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?

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