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The irreducible but solvable octic,

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

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

$$y = y_k = \omega^{k}+\omega^{12k}+\omega^{17k}+\omega^{28k}\tag{2}$$

$$z_k = 4(y^3+y^2-9y-4)(y^2-2)(y-1)+9\tag{3}$$

then I found a pair of roots of $(1)$ as,

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

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

and the other pairs using appropriate signs of the square roots.

Note: Of course, $y_k$ and $z_k$ are roots of two different 7th-deg eqns with integer coefficients, while $(3)$ is the 6th-deg Tschirnhausen transformation between them. (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 $(1)$ has such a simple form, and if we can find other similar irreducible but solvable octics 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.)

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10  
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 4:24
    
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 4:27
    
Actually I asked maxima to $$ \text{factor(X^8-X^7+29*X^2+29, x^8-4*x^7+8*x^6-6*x^5+2*x^4+6*x^3-3*x^2+x+3);} $$ 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 4:35
    
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 4:39
5  
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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1 Answer 1

up vote 6 down vote accepted

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 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.

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Thanks. I will check out that Magma online calculator. Didn't know one existed. –  Tito Piezas III Oct 19 '13 at 23:23
    
Elkies has a nice article on trinomials with interesting Galois groups, including the Trinks polynomial. –  Tito Piezas III Oct 21 '13 at 13:08
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