# Multiplicative formula for order of automorphism group

I am reading a proof of the following proposition from Dummit and Foote:

Let $E$ be the splitting field over $F$ of the polynomial $f(x) \in F[x]$. Then $$|\textrm{Aut}(E/F)|\leq [E:F]$$ with equality if $f(x)$ is separable.

Now the proof of this proposition is by induction, for $[E:F] = 1$ this is clear, whereas if $[E:F] > 1$ then $f(x)$ has some irreducible non-linear factor $p(x)$. Then one considers the number of ways in which the identity map from $F$ to itself can be extended to an isomorphism between $F[x]/(p(x)) \cong F(a)$ and $F[x]/(p(x)) \cong F(a')$ where $a$ and $a'$ are two different roots of $p(x)$ in $E$.

Now for the rest of the proof they seem to use the fact that

$$\left|\textrm{Aut}(E/F(a))\right||\textrm{Aut}(F(a)/F)| = |\textrm{Aut}(E/F)|.$$

Now we know $|\textrm{Aut}(E/F(a))| \leq [E:F(a)]$ by the induction hypothesis and that $|\textrm{Aut}(F(a)/F)| \leq [F(a):F]$. The reason why I think they use the above fact is because they mention the formula $[E:F] = [E:F(a)][F(a):F]$ and finish the proof of the proposition from there. Where can I find a proof of the fact I stated above, if it is true?

Thanks.

Edit: I just got an idea on how to prove the fact above. Consider the the map

\begin{align*} L : \textrm{Aut}(E/F)& \longrightarrow \textrm{Aut}(F(a)/F) \\ \phi& \longmapsto \phi|_{F(a)} \end{align*}.

It is easily seen that $L$ is a group homomorphism, and that the kernel of $\phi$ is those elements in $\textrm{Aut}(E/F)$ that are already the identity on $F(a)$, which means that

$$\ker L = \textrm{Aut}(E/F(a)).$$

As for the image of $L$, I think that $L$ is actually surjective. This is because the automorphisms from $E$ to itself were simply obtained by extending those from $F(a)$ to $F'(a)$, so going in the reverse direction any $\sigma \in \textrm{Aut}(F(a)/F)$ is the result of restricting an automorphism on $E$. Since

$$\begin{eqnarray*} |\textrm{Aut}(E/F)| &=& |\textrm{im} L||\ker L| \\ &=& |\textrm{Aut}(F(a)/F)|\left|\textrm{Aut}(E/F(a))\right| \end{eqnarray*}$$

this finishes my claim. Does this seem right?

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Your $L$ is not well-defined: the restriction of an automorphism of $E$ to $F(\alpha)$ need not map $F(\alpha)$ to itself. It is only well-defined if $F(\alpha)$ is normal, and you don't know that. –  Arturo Magidin Jun 5 '12 at 19:02
Hey Ben, Could you give a reference to where this happens in D&F? Cheers, –  Dylan Moreland Jun 5 '12 at 19:04
@Dylan: Page 541, Section 14.1, after Proposition 4, leading to Proposition 5, at least in my copy (2nd edition). –  Arturo Magidin Jun 5 '12 at 19:10
@Arturo Thanks! To Ben: We'd better explicitly use that $E$ is a splitting field if we want to prove your claim, since something like the example in this old answer of mine shows that it can't be true in general. –  Dylan Moreland Jun 5 '12 at 19:13
@ArturoMagidin Ah that is correct at the moment I don't know if $L(F(\alpha))$ will embed back to $F(\alpha)$.... –  user38268 Jun 5 '12 at 23:06

Given an automorphism $\phi$ of $E/F$, consider its restriction to $F(\alpha)$. This restriction must maps $F(\alpha)$ to some $F(\alpha')$, with $\alpha'$ another root of $f$ (it could be that $\alpha=\alpha'$, or it could be that $\alpha\neq\alpha'$). Each of the possible restrictions is the restriction of some automorphism $E/F$ (because $E$ is a splitting field), since we can extend any particular isomorphism $F(\alpha)\to F(\alpha')$ to an automorphism of $E$. So in order to count the number of possible automorphism of $E$ over $F$, it suffices to count how many homomorphisms restrict to a particular restriction, and how many possible restrictions there are.
The number of possible restrictions is, by the induction hypothesis, $[E:F(\alpha)]$. And the number of possible restrictions is the number of possible distinct roots of $f(x)$; this is bounded by $\deg(f)=[F(\alpha):F]$.
So we have that there are at most $[F(\alpha):F]$ possible restrictions (equality if all the roots of $f$ are distinct), and each of the restrictions extends in at most $[E:F(\alpha)]$ ways (equality if all the roots of $f$ are distinct). So in conclusion, the number of possible automorphisms is at most $[F(\alpha):F][E:F(\alpha)] = [E:F]$, with equality if and only if we have equality in each factor, if and only if all the roots of $f$ are distinct.