A method to generate solvable equations of degrees $p = 7, 13, 19, 31, 37,\dots$ using only cubics

Question

I've always wondered if the DeMoivre method to generate an algebraic number $x_p$,

$$x_p = u_1^{1/p}+u_2^{1/p}$$

of degree $p$ using only quadratic roots $u_i$ could be generalized using cubic roots $v_i$,

$$x_p = v_1^{1/p}+v_2^{1/p}+v_3^{1/p}$$

I serendipitously found a method which works for prime $p=6m+1$, the clue being the Klein quartic for $p=7$. The surfaces I used starts with that,

$$a^3b+b^3c+c^3a\quad \text{(deg 7)}$$ $$\;a^4b+b^4c+c^4a\;\quad \text{(deg 13)}$$ $$a^5b^2+b^5c^2+c^5a^2\quad \text{(deg 19)}$$

and so on. First, given the generic cubic for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

with discriminant $D=-(n^2-3n+9)^2,$ hence all roots $a,b,c$ are real.

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I. Level 7

Using the roots $a,b,c$ of the generic cubic, then,

$$x_7 = (a^3b)^{1/7}+(b^3c)^{1/7}+(c^3a)^{1/7}$$ $$x_7^{'} = (a^3c)^{1/7}+(b^3a)^{1/7}+(c^3b)^{1/7}$$

properly chosen, are the roots of,

\begin{align} x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= 28x - (n^2 + 5n + 9)\\ x^7 - 14x^4 + 7n x^3 - 14(n - 3)x^2 &= -7(n^2 - 3n + 5)x + (n^3 - 4n^2 + 4n - 9) \end{align}

The method generates pairs of solvable equations that curiously differ only by 2 coefficients for $p=7$, by 4 coefficients for $p=13$, etc.

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II. Level 13

Using the same $a,b,c$, then,

$$x_{13} = (a^4b)^{1/13}+(b^4c)^{1/13}+(c^4a)^{1/13}$$ $$x_{13}^{'} = (a^4c)^{1/13}+(b^4a)^{1/13}+(c^4b)^{1/13}$$

are the roots of,

$$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = A$$ $$x^{13} + 26x^{10} - 13n x^8 + 143x^7 - 65(3 - n)x^6 - 65n x^5 - 13 (39 - 13 n - 2 n^2) x^3 = B$$

where,

$$A = -442x^4 + 13 (9 + 15 n + n^2) x^2 - 13 (36 - 3 n + n^2) x + (189 - 72 n + 15 n^2 - n^3)$$ $$B = 26 (1 - 6 n + 2 n^2) x^4 + 13 (9 + 6 n + 4 n^2 - n^3) x^2 + 13 (-3 + 2 n)^2 x + (18 + n^2) (12 - 5 n + n^2)$$

hence differs only by 4 coefficients.

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III. Level 19

Still using the same $a,b,c$, then,

$$x_{19} = (a^5b^2)^{1/19}+(b^5c^2)^{1/19}+(c^5a^2)^{1/19}$$ $$x_{19}^{'} = (a^5c^2)^{1/19}+(b^5a^2)^{1/19}+(c^5b^2)^{1/19}$$

are the roots of two $19$-deg equations that differ only by several coefficients. (A parametric example can be given, but will clutter up the post.)

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IV. Levels 31 and 37

$$x_{31} = (a^6b)^{1/31}+(b^6c)^{1/31}+(c^6a)^{1/31}$$

$$x_{37} = (a^7b^3)^{1/37}+(b^7c^3)^{1/37}+(c^7a^3)^{1/37}$$

and so on.

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V. Questions

1. Why, starting with the Klein quartic and the roots of the generic cubic, does the method work?
2. By trial and error, I've found the surfaces for $p=43,61,67,73$. If the method works, can we predict these in advance from first principles?

Answer 1

The following can explain the shape of the expression and also why you always get pairs (why the minimal polynomials for the two elements in a pair differ by few coefficients, I don't know).

Let $p$ be a prime congruent to $1$ mod $6$, and consider the rational function $f(X):=g(X^p)$, where $g(X):=\frac{X^3-3X+1}{X-X^2}$ (note that setting $t:=g(X)$ and clearing denominators gives exactly your generic cubic, except that I like $t$ better than $n$ as a parameter). Then $X\mapsto f(X)$ yields a cover $\mathbb{P}^1\to \mathbb{P}^1$, and we can try to compute its monodromy group (i.e., the Galois group of $f(X)-t=0$ over $\mathbb{C}(t)$ - and, as a matter of fact, still over $\mathbb{Q}(\zeta_p)(t)$). Since $Gal(g(X)-t/\mathbb{C}(t)) = C_3$ and $Gal(X^p-t/\mathbb{C}(t))=C_p$, this monodromy group will be a subgroup of $C_p\wr C_3 (=(C_p\times C_p\times C_p)\rtimes C_3) \le S_{3p}$. We can be more explicit: IF $a,b,c$ denote the three roots of $g(X)-t=0$, and $\alpha:=a^{1/p}$, $\beta:=b^{1/p}$, $\gamma:=c^{1/p}$, then the roots of $f(X)-t=0$ fall into three blocks $\Delta_1, \Delta_2, \Delta_3$, with $\Delta_1:=\{\alpha, \zeta\alpha, \dots, \zeta^{p-1}\alpha\}$ (and analogously for $\Delta_2,\Delta_3$).

The inertia groups at the branch points of the cover $X\mapsto f(X)$ will now give more precise information about the Galois group. It's easy to see that $X\mapsto g(X)$ has exactly two (non-rational) branch points (namely the roots of your discriminant $D$), whereas $X\mapsto X^p$ has of course exactly the branch points $0$ and $\infty$, and these two get mapped to the same point $\infty$ under $X\mapsto g(X)$. So in total, the composition has exactly three branch points: the two irrational ones (whose inertia group generators are of order $3$ (permuting $a$, $b$, $c$ cyclically and thus permuting the three blocks $\Delta_i$ in the same way), and $t=\infty$ at which the inertia group is a double-$p$-cycle inside the block kernel (since unramified under $X\mapsto g(X)$, but with two totally ramified preimages on the "upper" level). The permutation group generated by these elements can then be identified as $G:=K\rtimes C_3$, where $K:=\{(x_1,x_2,x_3)\mid x_1,x_2,x_3\in \mathbb{F}_p; \sum x_i=0\}$ is the ``augmentation ideal". This group $G$ has two normal subgroups $N_1$ and $N_2$ isomorphic to $C_p$, both with quotient group $G/N_i\cong C_p\rtimes C_3$: namely, let $y_1,y_2$ be the two primitive third roots of unity in $\mathbb{F}_p$ and let $N_i:=\{(x,y_i x, y_i^2 x): x \in \mathbb{F}_p\}\trianglelefteq K$. We want to find an expression in $\alpha,\beta,\gamma$ which is fixed by $N_i$, and claim that this can be done in the form $\alpha^j\beta^k$ for $j,k \in \{1,\dots, p-1\}$ suitable, so that the minimal polynomial of $\alpha^j\beta^k (=a^jb^k)^{1/p}$ over $\mathbb{C}(t)$ will have Galois group $G/N_i\cong C_p\rtimes C_3$ (the remaining symmetrization $\alpha^j\beta^k + \beta^j\gamma^k + \gamma^j\alpha^k$ the only drops the degree by a factor $3$, reaching a degree-$p$ polynomial, but still with the same splitting field; the existence of such a polynomial is clear anyway, since $C_p\rtimes C_3 \le S_p$, so from the theoretical point of view this last step is not so important anymore). So let $\sigma_i:=(1,y_i,y_i^2)\in N_i$ be a generator of $N_i$. This acts quite explicitly on the three blocks $\Delta_1,\Delta_2,\Delta_3$ by $\sigma_i(\alpha) = \zeta_p \alpha$, $\sigma_i(\beta) = \zeta_p^{y_i}\beta$, $\sigma_i(\gamma) = \zeta_p^{y_i^2}\gamma$.

Hence, $\sigma_i(\alpha^j\beta^k) = \zeta_p^{j+k\cdot y_i} \alpha^j\beta_k$, and we're laughing if we can get $j+k\cdot y_i$ divisible by $p$. There's obviously some freedom here in choosing $j$. E.g., for $p=7$, the third roots of unity in $\mathbb{F}_7$ are $y_1=2$ and $y_2=4$; now if you choose $j=3$ with $y_2=4$, you get $k=1$, which after the symmetrization yields your first solution $x_7$; whereas $j=1$ with $y_1=2$ gives $k=3$, and that, after symmetrization is your second solution $x_7'$.

Just to check one more case, for $p=37$, the third roots of unity are $10$ and $26$. Choosing $j=7$ with $y_1=10$ gives $k=3$. You could of course choose other exponents $j$ (whether choosing to increase $j$ by $1$ for each new prime is actually a natural choice, I don't know; it happens to give small $k$ for the first few terms, but if I made no mistake, choosing $j=8$ for $p=43$ should give you $k=13$; instead always choosing $j=1$ would give the rather systematically-looking $k=y_i+1$!), although these would of course not generate new field extensions.

Answer 2

(Not an answer but an addendum that addresses three issues.)

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Issue 1

While a detailed answer has explained the method works for the generic cubic, numeric testing suggests any cubic $x^3+\alpha x^2+\beta x+1=0$ will work as long as its discriminant $D$ is a negative square. Is this true?

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Issue 2

To find $j,k$ with $j>k$ of the appropriate "surface" $a^jb^k+b^jc^k+c^ja^k$ for a level $p$, what I did was actually solve the system,

$$[a^jb^k,\;b^jc^k,\;c^ja^k] = [u_1,\; u_2,\; u_3]$$

for the "seed" cubic roots $a,b,c$ (easily done in Mathematica). For example, for $p=37$ and using $j=7$, only for $k = 3$, we get,

$$a = \left(\frac{u_1^{49}\,u_3^9}{u_2^{21}}\right)^{1/370}$$

So I figure if the fractional power contains a multiple of $p=37$, then this is the correct $j,k$. It seems then,

$$a^7b^3+b^7c^3+c^7a^3\quad\text{(deg 37)}$$ $$\; a^7b+b^7c+c^7a\quad\quad \text{(deg 43)}$$ $$a^9b^4+b^9c^4+c^9a^4\;\quad \text{(deg 61)}$$ $$a^9b^2+b^9c^2+c^9a^2\quad \text{(deg 67)}$$ $$a^9b+b^9c+c^9a\quad\quad \text{(deg 73)}$$

Using $j=8$ with $k<j$ just yields multiples of lower primes.

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Issue 3

For p = 7, constructing the two septics was easy. The Mathematica command Table[Mod[k^2, 7], {k, 1, 3}] gives ($1,4,2$) so with the roots $u_i$ of the first cubic,

$$P(x) = \prod_{k=0}^6 \Big(x-\big(\zeta_7^{k}\,u_1^{1/7}+\zeta_7^{4k}\,u_2^{1/7}+\zeta_7^{2k}\,u_3^{1/7}\big)\Big)$$

yields the first septic, while the roots $v_i$ of the second cubic is for the partner septic.

For p = 13, it gives ($1, \color{blue}4, 9, 3, \color{blue}{12}, \color{blue}{10}$) so one has to find the correct triplets. By trial-and-error,

$$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{k}\,u_1^{1/13}+\zeta_{13}^{9k}\,u_2^{1/13}+\zeta_{13}^{3k}\,u_3^{1/13}\big)\Big)$$ $$P(x) = \prod_{k=0}^{12} \Big(x-\big(\zeta_{13}^{\color{blue}{4k}}\,u_1^{1/13}+\zeta_{13}^{\color{blue}{10k}}\,u_2^{1/13}+\zeta_{13}^{\color{blue}{12k}}\,u_3^{1/13}\big)\Big)$$

so now there are two ways to generate the same 13-deg. (Using $v_i$ gives the partner 13-deg.)

For p = 19, it gives ($1, \color{blue}4, \color{blue}9, \color{red}{16}, \color{blue}6, \color{red}{17}, 11, 7, \color{red}5$) so there are three ways to generate the same 19-deg. And so on. Is there a way to find the correct triplets without trial-and-error?