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

I want to understand the Primitive Element Theorem in a form: every finite seaparable extension is simple. I don't want to use ideas from Galois theory (at least a biection between subgroups of Galois group and intermediate subfields). I consider Serge Lang's proof (Th. 4.6, p.243 in 2002 ed.).

Let $s_i$ are $n$ different embeddings of outer field $E=F(a,b)$ in algebraic closure $F^{alg}$. Let

$$P(X) = \prod_{i\neq j} (s_i(a) - s_j(a) + Xs_i(b) - Xs_j(b)).$$

This polynomial is obviously not equal to zero polynomial. But this leads Lang to conclude that there is a $c \in F$ such that $P(c) \neq 0$. I can't see how this can be true in any field $F$. That is obvious if $|F| = \infty$. But it is not clear in the case of finite fields (there are non-zero polynomials over finite fields which evaluate to zero everywhere, e.g. $x^{p^n} - x$ over $F_{p^n}$).

I'm aware about separate proof of the Primitive Element Theorem for finite fields. But the question is: can we make above mentioned conclusion without considering cardinality of $F$? If we can't, does it mean that Lang's proof has this shortcoming.

• The first paragraph of Lang's proof in my copy says that everything is easy when $k$ is finite so assume it is not. – Hoot Jul 15 '16 at 16:58
• Also my copy says that. – DonAntonio Jul 15 '16 at 17:04
• My bad: I though this remark was in action only for proof of first part concerning intermediate fields. – Artem Pelenitsyn Jul 15 '16 at 17:26
• If the field is finite, any generator of the cyclic group of nonzero elements is a primitive elements. – Pedro Tamaroff Jul 15 '16 at 17:33
• You should assume it, then. – Pedro Tamaroff Jul 15 '16 at 19:57

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.

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 $E = k(\alpha)$.