This is Exercise 26 from Chapter V Algebraic Extensions in Lang's Algebra.
$F$ is finite normal extension over $k$ and $f(x)$ is irreducible in $k[x]$. If $f(x)=g(x)h(x)k(x) \in F[x]$ where $g(x),h(x)$ are monic irreducible factors in $F[x]$, I need to show that there exists an automorphism of $F$ which takes $g(x)$ to $h(x)$.
Any hints would be appreciated.
Sending one root to another is okay but how to do that for a collection of roots?
Let $L/k$ be the splitting field of $f$, and $E$ the compositum of $L$ and $F$. Let $\alpha \in L$ be a root of $g$, and $\beta \in L$ be a root of $g$. Since $g, h$ are monic and irreducible in $F[x]$, they are the minimal polynomials over $F$ of $\alpha$ resp. $\beta$.
Since $f \in k[x]$ is irreducible, there is an automorphism of $L/k$ that takes $\alpha$ to $\beta$. This can be extended to an automorphism $\phi$ of $E/k$, as $F/k$ is finite and normal.
Since $F/k$ is normal, this automorphism restricts to an automorphism of $F/k$, which we still call, abusing notation, $\phi$.
Applying $\phi$ to $g(\alpha) = 0$ we obtain $0 = \phi(g) (\phi(\alpha)) = \phi(g) (\beta)$. Now $\phi(g) \in F[x]$ is monic and irreducible, and has $\beta$ as a root. Hence it is the minimal polynomial of $\beta$ over $F$, so $\phi(g) = h$.