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This is, at least spiritually, a follow-up to this question. But you don't need to have read it in order to understand this question.

I'm working on implementing an algorithm to solve a cyclic extension by radicals. At a high level, the algorithm is:

  1. Let $\zeta$ be a $n$th root of unity
  2. Let $\Theta(x_0, \ldots, x_{n-1}) = x_0 + \zeta x_1 + \zeta^2 x_2 + \ldots + \zeta^{n-1} x_{n-1}$
  3. Let $\psi = \Theta^n$. Note this is fixed by the galois group of $f$, which is $C_n$ by assumption, so lies in the base field $\mathbb{Q}(\zeta)$.
  4. Build the galois resolvent $\mathcal{L}_{f,\psi}(t)$, following the algorithm in Lehobey's Resolvent Computations By Resultants Without Extraneous Powers. The roots of this polynomial (in $\mathbb{Q}(\zeta)$) are exactly the conjugates of $\psi$.
  5. Then if we write $\psi_1, \ldots, \psi_{n-1}$ for the $n-1$ roots of $\mathcal{L}_{f,\psi}$, and $\psi_0$ for the negation of the second coefficient of $f$, we (should) find
  6. A root of $f$ is $$\frac{\psi_0 + \sqrt[n]{\psi_1} + \sqrt[n]{\psi_2} + \ldots + \sqrt[n]{\psi_{n-1}}}{n}$$

I've tried this on a variety of polynomials, of degree $3$, $4$, and $5$.

  • The algorithm always produces a root for degrees $3$ and $4$ (at least on all the examples I've tried)
  • The algorithm correctly produces a root for $x^5 + x^4 - 12x^3 - 21x^2 + x + 5$
  • The algorithm incorrectly produces non-roots for most cyclic quintics, eg $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$

If it's helpful, I'll attach the output of my script in a pastebin, and the contents of the script itself in another, but the tldr is that sage gives me a root of the first quintic, and decidedly not a root for the second.

Does anybody know of any theoretical mistakes I might be making? Are there differences between these two quintincs that I'm not aware of? They both have galois group $C_5$, and every reference I've found has said that I should be able to treat them the same... I don't think it's a bug in my code, because it works on every degree $3$ and $4$ polynomial I've tried it on, so I'm confused as to what's going wrong. If it's helpful, I've been pulling the quintic equations from this list

Thanks in advance!

HallaSurvivor
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  • My guess would be that it might be a situation similar to this https://math.stackexchange.com/a/604738/581023 . Where you have to pick specific choices of nth root that are compatable for the formula to work. This is hidden by the simple nth root notation –  Aug 05 '21 at 07:43
  • @river -- that's a fantastic idea, that's really obvious in hindsight. I'll try the other roots today and see if that solves the problem ^_^ – HallaSurvivor Aug 05 '21 at 18:28
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    You're exactly right! Now if I can just figure out which $n$th roots to choose... – HallaSurvivor Aug 05 '21 at 19:55
  • https://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf check the very end part of this "This is the Casus Irreducibilis for quintic polynomials". –  Aug 07 '21 at 12:53

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