The number of isomorphisms $\psi$ extending $\phi$ is less than or equal to $[E:F]$ [Theorem] Let $\phi: F \rightarrow F_1$ be a field isomorphism and $f(x) 
\in F[x]$. Let $\Phi: F[x] \rightarrow F_1[x]$ be the unique ring isomorphism which extends $\phi$ and $\Phi(x)=x$. Let $f_1(x)=\Phi(f(x))$ and $E/F$ and $E_1/F_1$ be splitting fields of $f(x)$. Then there exists an isomorphism $\psi:E \rightarrow E_1$ which extend $\phi$.
From the proof of this theorem using induction on $[E:F]$, I could show that the number of such $\phi$ is equal to $[F(\alpha):F]=deg(p)$, where $p(x)$ is the irreducible factor of $f(x)$ and $\alpha$ is the root of $p(x)$ used in proof of the theorem, and $deg(p)$  less than or equal to $[E:F]$ since $[E:F]=[E:F(\alpha)[F(\alpha):F]$ and $[F(\alpha):F]=deg(p(x))$. But I can't show why the equality holds if and only if $f(x)$ is separable. When the equality holds, $[E:F(\alpha)]=1$ which implies $E=F(\alpha)$. How do I proceed from this?
Can anyone give me some ideas? Thanks.
 A: If you know the primitive element theorem, then it should be easy to see that separability implies there are as many extensions as the degree: you just have $ [E : F] $ choices of where to map your primitive element, and all of these choices correspond to a unique isomorphism $ E \to E' $. For the other direction, note that inseparability of $ E/F $ implies that there's some $ \beta \in F $ whose minimal polynomial is inseparable, and in this case an isomorphism $ F \to F' $ extends to an embedding $ F(\beta) \to E' $ in (strictly) less than $ [F(\beta) : F] $ ways - otherwise, it would mean $ \beta $ has $ [F(\beta) : F] $ distinct conjugates in a splitting field, which is equivalent to separability. Then, each of these embeddings extend to an isomorphism $ E \to E' $ in $ \leq [E : F(\beta)] $ ways by the result you proved. ($ E $ and $ E' $ are splitting fields of $ f $ over $ F(\beta) $ and the image of $ F(\beta) $ under the embedding into $ E' $ as well, so you can use the result you proved.) The total number of embeddings is then $ < [E: F(\beta)] [F(\beta) : F] = [E : F] $.
