Some question on the separable extension.

Question

Let $E/F$ be a separable extension.

If $|F(\alpha):F|\le n$ for every $\alpha \in E$, show that $E/F$ is a finite extension and $|E:F|\le n$.

I know it can be easily proved using primitive element theorem. But I want to prove this without using that theorem.

Answer 1

Wouldn't proving this prove the 'primitive element theorem'?

You say $[F(\alpha):F] \leq n $ $\forall \alpha \in E$

But if $m = max\{[F(\alpha):F] | \alpha \in E\}$ then $m \leq n$ and your theorem applies to $m$ as well.

Now, $m = [F(\alpha_j):F]$ and we have $[E:F] = [E:F(\alpha_j)][F(\alpha_j):F] = [E:F(\alpha_j)]m \leq m$ implying $E = F(\alpha_j)$

Then the two theorems are equivalent.

I know this isn't a direct answer to your question but if the two are equivalent then the proof for one is good for the other, and most likely there isn't an easier proof then the one for the primitive element theorem (and this was too long for a comment)

** $m$ is well defined since the extension is algebraic by assumption.

Hope I didn't make a mistake in my reasoning.