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Does the octic,

$\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$

for any constant n have Galois group of order 1344? Its discriminant D is a perfect square,

$D = (1728n^4-341901n^3-11560361n^2+3383044089n+28121497213)^2$

Surely (1) is not an isolated result. How easy is it to find another family with the same Galois group and same form,

$\tag{2} \text{octic poly in}\ x = nx^2$

Perusing Kluener's "A Database For Number Fields" this seems to be the only one. (Though I was able to find a 2-parameter family of a different form.)

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  • $\begingroup$ What is (1)...? $\endgroup$
    – Steven Landsburg
    Apr 25, 2012 at 20:52
  • $\begingroup$ I believe (1) is the family described in equation (1), which apparently (as per @Noam's and @Mark's guesses) has a fixed Galois group of order 1344, except for the exceptional values when the polynomial is reducible. I am not at all sure what the significance of the squareness of the discrimant is (I mean, the Galois group is then a subgroup of $A_8,$ but so?) $\endgroup$
    – Igor Rivin
    Apr 26, 2012 at 2:15
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    $\begingroup$ For your second question, Malle in "Multi-parameter Polynomials with Given Galois Group" (citeseerx.ist.psu.edu/viewdoc/…) constructs a 3-parameter family of polynomials with Galois group $PSL(2,7)=PSL(3,2)$ (section 4). Then in section 6, he explains that, essentially by taking generic square roots you can construct a 4-parameter family with Galois group $2^3.PSL(3,2)$, which is your group. Then by fixing some of the parameters, you might be able to kill of the dependence on them in all terms but $x^2$. $\endgroup$
    – Tim Dokchitser
    Apr 26, 2012 at 8:49
  • $\begingroup$ That was a very good reference, Tim. $\endgroup$
    – Tito Piezas III
    Apr 26, 2012 at 14:10

3 Answers 3

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Maple gives Galois group $H$ of your polynomial $f=f_n$ as the subgroup of $S_8$ generated by these permutations: $$(1 2)(5 6), (1 2 3)(4 6 5), (1 2 6 3 4 5 7), (1 8)(2 3)(4 5)(6 7), (2 8)(1 3)(4 6)(5 7), (4 8)(1 5)(2 6)(3 7).$$ That can be easily proved by using the standard technique of Galois theory assuming that $n$ is such that your polynomial is irreducible (I think that the only exceptional values are 17 and 145 as in Alastair's answer). See Van der Waerden's book, for example: consider the polynomial in $9$ variables $$g(x,x_1,...,x_8)=\prod_{\sigma\in S_8} (x-x_{\sigma 1}a_1-...-x_{\sigma 8}a_8) $$ where $a_1,...,a_8$ are the roots of your polynomial $f=f_n$, $\sigma$ runs over all permutations of $S_8$. The coefficients of $g$ are symmetric polynomials in $a_i$, hence polynomials in the coefficients of $f_n$. Now show that the products of factors corresponding to permutations from $H$ form an irreducible factor of $g$. Hence $H$ is the Galois group of $f_n$ (again, assuming $f_n$ is irreducible).

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  • $\begingroup$ Thanks, Mark. I believe this takes care of the first part of the question. But there was a second part, namely: "how easy is it to find, for the same generic Galois group as above, another family where all coefficients remain constant except one?". Kluener's database gives more than a thousand examples for G1344, but this was the only family, among many I found, with a form like that. $\endgroup$
    – Tito Piezas III
    Apr 26, 2012 at 3:10
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    $\begingroup$ @Tito: I have no idea about the second part. In fact I would like to know how you got the first family (and also why is it interesting, what is the role of 1344, are there other groups except this, $S_n$ and $A_n$ with this property, etc.?) $\endgroup$
    – user6976
    Apr 26, 2012 at 7:31
  • $\begingroup$ @Mark: My interest is really more on solvable groups and, for the octic, especially those with order divisible by 7, such as Schein's nice $x^8-x^7+29x^2+29=0$ which can be solved using the 29th root of unity. While browsing Kluener's database for the non-solvable group 8T48, it was easy to notice a series of polynomials that differed only by a single coefficient. Further experimentation with other values of $n$ confirmed that certain expressions of the octic roots were 7th deg algebraic numbers. As to your question, it seems the family $x^8-x^6-2x^5+2x^3+x^2+n^2=0$ has group $S_8$. $\endgroup$
    – Tito Piezas III
    Apr 26, 2012 at 15:16
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More of a comment than an answer, but there are exceptions for $n = 17$ and $n = 145$, where the Galois group is simple of order 168. This is all the exceptions for $|n| < 2\times10^5$.

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  • $\begingroup$ Exceptions for what? This is rather cryptic. $\endgroup$
    – Igor Rivin
    Apr 25, 2012 at 20:50
  • $\begingroup$ Presumably the generic Galois group is the group of order 1344, so these are the exceptional values where the Galois group becomes smaller. $\endgroup$
    – Noam D. Elkies
    Apr 25, 2012 at 21:06
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    $\begingroup$ These are the cases when the polynomial is reducible. $\endgroup$
    – user6976
    Apr 26, 2012 at 1:37
  • $\begingroup$ @Igor: Apologies, I should have qualified this with 'to the initial question'. @Mark: Of course, I can't believe I forgot to check something so obvious in my calculations! $\endgroup$
    – Alastair Litterick
    Apr 26, 2012 at 7:43
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In arxiv.org/abs/1209.5300 I give the following polynomial with the same Galois group over Q(t): $(y+1)(y^7-y^6-11y^5+y^4+41y^3+25y^2-34y-29)-t(2y+3)^2$. This has $(6912t^4-3456t^3-95472t^2+23976t-1417)^2$ for a discriminant. This doesn't quite answer the second question, but substituting $y \mapsto (x-3)/2$, $t \mapsto s/256$ gives $(x-1) (x^7-23 x^6+181 x^5-547 x^4+515 x^3-325 x^2+71 x-1)-s x^2$, with discriminant $256 (27 s^4-3456 s^3-24440832 s^2+1571291136 s-23773315072)^2$, which does.

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