Looping through zeroes in polynomials

For any sequence of $n$ distinct (complex) numbers $\alpha_1 \ldots \alpha_n$, there is a unique interpolating polynomial $Q$ of degree $n-1$ with $P(\alpha_i) = \alpha_{i+1}$.

In the generic case, the coefficients of $Q$ are in $\Bbb Q[\alpha_i, \delta^{-1}]$, where $\delta = \prod_{i<j} (\alpha_i - \alpha_j)$ (to be more precise, $\delta Q \in \Bbb Z[\alpha_i]$)

In the case where the $\alpha_i$ are roots of a polynomial $P$ with no repeated roots, then for any $n$-cycle $\sigma$, $Q_\sigma$ is defined over the splitting field of $P$, but sometimes it is defined over a smaller field.

If $G$ is the Galois group of $P$, the action of $G$ on the $Q_\sigma$ corresponds to its action on the set of $n$-cycles by conjugation.

If $K = \Bbb Q(a_i)$ is the field of definition of $P$ and $L_\sigma$ is the field of definition of $Q_\sigma$ for an $n$-cycle $\sigma$, then $[ L_\sigma : K ]$ is exactly the size of the orbit of $\sigma$ under this conjugation action. Since the normalizer of an $n$-cycle in $S_n$ is the subgroup of $S_n$ generated by the $n$-cycle, we have that the subgroup of $G$ fixing the $n$-cycle is $G_\sigma = G \cap \langle \sigma \rangle$, and then $L_\sigma = \Bbb Q(\alpha_i)^{G_\sigma}$, and finally $[L_\sigma : K] = |G| / |G_\sigma|$

For example, if $n=2$, there is only one $2$-cycle, so there is only one $Q$, and $L= K$ which means that there are formulas for the coefficients of $Q$ in terms of the $a_i$.

If $n=3$, there are two $3$-cycles. If $G \not \subset A_3$ then $G$ will switch them so you will get an orbit of size $2$, so the coefficients of $Q$ will be in a degree $2$ extension of $K$. If $G \subset A_3$ then $G$ won't switch them, which tells you that in this case $L_\sigma = K$. In fact you have formulas for the two $Q$ in terms of the $a_i$ and $\delta = \pm \sqrt \Delta$ (and you switch between the two $Q$ by switching the sign of $\delta$)

If $n=4$, there are $3! = 6$ $4$-cycles, living in $6/\phi(4) = 3$ different cyclic subgroups of order $4$, so lots of stuff can happen. In the worst case, $G = S_4$ and they are all defined in $3$ different (non-normal) extensions of order $6$ of $K$. $S_4$ is still solvable so if you really want to you can get formulas for $Q$ with square roots and cube roots.

If $n \ge 5$ well then $G$ may not even be solvable anymore so you won't have those formulas anymore.

The only way you can have $L_\sigma = K$ is when $|G_\sigma| = |G|$, which means $G \subset \langle \sigma \rangle$, and for this to happen $G$ has to be cyclic too (not necessarily of order $n$ if $P$ is not irreducible, though).