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$\DeclareMathOperator{\aut}{Aut}$ $\newcommand{\vp}{\varphi}$Let $p(x)$ be an irreducible and separable polynomial over a field $F$ and let $E$ be the splitting field for $p(x)$ over $F$.

Question. Is is true that $[E:F]=|\aut(E:F)|$?

I think the answer to this question is YES. Here is my argument for thinking so.

Say $\deg p=n$ and let $a_1, \ldots, a_n$ be all the roots of $p$ in $E$. Write $F_i$ to denote $F(a_1, \ldots, a_i)$. We claim that each $F$-embedding of $F_i$ in $E$ admits exactly $[F_{i+1}:F_i]$ extensions to an embedding of $F_{i+1}$ into $E$.

Suppose $\vp:F_i\to E$ be an $F$-embedding of $F_i$ in $E$ and let $d_i=[F_{i+1}:F_i]$. We need to show that there are exactly $d_i$ different extensions of $\vp$ to an embedding of $F_{i+1}$ into $E$. Let $p_i(x)\in F_i[x]$ be the minimal polynomial of $a_i$ over $F_i$. Since $p_i$ and $p$ both have $a_{i+1}$ has a root, we have $p_i|p$ and therefore $p_i$ has $d_i$ distinct roots in $E$. Now we may extend $\vp$ to $F_{i+1}$ by sending $a_{i+1}$ to any root of $p_i$ we wish. Further, all extensions of $\vp$ send $a_{i+1}$ to some root of $p_i$. Lastly, any extension of $\vp$ is determined completely by its action on $a_{i+1}$. Therefore there are precisely $d_i=[F_{i+1}:F_i]$ different extensions of $\vp$ to $F_{i+1}$. This proves our claim.

From the claim, we see that there are at least $[F_n:F_{n-1}]\cdots [F_1:F]=[E:F]$ distinct elements in $\aut(E:F)$. There cannot be any more and thus we are done.

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  • $\begingroup$ I don't know about the notation you were taught but if $Aut(E:F)$ is a set as your last sentence suggests, then your question should be whether $[E:F] = \#(Aut(E:F))$. $\endgroup$ – user21820 Feb 19 '16 at 6:26
  • $\begingroup$ Right. I mean the number of automorphisms. $\endgroup$ – caffeinemachine Feb 19 '16 at 8:49
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I don't see anything wrong with your proof except for the last paragraph, for which I'm not sure what you mean. Prior to that you have already established that, at each step of adjoining a root of $p$, the number of extensions of an embedding is exactly the degree of the field extension. From this alone we can see that the total number of embeddings fixing $F$ is just the product of these degrees, because any embedding is in a 1-to-1 correspondence with the sequence of its restrictions to the chain of field extensions.

I may be mistaken but I do not see any other more obvious reason why $[E:F]$ is an upper bound on $\#(Aut(E:F))$.

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