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Apr 9, 2022 at 20:18 comment added KConrad For $n \geq 7$ there is different partition of $n$ that can't occur for separable $f$. By Bertrand's postulate, there is a prime $p$ s.t. $(n+1)/2 < p \leq n-2$. Use the partition $p + 2 + 1 + \cdots + 1 = n$; it has $k := n-p$ terms. The Galois group $G$ of $f$ over $F$ has a permutation $\sigma$ of the $n+1$ roots of type $(p,2,1,\ldots,1)$. Then $\sigma^2$ is a $p$-cycle on the roots and $\sigma^p$ is a $2$-cycle on the roots. A transitive subgroup of $S_{n+1}$ having a $2$-cycle and $p$-cycle is $S_{n+1}$, so Bjorn's argument shows $f/(x-\alpha)$ is irred. of degree $n$: contradiction.
Apr 9, 2022 at 18:11 vote accept Minseon Shin
Apr 9, 2022 at 4:46 comment added KConrad You could replace $S_5$ by $S_p$ for each prime $p\geq 5$ and the partition having one $2$ and all other terms equal to $1$.
Apr 9, 2022 at 4:33 history answered Bjorn Poonen CC BY-SA 4.0