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I have a sextic polynomial $P(x)$ which is known to be solvable by radicals according to Galois theory. The polynomial has the following form:

$$ P(x)=49 x^6 - 588 x^5 + 2638 x^4 - 5536 x^3 + 5572 x^2 - 2480 x + 368 $$

I verified its solvability using Magma, and it is confirmed to be solvable. However, I am struggling to explicitly express its roots in terms of radicals. I solved it only numerically using fsolve in Maple and provided $6$ positive real roots (that's what I wanted).

I have searched a lot though I did not quite understand how to solve it using one of those modular functions. I did find that the order of the Galois group of P is $72$ which stabilizes a partition of the set of the roots into two subsets of three roots, does that mean that the polynomial can be reduced into two cubic polynomials?

I'm puzzled by the Galois theory and the several methods used to successfully obtain solutions in radicals of a solvable sextic. Any help or suggestions would be greatly appreciated.

Thank you!

A. Brik
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    The proof that solvability by radicals corresponds to the Galois group being solvable is explicit, it actually produces an algorithm for constructing the splitting field as a sequence of Kummer extensions. The first step involves finding a normal subgroup $H$ of the Galois group $G$ such that the quotient $G/H$ is cyclic of some order $n$, then constructing the extension $L^H/\mathbb{Q}$ (where $L/\mathbb{Q}$ is the splitting field) as a Kummer extension, possibly after adjoining $n^{th}$ roots of unity; are you familiar with this? – Qiaochu Yuan Jul 30 '24 at 19:30
  • No, unfortunately. I'm far from being able to dive into Abstract Algebra and understand all about Galois theory (which is so interesting) in a short amount of time. I wish I studied it before. – A. Brik Jul 30 '24 at 19:40

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Along the lines of this post, you can check that $$P(x) = \left(7 x^3-42 x^2+74 x-44\right)^2-2 \left(9 x^2-36 x+28\right)^2\,.$$ This, using $a^2-b^2=(a+b)(a-b)$, leads to the factorization $$P(x) = \Bigl(7 x^3-\left(9 \sqrt{2}+42\right) x^2+\left(74+36 \sqrt{2}\right) x-28 \sqrt{2}-44\Bigr)\Bigl(7 x^3+\left(9 \sqrt{2}-42\right) x^2+\left(74-36 \sqrt{2}\right) x+28 \sqrt{2}-44\Bigr)\,.$$

Fabian
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    Abstractly this comes from / is roughly equivalent to the fact that the Galois group is the wreath product $S_3 \wr C_2$, or slightly more explicitly $(S_3 \times S_3) \rtimes C_2$, where the $C_2$ acts by exchanging the two $S_3$ factors. The two $S_3$ factors correspond to the two cubics while the $C_2$ exchanges them. So, step 1 of the abstract solvability algorithm would have been to construct the quadratic extension corresponding to this $C_2$, which this calculation reveals to be $\mathbb{Q}(\sqrt{2})$, and over this extension the polynomial splits into two cubics. – Qiaochu Yuan Jul 30 '24 at 20:04
  • I was tricked into thinking that the roots were real which I verified with maple numerically, it turns out that the roots are complex with very small imaginary parts. – A. Brik Jul 30 '24 at 20:38
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    @A.Brik: all the roots are in fact real (no imaginary part). The discriminant of the first factor is $4 (16234+7992 \sqrt{2})>0$; the discriminant of the first factor is $4 (16234-7992 \sqrt{2})\approx 19726>0$. – Fabian Jul 31 '24 at 06:03
  • @Fabian Well that's weird, cause I have solved the two equations in Maple and got Complex roots!! – A. Brik Jul 31 '24 at 20:05
  • @A.Brik: this sounds like a problem with the numerical accuracy. The solutions are approximately given by $0.289679, 0.674756, 1.21788, 2.25844, 3.21729, 4.34196$. You can plot the curve and convince yourself that this makes sense. – Fabian Aug 02 '24 at 07:26
  • @Fabian That's the issue. When you try to extract the roots in radicals you'll see that they are complex, I've converted them into float numbers using evalf and got the same real parts with small imaginary values. More confusion, when I tried to extend them to 100 and even 1000 digits the imaginary part got smaller with degree 10^(-100) and 10^(-1000) for the 1000 digits extension. I cared for the roots in radicals and real but Maple wants them complex Lol. – A. Brik Aug 02 '24 at 15:38