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 $\tilde{q}_n(x)$.

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 $\big|\text{Gal}(p/\mathbb{K})\big|=n$.

  • $\begingroup$ I don't understand why you state your divisibility condition with a fancy product involving the Moebius function. Couldn't the divisibility part of the condition be simply that $p$ (monic irred. of degree $n$) divides $(q^{[n]}(x)-x)$ but not $(q^{[d]}(x)-x)$ for all $d \lt n$ ? This would rule out the Z2xZ2 Galois group counterexample you gave. $\endgroup$ Apr 30, 2020 at 15:09
  • $\begingroup$ @EwanDelanoy It would. However, how do I know that there aren't distinct monic irreducible polynomials $p_j(x)$ of degree $n$ that divide $\tilde{q}_n(x)$, where $j=1,2,\ldots,n$, such that if $r_1,r_2,\ldots,r_n$ are the roots of $p_1(x)$, then the roots of $p_j(x)$ are $q^{[j-1]}(r_i)$? If this happens, I can't say anything about the relations between the roots of an individual polynomial $p_j(x)$. $\endgroup$ Apr 30, 2020 at 15:12
  • $\begingroup$ @EwanDelanoy Oh, I misunderstood your question, I think. The irreducible factors of this polynomial $\tilde{q}_n(x)$ are precisely the polynomials whose roots cannot be permuted by $q$ with a smaller orbit than $n$. That is basically the same as your characterization. $\endgroup$ Apr 30, 2020 at 15:20
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    $\begingroup$ Batominovski, it would seem your credit to lhf is this answer and following comments, math.stackexchange.com/questions/3646675/… Note how Jack pointed out, for cubics, this is just square discriminant. Anyway, recommend these chapters by Cox in his Galois Theory book: zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf $\endgroup$
    – Will Jagy
    Apr 30, 2020 at 17:27
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    $\begingroup$ @Batominovski I might misunderstand your comment and question, but I think you're making a mistake about the roots of $\tilde q_n$. That polynomial is the relatively well-known "dynatomic polynomial", whose roots only have formal period $n$. You write in your comment that "The irreducible factors of this polynomial $\tilde q_n(x)$ are precisely the polynomials whose roots cannot be permuted by q with a smaller orbit than n" (i.e. they have exact period $n$). In fact the roots of this polynomial can have period strictly smaller than $n$ although this is rare. $\endgroup$
    – user208649
    Apr 30, 2020 at 17:42

1 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 $n$.

  • $\begingroup$ I may have misunderstood your question or comments, please let me know! In any case I think looking into dynatomic polynomials may help you with your question. $\endgroup$
    – user208649
    Apr 30, 2020 at 18:10
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    $\begingroup$ I don't think this contradicts my conjecture, because I only look at irreducible factors of $\tilde{q}_n(x)$ of degree $n$. However, please do not delete this answer even if it doesn't answer the main question. I didn't expect that $\tilde{q}_n(x)$ can have an irreducible factor of degree less than $n$. Thank you very much. Your example is great! Also, I will take a look at your reference, thank you again. (You didn't misunderstand my question or comments.) $\endgroup$ Apr 30, 2020 at 18:12
  • $\begingroup$ It's entirely possible (maybe even likely) that this polynomial doesn't admit a multiple factor of degree $n$. I tried for a while to come up with one before posting my comment and got essentially nowhere. $\endgroup$
    – user208649
    Apr 30, 2020 at 19:04
  • $\begingroup$ Sorry for this. I just wanted to correct what I had said in my previous comment. I meant to say "I didn't expect $\tilde{q}_n(x)$ to have a root whose orbit under $q$ is of size less than $n$." Clearly, $\tilde{q}_n(x)$ can have an irreducible factor of degree less than $n$ (e.g., when $\mathbb{K}$ is algebraically closed). $\endgroup$ Apr 30, 2020 at 19:43
  • $\begingroup$ Even though you didn't answer the main question, your answer enlightens me regarding the behavior of the roots of dynatomic polynomials. Plus, you provided me with a good reference. The bounty award goes to you. Thank you very much for your contribution. $\endgroup$ May 7, 2020 at 13:21

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