Factorization of an irreducible polynomial in the field extension it defines

Question

In field theory, the following fact is used in the construction of splitting fields: Given a field $F$ and an irreducible polynomial $f \in F[x]$, the quotient $F[\alpha]/(f(\alpha))$ is a field extension of $F$ which contains a root of $f$ (namely the congruence class of $\alpha$).

Let $n$ be a positive integer and let $\lambda_{1} + \dotsb + \lambda_{k} = n$ be a partition of $n$. Does there exist a field $F$ and an irreducible polynomial $f \in F[x]$ of degree $\deg f = n+1$ such that, if we define $K := F[\alpha]/(f(\alpha))$, the factorization of $f$ in $K[x]$ is of the form $f = (x-\alpha) \cdot f_{1} \dotsb f_{k}$ where each $f_{i} \in K[x]$ is irreducible and $\deg f_{i} = \lambda_{i}$?

Answer 1

Let us show that for the partition $2+1+1=4$, there is no such $f$.

If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group $G$ would be a transitive subgroup of $S_5$ containing a transposition, so $G=S_5$, but then the stabilizer of a point would act transitively on the other four points, so $f$ would factor over $K$ into $x-\alpha$ and an irreducible polynomial of degree $4$.