# In a field extension, if each element's degree is bounded uniformly by $n$, is the extension finite?

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 $F$.

## 1 Answer

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.