Prove that $E$ is a finite extension of $F$ and $[E:F] \leq n$.

Question

Suppose that $E$ is a separable extension of $F$ and there exists a positive integer $n$ such that $\deg(a/F) \leq n$ for all $a$ in $E$. Prove that $E$ is a finite extension of $F$ and $[E:F] \leq n$.

I'm trying to solve this question from Abstract Algebra-A First Course, and I'm confused about where to even start. Does anyone has any idea?

Answer 1

You cannot use the primitive element theorem for $E/F$, because you do not know that it is finite a priori.

So you should assume that $[E:F] > n$ and thus you can take $n+1$ $F$-linear independent elements $e_0, \dotsc, e_n \in E$. The extension $F(e_0, \dotsc, e_n)/F$ is finite and of degree $\geq n+1$. Now you can use the primitive element theorem for this extension to get a contradiction.

Answer 2

Let $m:= \sup\{[F(a):F]: a \in E\}$, then $m\leq n < \infty$, and $\exists b\in E$ such that $m = [F(b):F]$. We claim that $E = F(b)$. Suppose not, then $\exists c\in E\setminus F(b)$. Consider $K=F(b,c)$, then $$ [K:F] > m $$ However, $[F(c):F]\leq n$, so $[K:F] \leq mn$ by the tower law. In particular, $F\subset K$ is a finite extension. By primitive element theorem, $\exists d\in K$ such that $K=F(d)$. But then $$ [F(d):F] > m $$ contradicting the definition of $m$. Hence, $E = F(b)$, which proves what you want.