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?