# Finding minimal polynomial of generator of quadratic subfield of cyclotomic extension $\mathbb{Q}(\xi_{17})$

I will prove that $$K_2=\mathbb{Q}(\sqrt{17})$$ skipping a few 'trivial' steps to keep the post short. There is a theorem that says that Gal$$(\mathbb{Q}(\xi_{17}))\cong\left(\mathbb{Z}_{17}\right)^{\times}\cong\mathbb{Z}_{16}$$. We claim that $$3$$ is a generator of the multiplicative group $$\left(\mathbb{Z}_{17}\right)^{\times}$$ of order $$16$$; this I have shown.

Thus $$\langle3\rangle=\left(\mathbb{Z}_{17}\right)^{\times}$$. We define the following maps: \begin{align*} \text{Gal}(\mathbb{Q}(\xi_{17}))&\to\left(\mathbb{Z}_{17}\right)^{\times},\\ \left(\sigma_k:\xi_{17}\mapsto\xi_{17}^k\right)&\mapsto k\text{ mod }17\\ \varphi(n):3&\mapsto3^n\text{ mod }17. \end{align*} We also know that there are subfields $$K_d$$ of $$\mathbb{Q}(\xi_{17})$$ such that $$[K_d:\mathbb{Q}]=d$$ with $$d\in\{1,2,4,8\}$$. We now list subgroups $$A,B,C$$ of order $$2,4,8$$ respectively. \begin{align*} A&=\{0,8\}\\ B&=\{0,4,8,12\}\\ C&=\{0,2,4,6,8,10,12,14\}\\ \end{align*} We now apply $$\varphi$$ to the three groups $$A,B,C$$: \begin{align*} A'&=\varphi({A})=\{1,-1\}\\ B'&=\varphi({B})=\{1,-1,4,-4\}\\ C'&=\varphi({C})=\{1,-1,2,-2,4,-4,8,-8\}\\ \end{align*} We define \begin{align*} \tau_{A}&=\sum_{k\in A'}\xi_{17}^k\\ \tau_{B}&=\sum_{k\in B'}\xi_{17}^k\\ \tau_{C}&=\sum_{k\in C'}\xi_{17}^k. \end{align*} The subfields of $$\mathbb{Q}(\xi_{17})$$ are $$\mathbb{Q}(\tau_A),\mathbb{Q}(\tau_B),\mathbb{Q}(\tau_C)$$. The minimal polynomials are \begin{align*} m_{\tau_Z}=\prod_{k\in G}(x-\sigma_k(\tau_{Z})) \end{align*} where $$Z\in\{A,B,C\}$$ and $$G$$ is a set of length $$\#Z$$ which does contain $$\#Z$$ elements from $$\mathbb{Z}_{16}-Z$$. This last sentence above is where I can not verify what I'm saying and I don't what should be correct. We compute: \begin{align*} m_{\tau_C}=\prod_{k\in \{3,-3\}}(x-\sigma_k(\tau_{C}))=x^2+x-4. \end{align*} Its roots are $$\frac{-1\pm\sqrt{17}}{2}$$ and $$m_{\tau_C}$$ is irreducible since it has no rational roots (only linear factors would have been possible), so $$K_2=\mathbb{Q}(\tau_C)=\mathbb{Q}(\sqrt{17})$$.

So, partly I know what I'm doing but at the point where I try to compute the minimal polynomials, I have no idea which indices to take in the product... The minimal polynomial I just bluffed by letting Mathematica compute it via the minimal polynomial finder on Wolfram online.

Extra: My Mathematica attempt reads the following

xi = Exp[2*Pi*I/17];
sigma[t_, n_] :=t^n;
tauC = xi^1 + xi^(-1) + xi^2 + xi^(-2) + xi^4 + xi^(-4) + xi^8 + xi^9;
Product[(x - sigma[xi, i]), {i, 2}] // Expand // N // Rationalize

(0.445738 + 0.895163 I) - (1.67148 + 1.03494 I) x + x^2


Btw, the minimal polynomimal of degree $$6$$ of the generator of the subfield $$K_6$$ of the cyclotomic field $$\mathbb{Q}(\xi_{13})$$ I did find this way taking the product from $$1$$ to $$6$$. Replacing $$13$$ by $$17$$ gets me trouble. How is that possible?

Any help or suggestion is appreciated!

• @JyrkiLahtonen Actually they do. But for computing the six degree polynomial for $\xi_{13}+\xi_{13}^{-1}$, it actually worked somehow after taking the product then from 1 to 6... I will replace the code for what I think is true^ – Algebear May 3 at 19:21
• Also, more importantly I think, aren't you trying to get the quadratic with zeros $\tau_C$ and $\sigma_3(\tau_C)$ as the other zero instead of $\sigma_2(\tau_C)$? I mean, $\tau_C$ is invariant under $\sigma_2$, no? – Jyrki Lahtonen May 3 at 19:23
• @JyrkiLahtonen That's exactly my question. Should I just compute $\sigma_k(\tau_C)$ for some $k$ to see what happens? And then what is the procedure for selecting those $k$'s? I really don't know what roots I want in the minimal polynomial next to $\tau_C$ of course. – Algebear May 3 at 19:28
• So you want the minimal polynomials of $\tau_A,\tau_B$ and $\tau_C$ over $\Bbb{Q}$. Gotcha! – Jyrki Lahtonen May 3 at 19:33
• I hope the answer helps you. I checked that this way Mathematica gave the correct minimal polynomia $x^2+x-4$ for $\tau_C$. But, it's getting late here. I gotta go sleep. I'll be back in 9 hours or so. – Jyrki Lahtonen May 3 at 19:58

What you seem to be having trouble with is the following.

You have correctly verified that the big Galois group $$G=Gal(K_{16}/K_1)$$ is of order $$16$$, and generated by $$\sigma_3$$. It has subgroups \begin{aligned} Gal(K_{16}/K_2)&=\langle\sigma_3^2\rangle=\langle \sigma_9\rangle&=C',\\ Gal(K_{16}/K_4)&=\langle\sigma_3^4\rangle=\langle\sigma_{-4}\rangle&=B',\\ Gal(K_{16}/K_8)&=\langle\sigma_3^8\rangle=\langle\sigma_{-1}\rangle&=A'. \end{aligned} As $$G$$ is abelian, these are all normal subgroups of $$G$$. You get the Galois groups for the extensions $$K_d/K_1,d\in\{2,4,8\}$$, as quotient groups of $$G$$. So \begin{aligned} Gal(K_2/K_1)&=G/C',\\ Gal(K_4/K_1)&=G/B',\\ Gal(K_8/K_1)&=G/A'. \end{aligned} Because $$G$$ is generated by $$\sigma_3$$, these are generated by the cosets $$\sigma_3C'$$, $$\sigma_3B'$$ and $$\sigma_3A'$$ respectively.

Therefore the sought after minimal polynomials are \begin{aligned} m_{\tau_A}(x)=\prod_{j=0}^{8-1}\left(x-\sigma_3^j(\tau_A)\right),\\ m_{\tau_B}(x)=\prod_{j=0}^{4-1}\left(x-\sigma_3^j(\tau_B)\right),&\ \text{and}\\ m_{\tau_C}(x)=\prod_{j=0}^{2-1}\left(x-\sigma_3^j(\tau_C)\right).\\ \end{aligned} You stop the products just before you reach the lowest positive power of $$\sigma_3$$ that belongs to the relevant subgroup. After all, the zeros of the minimal polynomials are simple, $$\tau_A$$ is invariant under the powers of $$\sigma_3^8$$, $$\tau_B$$ under the powers of $$\sigma_3^4$$, and $$\tau_C$$ already under $$\sigma_3^2$$. In yet other words, the powers of $$\sigma_3$$ in these product formulas range over the relevant quotient group.

The following Mathematica snippet does it

xi = Exp[2 Pi I/17];
tau[d_, j_] := Sum[xi^(3^(d k + j)), {k, 1, 16/d}];
minpol[d_] := N[Expand[Product[x - tau[d, j], {j, 0, d - 1}]]];


Here $$d$$ must be a factor of $$16$$ for the formulas to make sense. My definition of tau[d_,j_] calculates $$\sigma_3^j(\tau_X)$$. Here $$d=2$$ gives $$X=C$$, $$d=4$$ gives $$X=B$$ and $$d=8$$ gives $$X=A$$. The point is that $$Gal(K_d/\Bbb{Q})$$ consists of the restrictions of powers of $$\sigma_3^j=\sigma_{3^j}$$ with $$j$$ ranging from $$0$$ to $$d-1$$.

The use of $$3^j$$ comes from the isomorphism $$\phi:\Bbb{Z}_{16}\to Gal(K_{16}/\Bbb{Q})$$ that maps $$j$$ to $$\phi(j)=\sigma_3^j=\sigma_{3^j}$$. My definition for tau[d_,j_] applies this to the (additive) subgroup $$\langle d\rangle\le\Bbb{Z}_{16}, d=2,4,8$$. We get $$3^j$$ instead of $$3j$$ as the index of $$\sigma$$ because $$\phi$$ maps an additive structure to a multiplicative one. If you find this confusing, have Mathematica spell out tau[d,j] for various choices of $$d\in\{2,4,8\}, j=0,1,\ldots,d-1$$, and see which seventeenth root of unity appear in which conjugate. I'm sure you'll get the hang of it!

Anyway, this gives the predicted (-4. + 0. I) + (1. + 0. I) x + x^2 as minpol[2]. Leaving you the fun to check out minpol[4] and minpol[8].

• Nice explanation! Do you mean with that product the same as: xi = Exp[2*Pi*I/17]; sigma[t_, n_] := t^(3 n); tauC = xi^1 + 1/xi; Product[(x - sigma[tauC, i]), {i, 0, 7}] // Expand // N // Rationalize In other words, do you mean that $\sigma_3^j:\xi_{17}\mapsto\xi_{17}^{3j}$? And is $\sigma_k$ defined as $\sigma_k(\tau_C)=\tau_C^k$? Yes, I am still a bit lost and actually the only reason is now that Mathematica does not give me the right polynomials which the minimal polynomial finder gives me... – Algebear May 3 at 20:30
• Not quite @Algebear. The product in the Galois group is $\sigma_k\circ\sigma_\ell=\sigma_{k\ell}$, so $\sigma_3^j=\sigma_{3^j}$ with the arithmetic of the subscripts being done modulo $17$ (the Galois group is $\simeq\Bbb{Z}_{17}^*$). – Jyrki Lahtonen May 4 at 3:28
• Thanks a lot for your explanation. The pieces have fallen together why that generator $3$ plays a big role into computing these polynomials. The code works well too! However, if I were to ever compute such a polynomial again, there would be no way of doing that on paper in a reasonable amount of time ;-) – Algebear May 5 at 14:04
• Can this method also be adjusted to find the minimal polynomial for the generator of $K_4/K_2$? – Algebear May 5 at 14:39
• @Algebear Yes, it can. The Galois group $Gal(K_4/K_2)$ is cyclic of order two, and generated by $\sigma_3^2=\sigma_9$. Meaning that the minimal polynomial is $$m(x)=(x-\tau_B)(x-\sigma_9(\tau_B)).$$ Here $\sigma_9(\tau_B)$ is the sum of those powers of $\xi$ that appear in $\tau_C$ but don't appear in $\tau_B$. – Jyrki Lahtonen May 5 at 17:46