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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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  • $\begingroup$ Have you tried running through the proof with the specific example, to see if the argument breaks down, and where? $\endgroup$
    – Arturo Magidin
    Sep 9, 2015 at 18:53
  • $\begingroup$ Purely by group theory, one can see that when the Galois group is $F_20$, that polynomial is irreducible, as the group acts transitively on the set of pairs. Furthermore one can check that the discriminant is not a perfect square. I can't find the paper, but something must be wrong. $\endgroup$
    – Will Sawin
    Sep 9, 2015 at 18:57
  • $\begingroup$ The full reference is Ryan Kavanagh, "On Irreducible Rational Quintics", School of Computing, Queen's University At Kingston, Kingston, Ontario, Canada. You need to use quotes for the title for internet searches. $\endgroup$
    – Lorenz H Menke
    Sep 9, 2015 at 19:51
  • $\begingroup$ Yes I have checked the discriminant, not a perfect square, verified the group by other means or using known theorems for the ${F}_{20}$ cases. The other groups determination of irreducible quintics by this method work. $\endgroup$
    – Lorenz H Menke
    Sep 9, 2015 at 19:54
  • $\begingroup$ Yes I have tried with various examples. The theorem holds for the test cases. The first part of the Theorem is if ${\Delta}_{5} \in \mathbb{Q}^{2}$ and 1 ${P}_{10} \left({x}\right)$ is irreducible over $\mathbb{Q}$ if and only of $\text{Gal} \left({K/\mathbb{Q}}\right) \cong {A}_{5}$. Otherwise, ${P}_{10} \left({x}\right)$ is the product of two quintics irreducible over $\mathbb{Q}$. 2 if the quintic has a complex root then $\text{Gal} \left({K/\mathbb{Q}}\right) \cong {D}_{5}$ and 3 the quintic has five real roots then $\text{Gal} \left({K/\mathbb{Q}}\right) \cong {C}_{5}$. $\endgroup$
    – Lorenz H Menke
    Sep 9, 2015 at 20:35

2 Answers 2

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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

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  • $\begingroup$ The book Mathematical Sciences Research Institute Publications #45, Generic Polynomials Constructive Aspects of the Inverse Galois Problem by Christian U. Jensen uses these polynomials, ${P}_{n} \left({x}\right) = \prod\limits_{1 \le i < j < k < ... \le m} \left({x - \left({{\alpha}_{i} + {\alpha}_{j} + ... + {\alpha}_{m}}\right)}\right)$ to describe the Galois classifications. See p52-54 for treatment of 7th degree and p58-60 for the 11th degree. The 5th degree is not fully treated in this manner. $\endgroup$
    – Lorenz H Menke
    Sep 10, 2015 at 17:27
  • $\begingroup$ @LorenzHMenke Indeed, if I remember correctly, the degree 5 treatment / $P_10$ came from Jensen and Yui's 1980 paper, and partially their 1982 paper, though I think they focussed on the cases where the Galois group was $D_5$ or $\mathbb{Z}_5$. (Naturally, all mistakes in my paper are my own.) You should be able to find the 1982 paper online via sciencedirect, but the 1980 paper might be a bit harder to get a hold of (I got a print copy directly from the second author, who happened to be teaching my course). Hope this helps. $\endgroup$
    – Ryan Kavanagh
    Sep 11, 2015 at 13:14
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As pointed out by Will Sawin in the comments, the result is just wrong. However, the Monthly Paper by Dummitt (1991) Kavanagh cites (where he uses a sextic resolvent) is correct, so you might want to use that instead.

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    $\begingroup$ I have been using Dummitt's paper for this problem. I was using Ryan Kananagh's paper as an interesting alternative since I read Christian U. Jensen's above book and series of papers. $\endgroup$
    – Lorenz H Menke
    Sep 10, 2015 at 17:35

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