If $(F:E)<\infty$, is it always true that $\operatorname{Aut}(F/E)\leq(F:E)?$

If I understand correctly, Arturo Magidin says in this comment that the following is true.

If $E\subset F$ is a finite field extension, then $|\operatorname{Aut}(F/E)|\leq (F:E).$

I understand why this is true when the extension is simple, but the general case eludes me. (I also understand that my particular extension in that question is simple, so this is irrelevant to the problem, but I would like to understand this.)

Let $E\subset F$ be a finite extension. Any finite extension is finitely generated, so let $$F=E(a_1,\ldots,a_n).$$ Let $$E=F_0\subset F_1\subset\ldots\subset F_n=F,$$ with $F_i=F_{i-1}(a_i)$ for $i=1,\ldots,n.$ Let $f_i$ be the minimal polynomial of $a_i$ over $F_{i-1}$, and let $g_i$ be the minimal polynomial of $a_i$ over $E$. Then $$(F:E)=(F_n:F_{n-1})\cdots (F_1:F_0)=\deg(f_n)\cdots\deg(f_1).$$

I think I can't use $f_i$ to bound the order of $\operatorname{Aut}(F/E).$ Surely, $$|\operatorname{Aut}(F/E)|\leq \deg (g_n)\cdots\deg (g_1),$$ because the generators must be mapped to roots of their minimal polynomials over $E$ (taking non-minimal polnomials here would only make things worse), but this doesn't work because the inequality we have is $$\deg (f_i)\leq\deg (g_i),$$ not the other way around.

Solution 1:

Instead of counting automorphisms of $F$ over $E$, count embeddings $\sigma: F\hookrightarrow \bar E$ that act identically on $E$. If we denote the number of such sigmas by $(E:F)$, then one has $(K:F)=(K:F)(F:E)$ if $K\supset F\supset E$ is a tower of finite extensions. One checks immediately that $(K:F)\le [K:F]$ if $K=E(\alpha)$ is a simple extension, which implies the inequality $(F:E)\le[F:E]$ for arbitrary finite extensions.

Obviously, order of $\mathrm{Aut}(F/E)$ is less or equal to $(F:E)$.

Cf Lang's "Algebra" (3 ed.), Ch. 5, Secs. 2-4.