[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.
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] $.