Find the polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$.

Question

Let $z$ be a 13th root of unity $(z\neq 1)$. Find the polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$.

My idea: since $z$ such $$0=z^{12}+z^{11}+\cdots+z+1=\prod_{k=1}^{12}(z-z_{k})$$ maybe use $$\sum_{k=1}^{n}z_{k}=-1,\sum_{i<j}z_{i}z_{j}=1,\sum_{i<j<k}z_{i}z_{j}z_{k}=-1,\cdots$$ then we must find $$\sum_{k=1}^{12}(z_{k}+z^3_{k}+z^9_{k})=?$$ $$\sum_{k<i}(z_{k}+z^3_{k}+z^9_{k})(z_{i}+z^3_{i}+z^9_{i})$$ $$\cdots$$ I think it's too complicated, and it seems like an interesting question, how to think correctly

Answer 1

The hint.

Let $z+z^3+z^9=a$, $z^2+z^5+z^6=b$, $z^4+z^{10}+z^{12}=c$ and $z^7+z^8+z^{11}=d$.

Now, show that: $$a+b+c+d=-1,$$ $$ac=3+b+d,$$ $$bd=3+a+c,$$ $$bc=a+b+c$$ and get the polynomial:

The second equation gives $$ac=2-a-c$$ or $$c=\frac{2-a}{a+1}.$$ From the fourth equation we obtain: $$b(c-1)=a+c,$$ which gives $$b=\frac{a^2+2}{1-2a}.$$ Also, $$d=-1-a-b-c=\frac{a^3+3a-5}{(a+1)(1-2a)}.$$ Thus, from the third equation we obtain: $$\frac{a^2+2}{2a-1}\cdot\frac{a^3+3a-5}{(a+1)(2a-1)}=3+a+\frac{2-a}{a+1}$$ or $$(a-5)(a^4+a^3+2a^2-4a+3)=0$$ or $$a^4+a^3+2a^2-4a+3=0.$$ For example, $$ac=3+b+d$$ because $$ac=(z+z^3+z^9)(z^4+z^{10}+z^{12})=$$ $$=z^5+z^{11}+1+z^7+1+z^2+1+z^6+z^8=$$ $$=3+z^2+z^5+z^6+z^7+z^8+z^{11}=3+b+d.$$

Answer 2

Observe that $g=2$ is a generator of the unit group $G=(\mathbb{Z}/13\mathbb{Z})^\times\cong\mathbb{Z}/12\mathbb{Z}$, which is the Galois group of $\phi(x)=x^{12}+x^{11}+\ldots+x+1$ over $\mathbb{Q}$. Let $z=e^{2\pi i/13}$ be a primitive $13$th root of unity, which is a root of $\phi(x)$. The roots of $\phi(x)$ are $z^j$ for $j=1,2,\ldots,12$. The action of $g^s\in G$ on $z^j$ sends $z^j\mapsto z^{g^sj}$.

Define $$t_0=z^{g^0}+z^{g^4}+z^{g^8}=z+z^3+z^9.$$ The orbit of $t_0$ under the Galois group $G$ consists of $t_0$, $t_1$, $t_2$, and $t_3$, where $$t_k=\sum_{r=0}^2z^{g^{4r+k}}$$ for $k=1,2,3$. Hence $$t_1=z^2+z^5+z^6,$$ $$t_2=z^4+z^{10}+z^{12},$$ and $$t_3=z^7+z^8+z^{11}.$$ Therefore, the minimal polynomial of $t_0$ over $\mathbb{Q}$ is $$f(x)=(x-t_0)(x-t_1)(x-t_2)(x-t_3).$$ Note that the Galois group of $f(x)$ is the factor group $H$ of $G$ obtained by the quotient of $\langle g^4\rangle$. The group $H$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$.

Let $\mathbb{K}=\mathbb{Q}(t_0)$. Because $[\mathbb{K}:\mathbb{Q}]=4=|\mathbb{Z}/4\mathbb{Z}|=|H|$, $\mathbb{K}$ is Galois over $\mathbb{Q}$. Therefore, $t_0,t_1,t_2,t_3$ are all in $\mathbb{K}$. The group $H$ has a subgroup $N$ of order $2$ generated by $\langle g^2\rangle/\langle g^4\rangle$. If $\mathbb{L}$ is the fixed field of this subgroup $N$, then obviously $\mathbb{L}=\mathbb{Q}(t_0+t_2)=\mathbb{Q}(t_1+t_3)$. Therefore, we have a cascade of field extensions $$\mathbb{K}\supsetneq \mathbb{L}\supsetneq \mathbb{Q}$$ with each successive index being $2$.

Since the orbit of $t_0+t_2\in\mathbb{L}$ under $H$ consists of $t_0+t_2$ and $t_1+t_3$, the minimal polynomial of $t_0+t_2$ is $$h(x)=(x-t_0-t_2)(x-t_1-t_3).$$ Because $t_0+t_1+t_2+t_3=\sum_{j=1}^{12}z^j=-1$, and it is easy to see that $$(t_0+t_2)(t_1+t_3)=3\sum_{j=1}^{12}z^j=-3,$$ we have $$h(x)=x^2+x-3.$$ We can then conclude that $$t_0+t_2=\frac{-1+\sqrt{13}}{2}$$ and $$t_1+t_3=\frac{-1-\sqrt{13}}{2}.$$ (Hence, $\mathbb{L}=\mathbb{Q}(\sqrt{13})$.)

Intermission: You don't actually need to know which of $\frac{-1\pm\sqrt{13}}{2}$ is $t_0+t_2$. You can just assume $t_0+t_2=\frac{-1+a\sqrt{13}}{2}$ with $a=\pm1$. The next step remains almost identical, except you have a factor $a$ in some places. But at the end (when finding $f(x)$), you will get an expression in $a^2=1$.

The rest is just as Michael Rozenberg's solution. Observe that $$t_0t_2=3+t_1+t_3=\frac{5-\sqrt{13}}{2}$$ and $$t_1t_3=3+t_0+t_2=\frac{5+\sqrt{13}}{2}.$$ Therefore, $$(x-t_0)(x-t_2)=x^2+\frac{1-\sqrt{13}}{2}x+\frac{5-\sqrt{13}}{2}$$ and $$(x-t_1)(x-t_3)=x^2+\frac{1+\sqrt{13}}{2}x+\frac{5+\sqrt{13}}{2}.$$ Thus $$f(x)=(x-t_0)(x-t_2)\cdot (x-t_1)(x-t_3)=x^4+x^3+2x^2-4x+3.$$ The roots are $$t_0=\frac{-1+\sqrt{13}+i\sqrt{26-\sqrt{13}}}{4},$$ $$t_1=\frac{-1-\sqrt{13}+i\sqrt{26+\sqrt{13}}}{4},$$ $$t_2=\frac{-1+\sqrt{13}-i\sqrt{26-\sqrt{13}}}{4},$$ and $$t_3=\frac{-1-\sqrt{13}-i\sqrt{26+\sqrt{13}}}{4}.$$

Remark: We can then show that $$(x-z)(x-z^3)(x-z^9)=x^3-t_0x^2+t_2x-1.$$ This shows that $\frac{2\pi}{13}$ is a constructible angle via neusis construction. This means: the regular tridecagon is neusis-constructible (see also here). You can also show that $$x^3-\frac{-1+\sqrt{13}}{4}x^2-\frac14x+\frac{-3+\sqrt{13}}{16}$$ is the minimal polynomial of $\cos\frac{2\pi}{13}$, $\cos\frac{6\pi}{13}$, and $\cos\frac{18\pi}{13}$ over $\mathbb{L}=\mathbb{Q}(\sqrt{13})$.

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Incomplete attempt at generalisation

Here is the main idea of how this problem arises. This approach is also used here.

Let $p\geq 3$ be a prime number and $g\in\mathbb{Z}/p\mathbb{Z}$ a generator of the unit group $(\mathbb{Z}/p\mathbb{Z})^\times \cong \mathbb{Z}/(p-1)\mathbb{Z}$. If $w$ is a primitive $p$th root of unity, we want to construct the minimal polynomial $f(x)\in\mathbb{Z}[x]$ of $$\sum_{r=0}^{d-1}w^{g^{2^sr}}$$ where $p-1=2^sd$ with $d$ odd. Define $$t_k=\sum_{r=0}^{d-1}w^{g^{2^sr+k}}$$ for $k=0,1,2,\ldots,2^{s}-1$. Then let $$T_m^j=\sum_{k\equiv j\pmod{2^{s-m}}}t_k=\sum_{r\equiv j\pmod{2^{s-m}}}w^{g^r}$$ when $m=0,1,2,\ldots,s$, and $j=0,1,2,\ldots,2^{s-m}-1$. For example $T_0^j=t_j$ and $T_s^0=w+w^2+\ldots+w^{p-1}=-1$.

The minimal polynomial of $T_s^0$ is of course $f_s(x)=x+1$. We can construct the monic quadratic polynomial $f^0_{s-1}(x)\in\mathbb{Z}[x]$ with roots $T_{s-1}^0$ and $T_{s-1}^1$, which is $$f^0_{s-1}(x)=\left\{\begin{array}{ll}x^2+x-\frac{p-1}{4}&\text{if $p\equiv 1\pmod{4}$ (equivalently, $s\geq 2$)},\\ x^2+x+\frac{p+1}{4}&\text{if $p\equiv 3\pmod{4}$ (equivalently, $s=1$)}. \end{array}\right.$$ Then, we can compute $T_{s-1}^0$ and $T_{s-1}^1$.

Next $T_{s-1}^0=T_{s-2}^0+T_{s-2}^2$, and $T_{s-1}^1=T_{s-2}^1+T_{s-2}^3$. The values $T_{s-2}^0T_{s-2}^2$ and $T_{s-2}^1T_{s-2}^3$ can be written as linear combinations of $T_{s-1}^0$, $T_{s-1}^1$, and $1$. In this way, we get a monic quadratic polynomial $f^j_{s-2}(x)$ with roots $T_{s-2}^{j}$ and $T_{s-2}^{j+2}$, for $j=0$ and $j=1$. The quadratic formula would tell you the values $T_{s-2}^j$ and $T_{s-2}^{j+2}$ for $j=0$ and $j=1$. Therefore, $T_{s-2}^j$ for $j=0,1,2,3$ are known.

The process continues. Suppose that you known $T_m^0$, $T_m^1$, $\ldots$, $T_m^{2^{s-m}-1}$. Then, we have $$T_{m}^j = T_{m-1}^j+T_{m-1}^{j+2^{s-m}}.$$ We can write $T_{m-1}^jT_{m-1}^{j+2^{s-m}}$ as a linear combination of $T_\mu^\nu$ for $\mu\geq m$ and $0\le \nu < 2^{s-\mu}$. Therefore, we can find a monic quadratic polynomial $f^j_{m-1}(x)$ with roots $T_{m-1}^j$ and $T_{m-1}^{j+2^{s-m}}$ for $j=0,1,2,\ldots,2^{s-m}-1$. Hence, we can determine $T_{m-1}^j$ for $j=0,1,2,\ldots,2^{s-m+1}-1$.

From the previous computations, the polynomial $$F(x)=f_0^0(x)\cdot f_0^1(x)\cdot\ldots\cdot f_0^{2^{s-1}-1}(x)$$ is a polynomial with roots $t_0,t_1,t_2,\ldots,t_{2^s-1}$. It is a monic polynomial of degree $2^s$ with integer coefficients. The roots $t_0,t_1,t_2,\ldots,t_{2^s-1}$ can be written in terms of nested radicals (in particular, square roots) of rational numbers. This attempt at generalising is not as easy as I initially thought.