A finite separable extension and an inseparable extension with differentials

The separable case. By [F1] write $E=k(\alpha )$, and let $P\in k[X]$ be the monic minimal polynomial of $\alpha$, so that $E\cong k[X]\mathrm{/}(P)$ by [F2]. Since $E\mathrm{/}k$ is separable the element $\alpha$ is separable, so $P$ is separable and $\gcd (P,P^{\mathrm{\prime }})=1$ by [F4]; as $P(\alpha )=0$ and $P^{\mathrm{\prime }}(\alpha )=0$ would force $P\mid P^{\mathrm{\prime }}$ by [F2], impossible for $0\mathrm{\ne }{P}^{\mathrm{\prime }}$ of degree less than $\mathrm{deg}P$, we get $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$. By [F3] applied to the presentation $E=k[X]\mathrm{/}(P)$ the module ${\mathrm{\Omega }}_{E\mathrm{/}k}$ is $E\mathrm{/}(P^{\mathrm{\prime }}(\alpha ))$, and $P^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ is a unit of the field $E$, so ${\mathrm{\Omega }}_{E\mathrm{/}k}=0$.

The polynomial $t^p-a$ is irreducible. Let $Q$ be a monic irreducible factor of $t^p-a$ [F5] and put $F=k[t]\mathrm{/}(Q)$, a field with class $x$ of $t$ satisfying $x^p=a$ [F5]. By [F6] the identity $(t-x{)}^p=t^p-x^p=t^p-a$ holds in $F[t]$, so every root of $Q$ in a splitting field is a root of $(t-x{)}^p$, hence equals $x$: the polynomial $Q$ has exactly one distinct root. If $\mathrm{deg}Q=1$ then $x\in k$ and $a=x^p\in k^p$, contrary to the hypothesis, so $\mathrm{deg}Q\ge 2$; and $\mathrm{deg}Q\ge 2$ with $Q^{\mathrm{\prime }}\mathrm{\ne }0$ would make $Q$ separable with $\mathrm{deg}Q$ distinct roots by [F4], a contradiction. Hence $Q^{\mathrm{\prime }}=0$, which in characteristic $p$ means that $Q$ is a polynomial in $t^p$ and, since $0\mathrm{\ne }Q$ has degree at most $p$, forces $Q=t^p-a$; so $t^p-a$ is irreducible, $L=k[t]\mathrm{/}(t^p-a)$ is a field by [F5], and $t\in L$ satisfies $t^p=a\mathrm{\notin }{k}^p$ hence $t\mathrm{\notin }k$.

The differentials of the inseparable field. By [F3] applied to the single equation $t^p-a$ with $A=k$ and $B=L$, and by the derivative computation $P^{\mathrm{\prime }}(t)=p{t}^{p-1}=0$ of [F6], the module is ${\mathrm{\Omega }}_{L\mathrm{/}k}=L\mathrm{/}(0)=L$, generated by $\mathrm{d}t$; explicitly ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\cdot \mathrm{d}t$ is one-dimensional over the field $L$ and nonzero. Since $t^p=a$ with $a\in k\setminus k^p$, the element $t$ is not separable over $k$, so this is a field extension with a nonzero differential module, in contrast with clause 1.