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The irreducible but solvable octic,

$$x^8-x^7+29x^2+29=0\tag{1}\label{1}$$

was first mentioned by Igor Schein in this 1999 sci.math post (Wayback Machine). This does not factor over a quadratic or quartic extension, but over a 7th deg one. It can also be nicely solved using the $\color{blue}{29}$th root of unity. Let $\omega = \exp(2\pi i /29)$ then define,

\begin{gather} y_k = \omega^{k}+\omega^{12k}+\omega^{17k}+\omega^{28k}\tag{2}\label{2} \\ z_k = 4(y_k^3+y_k^2-9y_k-4)(y_k^2-2)(y_k-1)+9\tag{3}\label{3} \end{gather}

then I found a pair of octic roots as,

\begin{gather*} x = \frac{1\color{red}{-}\sqrt{z_{1}}+\sqrt{z_{2}}+\sqrt{z_{4}}+\sqrt{z_{8}}+\sqrt{z_{16}}+\sqrt{z_{32}}+\sqrt{z_{64}}}{8} \approx 1.79106+0.8286\,i\dots \\ x = \frac{1+\sqrt{z_{1}}\color{red}{-}\sqrt{z_{2}}\color{red}{-}\sqrt{z_{4}}+\sqrt{z_{8}}+\sqrt{z_{16}}\color{red}{-}\sqrt{z_{32}}+\sqrt{z_{64}}}{8} \approx 1.79106-0.8286\,i\dots \end{gather*}

and the other pairs using appropriate signs of the square roots. Note that the seven $y_k$ and $z_k$ are roots of two different 7th-deg eqns with integer coefficients, with the $z_k$ being,

$$\small{z^7 - 7z^6 - 2763z^5 - 19523 z^4 + 1946979z^3 + 34928043 z^2 + 119557031z - 3247^2=0}$$

and \eqref{3} is the 6th-deg Tschirnhausen transformation between the $y_k$ and $z_k$. (In an earlier edit, I used an alternative expression for $z_k$ by P. Montgomery found in the sci.math link, but I like this one better.)

Question: Does anyone know why \eqref{1} has such a simple form, and if we can find other irreducible but solvable octics with the same Galois group involving a $p$th root of unity for other prime $p$? (For some reason, this does not appear in the Kluener's database of number fields for 8T25.)

Oscar Lanzi
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Tito Piezas III
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    Actually the field does appear, just with a different generator polynomial $x^8-4x^7+8x^6-6x^5+2x^4+6x^3-3x^2+x+3$. If $x$ is a root of this octic then $X = (-5x^7+13x^6-15x^5-8x^4-11x^3+9x^2-37x+1)/17$ satisfies $X^8 - X^7 + 29X^2 + 29 = 0$. – Noam D. Elkies Oct 18 '13 at 04:24
  • Ah, a 7th deg Tschirnhausen transformation. How did you find the coefficients so quickly? An integer-relations algorithm, I bet? Mathematica has one too. :) – Tito Piezas III Oct 18 '13 at 04:27
  • Actually I asked maxima to $$ \text{factor(X^8-X^7+29X^2+29, x^8-4x^7+8x^6-6x^5+2x^4+6x^3-3x^2+x+3);} $$ and observed the linear factor in the numerator. I don't know what algorithm maxima* uses, but it might well be integer relations. – Noam D. Elkies Oct 18 '13 at 04:35
  • Based on that transformation, I think a candidate would be the one for $p = 239$, or $x^8-3x^7+20x^6+4x^5+39x^4+106x^3+304x^2+434x+187=0$. One can let $y = a_1x^7+a_2x^6+\dots+a_7$, and use the seven indeterminate $a_i$ to knock out coefficients in the new equation in $y$. – Tito Piezas III Oct 18 '13 at 04:39
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    You can indeed construct soluble octics with the same Galois group $C2^3:C7$ with other roots of unity. Say $p$ is 1 mod 7, and $\alpha$ generates a degree 7 extension in ${\mathbb Q}(\zeta_p)$. Then $\sqrt{\alpha}$ generates a field with Galois group $C2^7:C7$, and this group has two normal subgroups $C2^4$ that cut out a field of the type that you want - Galois group $C2^3:C7$, and solvable in radicals involving $p$th roots of unity and square roots. But, say, for $p=43$ I get the octic $x^8-172x^6-1204x^5-860x^4+16856x^3+59856x^2+65016x+7396$, so it's far from being as pretty as Schein's. – Tim Dokchitser Oct 18 '13 at 11:17
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    @TimDokchitser: I find a solution to your octic is $$\omega = \exp(2\pi i/43)$$ $$y = y_k = \omega^{k}+\omega^{6k}+\omega^{7k}+\omega^{36k}+\omega^{37k}+\omega^{42k}$$ $$z_k = y^6-18y^4-17y^3+59y^2+38y-35$$ $$x = \frac{\sqrt{z_{1}}+\sqrt{z_{2}}+\sqrt{z_{4}}+\sqrt{z_{8}}+\sqrt{z_{16}}+\sqrt{z {32}}+\sqrt{z{64}}}{2} = 15.7077\dots$$ – Tito Piezas III Oct 19 '13 at 03:33
  • @TimDokchitser: Does the method you use always result in an octic with (1) a square constant term? (Your example has $7396 = (2\cdot43)^2$.) (2) all real roots? – Tito Piezas III Oct 19 '13 at 04:40
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    I just realized the coefficients of your octic for $p=43$ have a common factor $x^8 - 4\times43(x^6 + 7x^5 + 5x^4 - 98x^3 - 348x^2 - 378x - 43)=0.$ – Tito Piezas III Dec 12 '22 at 06:39
  • The link at the end of the post seems to be dead. The website http://galoisdb.math.upb.de/ is described as "a database of number fields up to degree 19 created by Jürgen Klüners and Gunter Malle." I do not know whether this is the same think as the one that used to be at your link. – Martin Sleziak Dec 14 '22 at 04:02
  • @MartinSleziak Yes, it is the same one. I revisited this old question as it inspired a new question about trinomial octics and the tribonacci constant. Kindly see this MSE post. – Tito Piezas III Dec 14 '22 at 04:05
  • Since the question was bumped anyway, perhaps it would make sense to replace the old link (which no longer works) by a new one. (If there is something corresponding on the new website, too.) – Martin Sleziak Dec 14 '22 at 04:13
  • @MartinSleziak Thanks for the edits. I didn't check the sci.math link, so mea culpa. Sci.math is gone? Sigh, spent many happy hours on that site. – Tito Piezas III Dec 14 '22 at 04:16
  • Maybe $43$ gives a less elegant form because its Euler totient is a multiple of $3$ whereas that of $29$ has only prime factors of $2$ and $7$. The next prime with the latter property is $113$, what does that give? – Oscar Lanzi Dec 14 '22 at 17:34
  • @OscarLanzi The Magma calculator for $p=113$ gives $x^8 + 512x^6 + 8384x^5 + 55568x^4 - 1472000x^3 - 2752640x^2 - 42152576x + 623706240$. For $p=29$, it yields $x^8 + 3x^7 + 9x^6 + 15x^5 + 23x^4 + 21x^3 + 18x^2 + 7x + 2$, It would really be nice to know in what context Igor Schein found his octic. – Tito Piezas III Dec 14 '22 at 17:56

1 Answers1

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To answer your second question, there are soluble octics with the same Galois group involving other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$ the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta_p)$. E.g. take $\alpha=\sum_i \zeta_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F_p}$.

If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$. That is, its Galois group over ${\mathbb Q}$ is $$ G=C_2\wr C_7\cong C_2^7:C_7. $$ Viewing $C_2^7$ as a 7-dimensional representation of $C_7$ over ${\mathbb F}_2$, it decomposes as a 1-dimensional (trivial) representation plus two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out $C_2^4\triangleleft G$, which is one of those plus the trivial one. This gives a Galois group $C_2^3:C_7$ that you are after, and a subgroup $C_7$ in it cuts out the required octic field.

Here is a Magma code that can be used in the Magma online calculator to get such an octic:

p:=43;        // or some other p = 1 mod 7

K<z>:=CyclotomicField(p);
alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)];


R<x>:=PolynomialRing(Rationals());
f:=Evaluate(MinimalPolynomial(alpha),x^2);
K:=NumberField(f);
assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16};
F:=Field(Minimize(a));
DefiningPolynomial(F);

You can also stick in a Tschirnhaus transformation, say,

alpha:=alpha^3+alpha+1;

in the 5th line to vary the generator of $K$ — in this way you get all possible $C_2^3:C_7$-extensions involving $p$th roots of unity.

For your questions in the comments, the roots might be real or complex, and the constant term may or may not be a square — this depends on whether $\alpha$ is chosen to be totally real, and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation to modify the final output or Pari's polredabs function to try and get the coefficients smaller.

I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group $\operatorname{PSL}(2,7)$, and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.

LSpice
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Tim Dokchitser
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