A finite extension generated by elements all but possibly one of which are separable is simple

For $r=0$, one has $E=F=F(0)$, and for $r=1$ the displayed presentation is already simple. Assume $r\ge 2$. It is enough to combine two generators: if $F(\alpha ,\beta )=F(\gamma )$ whenever $\beta$ is separable, repeated combination leaves at most the originally exceptional generator as the first entry and a separable generator as the second. Finiteness of each intermediate extension follows from [L4].

If $F$ is finite, the two-generator extension is simple by [L3].

Suppose $F$ is infinite. In a common splitting field from [L2], list the distinct conjugates ${\alpha }_i$ of $\alpha$ and the pairwise distinct conjugates ${\beta }_j$ of the separable element $\beta$. Choose a nonzero $c\in F$ avoiding the finitely many values $({\alpha }_1-{\alpha }_i)\mathrm{/}({\beta }_j-{\beta }_1)$ with ${\beta }_j\mathrm{\ne }{\beta }_1$, and put $\gamma =\alpha +c\beta$.

In $F(\gamma )[x]$, the minimal polynomial of $\alpha$ and the translated minimal polynomial of $\beta$ have $\alpha$ as a common root. By the choice of $c$, any common root would give ${\alpha }_i+c{\beta }_j=\alpha +c\beta$ and hence must be $\alpha$; [L1] therefore makes their monic gcd $x-\alpha$. Thus $\alpha \in F(\gamma )$ and then $\beta =(\gamma -\alpha )\mathrm{/}c\in F(\gamma )$.

Hence $F(\alpha ,\beta )=F(\gamma )$ over either a finite or an infinite base. Iterating step 1.1 proves the theorem, and when every generator is separable it gives the usual finite separable primitive-element theorem.