Exercise 5.4.7

Proof. Let a finite extension $F\subset L=F({\alpha }_1,\cdots ,{\alpha }_n)$, where ${\alpha }_1,\cdots ,{\alpha }_{n-1}$ are separable over $F$ (but not ${\alpha }_n$). The Primitive Element Theorem (5.4.1) shows that $F({\alpha }_1,\cdots ,{\alpha }_{n-1})$ has a primitive element $\beta$ separable over $F$.

The extension $F\subset L=F(\beta ,{\alpha }_n)$ is such that $\beta$ is algebraic separable over $F$, and ${\alpha }_n$ algebraic over $F$.

If $F$ is infinite, by Exercise 6 this is sufficient to prove the existence of a primitive element of $F\subset L$ (but perhaps not separable).

If $F$ is a finite field, then $L$ also, and it has a primitive element by Exercise 2(b). □