I am computing the Galois groups of quintics using the theorems from Ryan Kavanagh paper "On Irreducible Rational Quintics" using the decic resolvent ${P}\_{10} \left({x}\right) = \prod\limits\_{1 \le i < j \le 5} \left({x - \left({{\alpha}\_{i} + {\alpha}\_{j}}\right)}\right)$. The relevant part is the second part of Theorem 2 where given the discriminant ${\Delta}\_{5} \not\in \mathbb{Q}^{2}$ of the quintic then ${P}\_{10} \left({x}\right)$ is irreducible over $\mathbb{Q}$ if and only if $\text{Gal} \left({f/\mathbb{Q}}\right) \cong {S}\_{5}$ and otherwise, ${P}\_{10} \left({x}\right)$ is the product of two quintics irreducible over $\mathbb{Q}$ and $\text{Gal} \left({f/\mathbb{Q}}\right) \cong {F}\_{20}$.


Now I have computed ${P}\_{10} \left({x}\right)$ both symbolically and numerically for a given test polynomial that I know has Galois group ${F}\_{20}$ such as ${x}^{5} + a$ or the two cases of ${x}^{5} + a\, {x}^{2} + b$. The problem is that the decic resolvent ${P}\_{10} \left({x}\right)$ does not factor as two irreducible quintics over $\mathbb{Q}$.


I am using Mathematica for the calculations. My computation of ${P}\_{10} \left({x}\right)$ compares correctly to the examples given. I am using two different methods which all agree. For the ${F}\_{20}$ Galois group cases the factoring is not correct. Is there an error in the statement of the theorem?



Theorems of the Galois groups of quintics appears not to work for the ${F}_{20}$ group determination
Alas, I have insufficient reputation to comment, so I'll submit my comment as an answer.


Thanks to everybody for the feedback; I'll look at working it into the paper and trying to salvage it. I wrote it during my third year of undergraduate studies for my Galois theory class, and to be honest, I never expected anybody else to ever look at it.


Here's a direct link to the paper: <http://ryanak.ca/files/papers/galois.pdf>


Best wishes,


Ryan


