8
$\begingroup$

Given a field extension $F\subset E$, if there exists a $n$ such that for each $\alpha\in E$ we have $|F[\alpha]: F| \leq n$, can we conclude that $|E:F|<\infty$?

This is the question here , but I would like to drop the char $0$ assumption on $F$.

$\endgroup$
5
$\begingroup$

Let $k=\Bbb F_p$ and consider a purely transcendental extension of countably infnite transcendence degree $F=k(t_1, t_2, \dots)$. Let $E=k(\sqrt[p]{t_1}, \sqrt[p]{t_2}, \dots)$

Given any element $\alpha \in E$, we can write $\displaystyle \alpha= \frac{f(\sqrt[p]{t_1}, \dots, \sqrt[p]{t_k})}{g(\sqrt[p]{t_1}, \dots, \sqrt[p]{t_k})}$. where $f,g \in k[x_1, \dots, x_k]$ and $k \in \Bbb N$ is large enough.

Applying the Frobenius, using that we're in characteristic $p$ and the Frobenius fixes $k$, we get $\displaystyle \alpha^p= \frac{f(t_1, \dots, t_k)}{g(t_1, \dots, t_k)} \in F$, so that $\alpha$ is a root of the polynomial $X^p-\alpha^p \in F[X]$, thus $[F[\alpha]:F] \leq p$.

Evidently, $E/F$ is not finite.

The statement is however true for separable extensions in any characteristic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.