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While browsing the Database of Number Fields, I came across 17T8. It only had four equations, one of which is,

$$\small{x^{17} - 5x^{16} + 40x^{15} - 140x^{14} + 610x^{13} - 1622x^{12} + 4870x^{11} - 10220x^{10} + 22720x^9 - 38080x^8 + 63500x^7 - 84100x^6 + 102200x^5 - 102400x^4 + 83000x^3 \color{brown}{- 55864x^2 + 24080x - 9400}=0}$$

It may not look much, but all four examples had the same coefficients except for the part in brown (the $x^2, x^1, x^0$ terms), so I assume there might be a parameterization. After some fiddling, I found the rather simple,

$$\frac{(x^5 - 2x^4 + 10x^3 - 10x^2 + 20x - 10)^3\,(x^2 + x + 4)}{(4x^2 - 5x + 25)} = -36m\tag1$$

The four were just the cases $m = -6, -2, -12, 9$. The discriminant of $(1)$ is,

$$D = 2^{36}\, 3^{52}\, 5^{18}\, m^{10}(16m + 81)^8$$

Update: Note that,

$$\frac{(x^5 - 2x^4 + 10x^3 - 10x^2 + 20x - 10)^\color{red}3\,(x^2 + x + 4)}{4x^2 - 5x + 25}-\frac{27^2}{2^2} = \frac{(x-1)(2x^8 - 4x^7 + 32x^6 - 40x^5 + 170x^4 - 136x^3 + 362x^2 - 166x + 185)^\color{red}2}{4x^2 - 5x + 25}\tag2$$

Note that the quintic and octic factors (both irreducible) have solvable Galois groups. Also, I've seen similar factoring behavior in formulas for the j-function like the well-known icosahedral equation,

$$j(\tau)-1728 =-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^\color{red}3}{r^5(r^{10} + 11r^5 - 1)^5}-12^3 = -\frac{(r^{30} + 522r^{25} - 10005r^{20} - 10005r^{10} - 522r^5 + 1)^\color{red}2}{r^5(r^{10} + 11r^5 - 1)^5}\tag3$$

Makes me wonder if $(1)$ is a formula for something.

Questions:

  1. Does the whole family, except for special $m$, belong to 17T8?
  2. Can one derive it from first principles, instead of a computer search? (Its simple form seem to suggest there might be others.)
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    $\begingroup$ Your link seems broken. Do you mean this link? galoisdb.math.uni-paderborn.de/groups/view/… $\endgroup$
    – Wolfgang
    Commented Dec 18, 2015 at 15:11
  • $\begingroup$ @Wolfgang: Hm, my link seems to work for me. Yours is close, but it goes to 17T5, not 17T8. There are only four results for 17T8. $\endgroup$
    – Tito Piezas III
    Commented Dec 18, 2015 at 15:19
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    $\begingroup$ Funny... your link tells me "- no search parameters found. Please go back to the search page -" while mine has the page title "Transitive Group 17T8 - Polynomials with signature 1". Are computers fuzzy? $\endgroup$
    – Wolfgang
    Commented Dec 18, 2015 at 15:22
  • $\begingroup$ @Wolfgang: I'm quite sure a minute ago I was directed to 17T5. :) But now your link works fine. Thanks. $\endgroup$
    – Tito Piezas III
    Commented Dec 18, 2015 at 15:25
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    $\begingroup$ A part of me really appreciates the accidental 'pun' that 17T8 'secretly' stands for "1 7 Two 8", or the same 1728 the appears in the j-invariant icosahedral equation. ;) $\endgroup$
    – Alex Meiburg
    Commented Mar 14, 2017 at 22:38

1 Answer 1

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Many of the polynomials in the Klueners-Malle database and also in my database with John Jones come from families in the way you correctly describe. So you have "reverse engineered" the source family.

This particular source family is the first of two similar families described in Section 13 my Galois number fields with small root discriminant with Jones. Your m and our t are related by m=-81/16 t. The generic Galois group of this family is 17T8 = SL_2(16).4 and so for almost all t the specialized Galois group will be 17T8. Section 13 discusses some details of the specialization process. In particular, specializing at t=-8 keeps the Galois group at 17T8 and has the unusually small discriminant 2^16 3^20 5^16.

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  • $\begingroup$ I had assumed they found them by computer search. By the way, is it relevant that when $m=0$, the quintic is solvable and has discriminant $2^4\cdot 3^8\cdot 5^3$? I mean, if you pick a random eqn with deg $> 4$, it most probably will have a non-solvable Galois group. But this quintic isn't really random, is it? $\endgroup$
    – Tito Piezas III
    Commented Dec 19, 2015 at 2:36
  • $\begingroup$ In general, when you have a family with generic Galois group G and you specialize at a perhaps degenerate point then the Galois group of the specialization is a subquotient of G. For the family from the original post, the degenerate points are 0 and -81/16. This comment by itself does not force the quintic at 0 to be solvable, but a closer analysis at the nature of the degeneration would. On the other hand, at -81/16 the polynomial factors as quintic (sextic)^2. The quintic has Galois group S5 and "pure thought" forces the sextic to be non-generic, with Galois group PGL2(5)=S5. $\endgroup$
    – Dave Roberts
    Commented Dec 19, 2015 at 15:38
  • $\begingroup$ I guess the degenerate points are $m(16m+81) = 0$, correct? Well, I tried plugging $m=-81/16$ into $(1)$ and I get a linear (octic)^2. The octic still has a solvable Galois group. How did you get the quintic (sextic)^2? $\endgroup$
    – Tito Piezas III
    Commented Dec 19, 2015 at 16:16
  • $\begingroup$ Oh, I get it. You were using the second family in your paper, $$3^3(x^4 + 2x^3 + 4x^2 + 28x - 4)^4(x - 2) + 2^{12} 5^5\,t(2x^2 - 3x + 18) = 0$$ The degenerate points of this is $t = 0,1$ and at $t=1$ I get your quintic (sextic)^2. However, the first family is linear (octic)^2 with a solvable octic. $\endgroup$
    – Tito Piezas III
    Commented Dec 19, 2015 at 16:27
  • $\begingroup$ Yes, indeed Tito! I accidentally was referring to the second family rather than the first in the last two sentences of my general comment on specialization at degenerate points. Your last two comments have it right. $\endgroup$
    – Dave Roberts
    Commented Dec 19, 2015 at 18:34

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