The Galois group of polynomial $p(x)\in\mathbb{K}[x]$ is cyclic and is generated by $q(x)\in\mathbb{K}[x]$.

Question

My question here is motivated by this question and that question. The subject of particular interest is when $n=3$ and $\mathbb{K}=\mathbb{Q}$.

Let $\mathbb{K}$ be a field with the algebraic closure $\bar{\mathbb{K}}$. For a given integer $n\geq 2$, suppose that $$p(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\ldots+a_{n-1}x+a_n$$ is an irreducible polynomial in $\mathbb{K}[x]$ such that the Galois group $G:=\text{Gal}(p/\mathbb{K})$ of $p(x)$ over $\mathbb{K}$ is isomorphic to the cyclic group $C_n$ of order $n$. We say that a polynomial $$q(x)=b_1x^{n-1}+b_2x^{n-2}+\ldots+b_{n-1}x+b_n$$ in $\mathbb{K}[x]$ augments $p(x)$ if $q$ generates $G$. That is, we can order the roots of $p(x)$ as $r_1,r_2,\ldots,r_n\in\bar{\mathbb{K}}$ in such a way that $$r_j=q(r_{j-1})$$ for $j=1,2,\ldots,n$ (where $r_0:=r_n$). Furthermore, we also say that $q(x)\in\mathbb{K}[x]$ of degree at most $n-1$ is an $n$-augmentation polynomial if $q(x)$ augments some monic irreducible polynomial $p(x)\in\mathbb{K}[x]$ of degree $n$ such that $\text{Gal}(p/\mathbb{K})\cong C_n$. What are $n$-augmentation polynomials $q(x)\in\mathbb{K}[x]$ for a given positive integer $n$?

A trivial example is when $n=2$. Let $a_1$ be an element of $\mathbb{K}$ such that $$p(x)=x^2+a_1x+a_2$$ is irreducible for some $a_1\in\mathbb{K}$. Then, $$q(x)=-x-a_1$$ is a $2$-augmentation polynomial. These are all possible $2$-augmentation polynomials.

Let now $n>2$. For a given $m\in\mathbb{Z}_{\geq 0}$, let $f^{[m]}(x)$ denote the $m$-time iteration of $f(x)\in\mathbb{K}[x]$. A necessary condition is that $\tilde{q}_n(x):=\prod\limits_{d\mid n}\big(q^{[d]}(x)-x\big)^{\mu\left(\frac{n}{d}\right)}$ has an irreducible factor of degree $n$ (here, $\mu$ is the Möbius function). I conjecture that this is also a sufficient condition, and any irreducible factor $p(x)$ of $\tilde{q}_n(x)$ such that $p(x)$ is monic and has degree $n$ is augmented by $q(x)$. Is my conjecture (restated below) true?

Conjecture. A polynomial $q(x)\in\Bbb{K}[x]$ of degree less than $n$ is an $n$-augmentation polynomial if and only if $$\tilde{q}_n(x):=\prod\limits_{d\mid n}\big(q^{[d]}(x)-x\big)^{\mu\left(\frac{n}{d}\right)}\in\mathbb{K}[x]$$ has an irreducible factor of degree $n$. Furthermore, $q(x)$ augments every monic irreducible factor $p(x)$ of degree $n$ of $\tilde{q}_n(x)$.

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Closing Remark.

If $G:=\text{Gal}(p/\mathbb{K})$ is cyclic, then it is generated by some $\gamma\in\text{Aut}_\mathbb{K}(\bar{\mathbb{K}})$. Fix a root $r_1\in\bar{\mathbb{K}}$ of $p(x)$. Because $G$ is cyclic (in particular, $|G|=n=\big[\mathbb{K}(r_1):\mathbb{K}\big]$), $p(x)$ splits into linear factors over $\mathbb{K}(r_1)$. Thus, $r_2:=\gamma(r_1)$ is in $\mathbb{K}(r_1)$. Therefore, there exists a polynomial $q(x)\in\mathbb{K}[x]$ of degree less than $n$ such that $$r_2=q(r_1)\,.$$ Define $r_j:=\gamma^{[j-1]}(r_1)$ for $j=1,2,\ldots,n$ (and $r_0:=r_n$ as before). Then, the equality above ensures that $r_j=q(r_{j-1})$ for $j=1,2,\ldots,n$.

Conversely, suppose that there exists $q(x)\in\mathbb{K}[x]$ such that the roots $r_1,r_2,\ldots,r_n\in\bar{\mathbb{K}}$ of $p(x)\in\mathbb{K}[x]$ satisfy $r_j=q(r_{j-1})$ for $j=1,2,\ldots,n$. Then, it follows that $\text{Gal}(p/\mathbb{K})$ is generated by $q$, whence it is a cyclic group. Therefore, the condition that $p(x)$ has a cyclic Galois group is both necessary and sufficient for this question.

However, that $p(x)$ splits into linear factors over $\mathbb{K}(r_1)$ for some root $r_1\in\bar{\mathbb{K}}$ is not sufficient for the setting of this question. For example, when $\mathbb{K}=\mathbb{Q}$ and $p(x)=x^4-10x^2+1$, then the roots of $p$ are $\pm\sqrt{2}\pm\sqrt{3}$, and $p(x)$ splits into linear factors over $\mathbb{Q}(r_i)$ for any root $r_i$ of $p(x)$. For any choice of indexes of the roots $r_i$ of $p(x)$, you can write $r_2=q(r_1)$ for some $q(x)\in\mathbb{K}[x]$, you cannot expect that $r_3=q(r_2)$, $r_4=q(r_3)$, and $r_1=q(r_4)$. This is because $\text{Gal}(p/\mathbb{K})$ is not cyclic. It is isomorphic to $C_2\times C_2$. (That is, if you have $r_2=q(r_1)$, then you will have $r_1=q(r_2)$.) Anyway, the condition that $p(x)$ with roots $r_1,r_2,\ldots,r_n\in\bar{\mathbb{K}}$ splits into linear factors over $\mathbb{K}(r_1)$ is necessary and sufficient for the existence of polynomials $q_j(x)\in\mathbb{K}[x]$ such that $r_j=q_j(r_1)$ for all $j=1,2,\ldots,n$, which is equivalent to the condition that $\big|\text{Gal}(p/\mathbb{K})\big|=n$.

Answer 1

If my understanding of your question and comments is correct, the polynomial $\tilde q_n(x)$ does not quite do "what you want" in that its roots are not points of exact period $n$ under $q$.

Your polynomial is the well-known dynatomic polynomial associated to $q$, and these polynomials can in general have roots which are not points of exact period $n$. A counterexample (from Silverman's book on arithmetic dynamics) is given by $$q(x) = x^2 - \frac 3 4$$ And so for $n=2$ we would have $$\tilde q_2(x) = \frac{q^{[2]}(x) - x}{q(x) - x} = \frac{(x-3/2)(x+1/2)^3}{(x - 3/2)(x + 1/2)} = \left(x + \frac12\right)^2$$

The root of this is $-\frac 12$ which we can see does not have exact period $2$.

The roots of this polynomial are said to have formal period n. It is known that for a point $\alpha$ which has formal period $n$ but actual period $m < n$ that $(x-\alpha)^2$ divides the dynatomic polynomial (Thm 2.4 of "The Galois Theory of Periodic Points of Polynomial Maps" by Morton & Patel). So I think possible place to start looking for a counterexample to your question (if there is one) is to find a dynatomic polynomial with a repeated irreducible factor of degree $n$.