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

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

• 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. Apr 30, 2020 at 15:09
• @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 $p_j(x)$. Apr 30, 2020 at 15:12
• @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. Apr 30, 2020 at 15:20
• 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 Apr 30, 2020 at 17:27
• @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.
– user208649
Apr 30, 2020 at 17:42

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 $$\tilde q_2(x) = \frac{q^{}(x) - x}{q(x) - x} = \frac{(x-3/2)(x+1/2)^3}{(x - 3/2)(x + 1/2)} = \left(x + \frac12\right)^2$$

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 $$n$$.

• 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.) Apr 30, 2020 at 18:12
• 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.
• 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). Apr 30, 2020 at 19:43