Primitive element theorem w/o Galois theory (as in Lang's Algebra)

Solution 1:

As noted by @Hoot, the first paragraph of Lang's proof says that everything is easy when $k$ is finite, so assume it is not.

Solution 2:

In my opinion, Lang should have stated the Primitive Element Theorem separately for finite fields, because the implications don't seem to make much sense otherwise. The following is the statement of the theorem taken from page 243 of the 3rd edition.

Theorem 4.6. (Primitive Element Theorem) Let $E$ be a finite extension of a field $k$. There exists an element $\alpha \in E$ such that $E = k(\alpha)$ if and only if there exists only a finite number of fields $F$ such that $k \subset F \subset E$. If $E$ is separable over $k$, then there exists such an element $\alpha$.

If $k$ is a finite field and $E$ is a finite extension of $k$, then the following statements are true:

1. There is an element $\alpha \in E$ such that $E = k(\alpha)$.
2. There are only a finite number of fields $F$ such that $k \subset F \subset E$.
3. $E$ is separable over $k$.

All three are quite independently true, which makes the various implications superfluous (but true nonetheless). I think this is quite confusing; the statement of Theorem 4.6 could instead be made only for infinite fields $k$, and a separate theorem could have been stated in the next section on finite fields, having the content:

Theorem x.y. Let $E$ be a finite extension of a finite field $k$. Then there exists an element $\alpha \in E$ such that $E = k(\alpha)$.