Frobenius on the affine line is finite flat but not smooth

The presentation. $B:=k[u]=k[t][u]\mathrm{/}(u^p-t)$ (the quotient map sends $u$ to the class of $u$, and $u^p=t$ holds in $B$). We claim that $1,u,\dots ,u^{p-1}$ is a $k[t]$-basis of $B$. Every power $u^n$ with $n\ge p$ equals $t{u}^{n-p}$, so the displayed elements span $B$ over $k[t]$ and every element of $B$ is ${\sum }_{i=0}^{p-1}{c}_i(u^p)u^i$ with $c_i\in k[t]$. If such a combination vanishes in $k[u]$, then its coefficients in the basis $\{u^n:n\ge 0\}$ of $k[u]$ over $k$ all vanish; the exponent $n$ receives contributions only from the $i$ with $i\equiv n(\mathrm{m}\mathrm{o}\mathrm{d}p)$, hence $c_i(u^p)=0$ for all $i$ and $c_i=0$ because $u$ is transcendental over $k$. This proves the claim.

Finiteness and finite presentation. The basis of step 1.1 exhibits $B$ as a module-finite $k[t]$-algebra, generated by $u$ (indeed by $1,u,\dots ,u^{p-1}$), so $F$ is finite by [F3]. Moreover $k[t][u]$ is a polynomial algebra over $k[t]$, hence a finitely presented $k[t]$-algebra by [F4], and $B$ is its quotient by the principal ideal $(u^p-t)$, so $B$ is a finitely presented $k[t]$-algebra and $F$ is locally of finite presentation by [F4].

Flatness. By step 1.1, $B$ is a free $k[t]$-module, hence flat over $k[t]$ by [F2]. The morphism $F$ has the single affine chart $\mathrm{Spec}B\to \mathrm{Spec}k[t]$, so flatness at every point follows from the affine-local criterion [F1].

The special fibre. Applying [F9] to $k[t]\to B$ and the residue map $k[t]\to k$, $t\mapsto 0$, the fibre over the prime $(t)$ is $B{\otimes }_{k[t]}k=k[u]\mathrm{/}(u^p)$. The prime $(u)\subseteq B$ lies over $(t)$ because $t=u^p\in (u)$, and its image in the fibre is the maximal ideal $(u)k[u]\mathrm{/}(u^p)$.

The fibre local ring is not regular. Write $R=(k[u]\mathrm{/}(u^p){)}_{(u)}$ for the local ring of the fibre at the image of $(u)$, with maximal ideal $\mathfrak{m}=(u)R$. Since $u^p=0$ in $R$, one has ${\mathfrak{m}}^p=0\subseteq P$ for every prime $P$ of $R$; as $P$ is prime this forces $\mathfrak{m}\subseteq P$, hence $P=\mathfrak{m}$ by maximality of $\mathfrak{m}$. Thus $\mathfrak{m}$ is the only prime of $R$ and $\dim R=0$ by [F10]. On the other hand $u\mathrm{\notin }{\mathfrak{m}}^2$ and $\mathfrak{m}=(u)$, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ is one-dimensional over $\kappa =k$, i.e. $\mathrm{edim}R=1$ by [F7]. Therefore $\dim R=0\mathrm{\ne }1=\mathrm{edim}R$ and $R$ is not regular.

Failure of smoothness. If the fibre were geometrically regular at the image of $(u)$, then by the case $K=\kappa ((t))=k$ of [F6] the local ring $R$ would be regular; step 3.1 shows it is not, so the fibre is not geometrically regular at that point. Since $F$ is locally of finite presentation (step 2.1) and flat (step 2.2) but its fibre fails geometric regularity, [F5] shows that $F$ is not smooth at the prime $(u)\in \mathrm{Spec}k[u]$. By [F8] $F$ is therefore not étale at $(u)$, and hence neither smooth nor étale; the relative derivative remark in the Statement is the observation that the Jacobian of the presentation $k[t][u]\mathrm{/}(u^p-t)$ is $p{u}^{p-1}=0$ in $k[u]$. [F5, F6, F8, step 3.1] $\mathrm{\square }$