A purely inseparable field has nonzero Omega

The class $\alpha$ satisfies ${\alpha }^p=a$ by construction, and $L=k(\alpha )$, since $L$ is generated as a $k$-algebra by $\alpha$. If $f=X^p-a$ is irreducible, then $L$ is a field of degree $p$ over $k$ and $\alpha$ is a primitive element; the next steps establish the irreducibility.

Assume for contradiction that $f$ is reducible. Since $k[X]$ is a principal ideal domain [F1], a reducible nonzero non-unit factors into irreducibles, so $f$ has a monic irreducible factor $m$ of degree $d$ with $1\le d\le p-1$. Let $A=k[X]\mathrm{/}(m)$, a field by [F2], and let $x$ denote the class of $X$ in $A$; then $m(x)=0$, and $x^p=a$ because $m$ divides $f=X^p-a$ in $k[X]\subseteq A[X]$.

In $A[X]$ the binomial theorem [F3] together with $p\mid \left(\genfrac{}{}{0px}{}{p}{i}\right)$ for $0<i<p$ [F4] gives the Frobenius identity $(X-x{)}^p=X^p-x^p=X^p-a$, the intermediate coefficients vanishing in $A$ of characteristic $p$; hence $m$, considered in $A[X]$, divides $(X-x{)}^p$.

On the other hand $m(x)=0$, so the division algorithm in the field $A$ [F5] gives $m=q(X-x)+m(x)=q(X-x)$, that is, $X-x$ divides $m$. In the principal ideal domain $A[X]$ [F1], the degree-one polynomial $X-x$ is irreducible (a factorization would have to split the degree $1$ into two nonnegative degrees, forcing a degree-$0$ factor, which is a unit of $A[X]$), so the only monic factor of $(X-x{)}^p$ of degree $d$ is $(X-x{)}^d$; since $m$ is monic of degree $d$, we get $m=(X-x{)}^d$.

Comparing the coefficient of $X^{d-1}$ in the identity $m=(X-x{)}^d$ of step 4.1 gives: the coefficient of $X^{d-1}$ in $(X-x{)}^d$ is $-dx$, and the coefficient of $X^{d-1}$ in $m\in k[X]$ lies in $k$, so $dx\in k$. Since $1\le d\le p-1$ and $k$ has characteristic $p$, the class of $d$ in $k$ is nonzero and invertible, so $x=d^{-1}(dx)\in k$; then $a=x^p\in k^p$, contradicting the hypothesis $a\mathrm{\notin }{k}^p$. Hence $f=X^p-a$ is irreducible, $L$ is a field with $[L:k]=p$, and $L=k(\alpha )$.

Now compute the differentials. Apply [F6] with $A=k$, $n=r=1$, $P=k[X]$, $q=f=X^p-a$ and $B=L$: the derivative is $f^{\mathrm{\prime }}=p{X}^{p-1}=0$, because $p=0$ in $k$, so the relation submodule $L\cdot f^{\mathrm{\prime }}(\alpha )$ is zero and ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\mathrm{/}(0)$ with the image of the basis vector written $\mathrm{d}\alpha$. Thus ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\mathrm{d}\alpha \cong L$ as $L$-modules, in particular ${\mathrm{\Omega }}_{L\mathrm{/}k}\mathrm{\ne }0$ because the field $L$ is nonzero.

The extension is finite, algebraic and purely inseparable: every element of $L=k(\alpha )$ is a polynomial ${\sum }_i{c}_i{\alpha }^i$ with $c_i\in k$ by step 5.1, and its $p$-th power is ${\sum }_i{c}_i^p{a}^i\in k$ because ${\alpha }^p=a$ and the Frobenius map is additive in characteristic $p$ [F3]. So every element of $L$ has its $p$-th power in $k$, and the elementwise criterion of [F7] makes $L\mathrm{/}k$ purely inseparable; by step 5.1 it is finite of degree $p$.

Combining steps 6.1 and 6.2: ${\mathrm{\Omega }}_{L\mathrm{/}k}$ is a free $L$-module of rank one and hence nonzero, while $L\mathrm{/}k$ is a finite algebraic purely inseparable extension. This refutes the displayed statement and shows that the vanishing of $\mathrm{\Omega }$ for finite separable extensions cannot be extended to all finite algebraic extensions; the obstruction is precisely the vanishing derivative $f^{\mathrm{\prime }}=0$ of the inseparable polynomial $X^p-a$.