Base change of an inseparable field extension is a thickening

The ring $L$ is a field. Choose a monic irreducible factor $Q$ of $T^p-a$ by [F1] and let $\xi$ be the class of $T$ in the field $F=k[T]\mathrm{/}(Q)$. Then ${\xi }^p=a$, and [F2] gives $T^p-a=(T-\xi {)}^p$ in $F[T]$. Thus $Q$ has only one distinct root in a splitting field. If $\mathrm{deg}Q=1$, then $\xi \in k$ contradicts $a\mathrm{\notin }{k}^p$. If $Q^{\mathrm{\prime }}\mathrm{\ne }0$, irreducibility and [F1] would make $Q$ separable with $\mathrm{deg}Q\ge 2$ distinct roots, also impossible. Thus $Q^{\mathrm{\prime }}=0$, so all exponents of $Q$ are divisible by $p$. Since $1\le \mathrm{deg}Q\le p$ and $Q$ is monic, it has degree $p$; as a monic divisor of $T^p-a$ of that degree it equals $T^p-a$. Hence $L$ is a field by [F1].

The tensor product. By [F3] there is an isomorphism $L{\otimes }_{k}L\cong L[T]\mathrm{/}(T^p-a)$, the second factor acting on coefficients; by [F2] one has $T^p-a=(T-\alpha {)}^p$ in $L[T]$, so substituting $u=T-\alpha$, an automorphism of $L[T]$, gives $L{\otimes }_{k}L\cong L[u]\mathrm{/}(u^p),$ under which $u$ corresponds to $T\otimes 1-1\otimes \alpha$.

The element $u$ is a nonzero nilpotent. In $L[u]\mathrm{/}(u^p)$ the classes of $1,u,\dots ,u^{p-1}$ are an $L$-basis, because $u^p$ is monic of degree $p$ and division with remainder is available: hence $u\mathrm{\ne }0$ while $u^p=0$. Consequently $L{\otimes }_{k}L$ is not reduced, so the base change of the field $L$ along $k\to L$ is a nonreduced local ring with residue field $L$, and the fibre is a thickening rather than a reduced point.