity detected in $k$ is not stable. No reduction or Frobenius twist is applied: the isomorphism of step 3.1 is an isomorphism of the actual tensor product $L{\otimes }_{k}L$ and retains the class $u$. The scheme $X$ is nonempty, since $L$ is a field with $1\mathrm{\ne }0$, and has a single point of residue field $L$; the empty-scheme case is therefore absent, while $X$ has Krull dimension zero and embedding dimension zero because its local ring is the field $L$. Its base change in step 3.1 still has dimension zero but has embedding dimension one, and the finite-type hypothesis of [F4] is met by construction. Perfectness of $k$ fails: $s\mathrm{\notin }{k}^p$ exhibits the Frobenius $x\mapsto x^p$ of $k$ as nonsurjective by [F7]. Choice is declared in [F8] and is used only through [F4]; the example exhibits one field, one element and one base change, so no simultaneous selection occurs.