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.
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.
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.
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.)
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.
V. Questions
- Why, starting with the Klein quartic and the roots of the generic cubic, does the method work?
- 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?
Best Answer
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.