Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
You should use Dedekind's theorem: see here. –  Zhen Lin May 6 '12 at 11:47
1  
In fact, you CAN say $|\mathrm{Aut}(F/E)| \leq \mathrm{deg}(f_n) \cdots \mathrm{deg}(f_1)$. Prove this by induction. If you know this is true for $n=k$, take any automorphism $\sigma$ of $F_k/E$. If $\tilde{\sigma}$ is an extension of $\sigma$ to $F_{k+1}$, then under $\tilde{\sigma}$ a root of $f_{k+1}$ must go to a root of $\sigma (f_{k+1})$ (this meaning $\sigma$ applied to the coefficients of $f_{k+1}$). There are thus at most $\mathrm{deg}(f_{k+1})$ such extensions, and so at most $\mathrm{deg}(f_{k+1}) \mathrm{deg}(f_k)\cdots \mathrm{deg}(f_1)$ automorpshims of $F_{k+1}/E$. –  Barry Smith May 6 '12 at 12:25
    
@BarrySmith Thanks. Could you clarify one thing for me? You're counting the extensions of automorphims of $F_k/E.$ How do we know it's the same as counting the automorphisms of $F_{k+1}/E?$ Can't there be an automorphism $\rho$ of $F_{k+1}/E$ such that $\rho|_{F_k}$ is not an automorphism of $F_k/E?$ I mean, can't this restriction have an image that's different from $F_k?$ Certainly, this image has to share the copy of $E$ with $F_k,$ but is this enough? –  user23211 May 6 '12 at 13:22
    
You're right --- I meant to say to take $\sigma$ to be any embedding of $F_k$ into $E$ (which is what must be induced by an automorphism of $E$). If $\tilde{\sigma}$ is an extension, then under $\tilde{\sigma}$, a root of $f_{k+1}$ must still go to a root of $\sigma (f_{k+1})$. This is essentially the idea Serge provided --- that counting embeddings is cleaner than counting automorphisms --- but I wanted to point out that you can apply this idea in the context of the work you gave above, and also to indicate that there is no need to involve an algebraically closed field. –  Barry Smith May 6 '12 at 14:02
    
@BarrySmith I think I understand what you are saying, but I'd like to make two points perfectly clear. Did you mean "embedding of $F_k$ into $\overline{E}?$ And when you said that $\sigma$ is what must be induced by an automorphism of $E$, what did you mean? Isn't $\sigma|_E=\operatorname{id}_E?$ –  user23211 May 6 '12 at 15:54
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
1  
Thanks for the answer. Did you mean "Obviously, order of $\operatorname{Aut}(F/E)$ is less or equal to $(F:E)$," with $(F:E)$ instead of $[F:E]$ at the end? –  user23211 May 6 '12 at 12:10
    
Yes, this is what I meant. Sorry for the typo. –  serge May 6 '12 at 13:17
    
No problem. :) I'll edit that. –  user23211 May 6 '12 at 13:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.