Let $\phi : F \rightarrow F_1$ be an isomophism of fields and $f(x) > \in F[x]$.
Let $\Phi : F[x] \rightarrow F_1[x]$ be the unique ring isomorphism which extends $\phi$ and maps $x$ to $x$.
Let $f_1(x) = \Phi(f(x))$ and $E/F$, $E_1/F_1$ be the splitting fields of $f(x)$ and $f_1(x)$ respectively.
Then there exists an isomorphism $\gamma : E \rightarrow E_1$ which extends $\phi$.
I wish to prove that the number of such $\gamma \leq [E:F]$.
My intuition is to use induction on $[E:F]$. Furthermore the proof for the above theorem uses the same induction strategy.
In the base case $[E:F] = 1$ and $\gamma$ must map the single root of $f(x)$ to the single root of $f_1(x)$, so there can only be a single isomorphism $\gamma$.
Considering the case $[E:F] = 2$, I notice that strict equality holds if (and only if?) $f$ is seperable.
I'm not sure how to proceed with the inductive step. Is this approach using induction on $[E:F]$ even correct? If so, what's the right direction to apply the inductive step?