The radical solution of a solvable 17th degree equation

(The question is at the bottom of the post.) Here's a "natural" solvable 17-th deg eqn with small coefficients:

$$\begin{align*} x^{17}-6 x^{16}&-24 x^{15}-42 x^{14}-31 x^{13}-23 x^{12}-7 x^{11}-x^{10}\\ &\quad-4 x^9-11 x^8-7 x^7-13 x^6-x^5+x^3+x^2+x-1 = 0 \quad\text{(eq.1)} \end{align*}$$

Its unique real root is exactly,

$$x = \frac{\zeta_{48} \eta(\tau)}{\sqrt{2}\,\eta(2\tau)} = 9.1630942 \dots$$

with root of unity $\zeta_{48} = \exp(2\pi i/48)$, the Dedekind eta function $\eta(\tau)$, $\tau = (1+\sqrt{-d})/2$, and $d = 383$. This $d$ has class number $h(-d) = 17$.

To solve this, depress eq.1 (get rid of its $x^{n-1}$ term), by letting $x = (y+6)/17$ to get, $$\begin{align*} y^{17}&-11832 y^{15}-1124346 y^{14}-55393735 y^{13}-1784741617 y^{12}\\ &\quad-41171464807 y^{11}-711423456455 y^{10}-9455898295636 y^9-99724287747103 y^8\\ &\quad 887992943070295 y^7-7665207188897171 y^6-70479807472769473 y^5\\ &\quad -592167373130143650 y^4-3496187093606980919 y^3-8695712981307573757 y^2\\ &\quad +68265051092799270505 y-427806967360317821039 = 0 \qquad \text{(eq.2)} \end{align*}$$ Its 16-deg resolvent, a polynomial with INTEGER coefficients, call this $R_{16}$, has roots,

$$\begin{align*} z_k &= [(y_1 + w^k y_2 + w^{2k} y_8 + w^{3k} y_7 + w^{4k} y_{16} + w^{5k} y_4 + w^{6k} y_{12} + \\ &\qquad + w^{7k} y_{15} + w^{8k} y_{11} + w^{9k} y_{10} + w^{10k} y_{14} + w^{11k} y_{13} + w^{12k} y_5 +\\ &\qquad + w^{13k} y_{17} + w^{14k} y_6 + w^{15k} y_9 + w^{16k} y_3)/17]^{17} \end{align*}$$

for $k = 1,\dots,16$ where w is any complex 17th root of unity.

Note the specific arrangement of the $y_n$. There are $16! \approx 2 \times 10^{13}$ possible permutations of the $y_n$, and out of that huge number, there are only 16 such that $R_{16}$ has integer coefficients, and we have given one of them. Of course, a short cut was used to find it, because even if your computer can check a million permutations a second, it would still take about 8 months to go through them all. The short cut took less than two hours to find $R_{16}$.

The $y_n$ follows the root object Root[poly, n] ordering in Mathematica. Approximately, these are,

$$\begin{align*} y_1 &= 149.7726\\ \{y_2, y_3\} &= -27.62 \mp 18.49i\\ \{y_4, y_5\} &= -21.61 \mp 7.52i\\ \{y_6, y_7\} &= -16.58 \mp 6.34i\\ \{y_8, y_9\} &= -10.57 \mp 15.32i\\ \{y_{10}, y_{11}\} &= -5.02 \mp 13.71i\\ \{y_{12}, y_{13}\} &= -2.34 \mp 13.15i\\ \{y_{14}, y_{15}\} &= 2.57 \mp 2.60i\\ \{y_{16}, y_{17}\} &= 6.31 \mp 7.04i \end{align*}$$

$R_{16}$ has extremely large integer coefficients, with the largest being the 248-digit constant term $429534618434587^{17}$ which, naturally enough, is a 17th power. (Note: $R_{16}$ can easily be formed using 500-digit precision or more on the $y_n$, and multiplying the 16 factors together to form the polynomial.)

The polynomial $R_{16}$ can then be factored into two octics over the radical extension $\sqrt{17}$. This, in turn, can be factored into 2 quartics over $\sqrt{2(17+\sqrt{17})}$. This can be factored further into 2 quadratics using an expression involved in the 17th root of unity. Apparently, to solve $R_{16} = 0$, only square roots of square roots of square roots, etc, are needed.

The real root of eq.2 in radicals is then,

$$y_1 = {z_1}^{1/17} + {z_2}^{1/17} + {z_3}^{1/17} + \dots + {z_{16}}^{1/17} = 149.7726 \dots$$

Problem: Express the roots of this particular $R_{16}$ purely in terms of the complex 17th root of unity. (If anyone knows how to contact the mathematician Peter-Lawrence Montgomery, he probably will know how, since he has done something similar with a septic root and the 29th root of unity.)

Since the Galois group of this polynomial is rather small, this problem is in fact very doable.

First, you want to determine the Galois group.
Factorizing enough modulo $p$ to get the cycle length decomposition of $Frob_p$ is done in $O(\log(p))$ time, so doing this for primes up to a few tens of thousands quickly tell you that the polynomial splits mod $p$ in $1/34$ cases, it's irreducible in $16/34$ cases and it has a $(1)(2)^8$ factorisation in the remaining $17/34$ cases.

Clearly, you are facing the diedral group of order $34$.

When you compute the complex roots, it automatically comes with complex conjugation, which is one of the $17$ octuple transpositions in the Galois group. The challenging part is then to discover the rest of the group, for example, by getting one of the $16$ $17$-cycles.
However, the fact the complex conjugation has to transform such a $17$-cycle into its own inverse greatly reduces the number of candidates. This leaves you with $16$ good cycles among $16\times 14 \times 12 \times \ldots \times 2$, or, if you fix the image of one root, one cycle among $14 \times 12 \times \ldots \times 2 = 645120$

Testing the validity of the cycle is checking that a candidate Galois group invariant is rational (even an integer if your invariant is polynomial in the root, because the roots are algebraic inteegrs), and so I get the $17$-cycle in a few seconds.

Once you have done that, it is essentially game over. Any integer quantity that you may want to know or that you are looking for in your factorizations can be obtained by carefully crafting a polynomial expression in the roots that is invariant by the group and then numerically evaluating it.

Say you number the roots $x_0,x_1,x_2, \ldots x_{16}$ where $x_0 \in \Bbb R$ and $x_0 \mapsto x_1 \mapsto x_2 \ldots$ is a $17$-cycle $\sigma$in the Galois group.

The subgroup $H$ generated by $\sigma$ is a normal index $2$ subgroup of $G$, and corresponds to a degree $2$ extension of $\Bbb Q$.

Let $y = \sum x^2 \sigma(x) - \sum x \sigma(x)^2$. Then any octuple transposition turns $y$ into $-y$, so this degree $2$ extension is $\Bbb Q(\sqrt{y^2}) $. It turns out $y^2 = -3447 = 9\times(-383)$ so we have $K^{H} = \Bbb Q(\sqrt{-383})$ as expected.

Then fix a primitive $17$th root of unity $\zeta_{17}$ and let $r_k = \sum \zeta_{17}^{jk} x_j$. Then $\sigma(r_k) = \zeta_{17}^{-k} r_k$.

As a result, expressions like $r_0,r_1^{17}, r_2 r_1^{15}, r_3 r_1^{14} \ldots$ are all invariant by $\sigma$ and therefore in $\Bbb Q(\sqrt{-383}, \zeta_{17})$.

To get them as polynomial expressions in $\zeta_{17}$ with coefficients in $\Bbb Q(\sqrt{-383})$, note that the ring endomorphisms $\tau_n : \zeta_{17} \mapsto \zeta_{17}^n$ of $\Bbb Q(\zeta_{17})$ extend to our expressions as $\tau_n : r_i \mapsto r_{ni}$ .
Given an element $x$ in $\Bbb Q(\sqrt{-383},\zeta_{17})$, if you know its images by those $\tau_n$ morphisms, you get its $\Bbb Q(\sqrt{-383})$-coefficients by the inverse Fourier transform :

$x = \frac 1{17} \sum_{n=0}^{16} \tau_{-n}(x) \zeta_{17}^n $

(this is almost trivial when $x = \zeta_{17}^k$ and you extend to the whole space because things are $\Bbb Q(\sqrt{-383})$-linear)

Then since everything is an algebraic integer, you end up with coefficients in the ring of integer $\Bbb Z[\frac {1+\sqrt{-383}}2]$

With my roots numbering (which is $x_0 \approx 9.16, x_1 \approx 0.72+0.41i$), I got :

$r_0 = 6$ (yay ! you would have $0$ instead because you decided to shift everyone but this should be the only difference)

$r_1^{17} = 15135336932252598174 + (12339796372897695943.5 -27829425390565964.5\sqrt{-383})\zeta_{17} + (5778331619368472661 -89788835135812214\sqrt{-383})\zeta_{17}^2 + \cdots $

$r_1^2 r_{15} = 836 + (180-6\sqrt{-383})\zeta_{17} +(78.5-15.5\sqrt{-383})\zeta_{17}^2 +(-77-11\sqrt{-383})\zeta_{17}^3 + \cdots$

Note that those expressions allow you to get $r_2,r_3,\cdots,r_{16}$ as polynomial expressions in $r_1$ and $\zeta_{17}$, so you only need to take one $17$th root to get every resultant. Also, you can apply $\tau_n$ to them and get the expressions for $r_2^{17},r_2^2 r_{13}$ and so on simply by shuffling those coefficients in the right way.

One last thing I want to point out, is that all this stuff with roots of unity is only important if you want to get the actual procedure with the $17$th root. But there is also some other stuff that the Galois group says, mainly that if you have one root, then the other $16$ get paired into $8$ distinguishable pairs, so there are rational polynomials $f_i,g_i$ for $i=1 , \ldots, 8$ such that $x_{j+i}+x_{j-i} = f_i(x_j)$ and $x_{j+i}x_{j-i} = g_i(x_j)$, and that given one root $x_j$ and one root $x_{j \pm i}$ of $t^2-f_i(x_j)+g_i(x_j)$, you deduce all of the rest with rational polynomials $h_i^1,h_i^2,\ldots,h_i^{15}$ of those two.

The data of those polynomial transformations is dual to the knowledge of the Galois group and they are sort of at opposite ends of each other. They are very concrete and instantly tell you how much relations there are between the roots. You can compute their coefficients by Lagrange interpolation and then you can very easily check numerically that you have got the right expressions.