Solving solvable septics using only cubics?

Question

After the satisfying resolution of my question on the Kondo-Brumer quintic, I decided to revisit my old post on septic equations.

I. Solution by eta quotients

The septic mentioned in that post may not look much,

$$h^2 -7^3h\,(x+5x^2+7x^3)\\ -7^4\,(x + 7 x^2 + 21 x^3 + 49 x^4 + 147 x^5 + 343 x^6 + 343 x^7) =0 $$

but has some surprises. It is solvable in radicals for any $h$, but also by eta quotients,

$$h = \left(\frac{\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4,\quad x=\left(\frac{\eta(49\tau)}{\eta(\tau)}\right)$$

II. Solution by radicals

If we do a change of variables $x = (y-1)/7$ and $h = -n-8$, we get a much simpler form,

$$y^7 + 14y^4 - 7n y^3 - 14(3 + n)y^2 - 28y - (n^2 - 5n + 9) = 0$$

Surprisingly, its solution needs only a cubic Lagrange resolvent,

$$y = u_1^{1/7} + u_2^{1/7} + u_3^{1/7}$$

so the $u_i$ are the three real roots of,

$$u^3 - (n^2 + 2n + 9)u^2 + (n^3 + 5n^2 + 14n + 15)u + 1 = 0$$

which has negative discriminant $d = -(n^2 + 3n + 9)^2 (n^3 + 2n^2 - 8)^2$ so always has three real roots.

III. Tschirnhausen transformation

While browsing the book "Generic Polynomials" (thanks, Rouse!), in page 30 I saw the generic cubic for $C_3 = A_3$,

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

Suspecting it was connected to the cubic I found, I verified they were indeed related by a quadratic Tshirnhausen transformation,

$$u = 2 v^2 + (n + 2) v - 1$$

Note that the discriminant of the septic (in $y$), resolvent cubic, and generic cubic have the common square factor $(n^2+3n+9)^2$.

IV. Questions

1. In general, a solvable septic has a sextic Langrange resolvent. So what are the Galois conditions such that this is reduced to a a cubic resolvent?
2. Would any parametric septic solvable just by a cubic resolvent share a common square factor with the generic polynomial for $C_3 = A_3$? Or is the one involved in $\frac{\eta(\tau)}{\eta(7\tau)}$ a "special" case?

Answer 1

Regarding question 1), of course the obvious (sufficient) answer is "When the Galois group is contained in $C_7\rtimes C_3$". That's not quite the case here, but "almost". To be precise, your septic has discriminant $-7\cdot f(h)^2$ (for a suitable polynomial $f(h)$, so the quadratic subextension of the splitting field over $\mathbb{Q}(h)$ is $\mathbb{Q}(h)(\sqrt{-7})\subset\mathbb{Q}(h)(\zeta_7)$. Making the expression $y=u_1^{1/7}+u_2^{1/7}+u_3^{1/7}$ well-defined requires picking the correct 7-th roots inside the splitting fields $\mathbb{Q}(\sqrt[7]{u_i}, \zeta_7)$, so I guess this is where the above quadratic subextension gets eaten up.

Answer 2

I finally figured out part of my second question, on whether this septic was a special case. The answer, perhaps not surprisingly, depends on Ramanujan's work. Given the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

Define the following theta quotients,

\begin{align} r_1 &= \frac{1}{q^{2/7}}\frac{f(-q^2,-q^5)}{f(-q,-q^6)} = \frac{1}{q^{2/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}\\ r_2 &= \frac{-1\;}{q^{1/7}}\frac{f(-q^3,-q^4)}{f(-q^2,-q^5)} \;=\; \frac{-1\;}{q^{1/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}\\ r_3 &= \frac{1}{q^{-3/7}}\frac{f(-q,-q^6)}{f(-q^3,-q^7)} = \frac{1}{q^{-3/7}}\prod_{n=1}^{\infty}\frac{(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})} \end{align}

Then the cubic formed by their $7$th powers

$$P(u) = (u-r_1^7)(u-r_2^7)(u-r_3^7) = 0$$

has coefficients in the Dedekind eta quotient $ m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,$

$$P(u) = u^3- (57 + 14 m + m^2) u^2-(289 + 126 m + 19 m^2 + m^3) u +1 =0$$

(This cubic in fact was also found by Ramanujan.) While the septic formed by the expression,

$$P(y)=\prod_{k=0}^6 \Big(y - (\zeta^k r_1 + \zeta^{4k}r_2 + \zeta^{2k}r_3)\Big) = 0$$

with $\zeta = e^{2\pi i/7}$ also has coefficients in $m$,

$$P(y) = y^7 + 14y^4 + 7 (8 + m) y^3 + 14 (5 + m) y^2 - 28y - (113 + 21 m + m^2) = 0$$

Of course, a root of $P(y)$ is then,

$$y = r_1 + r_2 + r_3 = u_1^{1/7} + u_2^{1/7} + u_3^{1/7}$$

and a minor change of variable $m \to -(n+8)$ will recover the cubic and septic in my question.

P.S. While I now know how to construct the septic from first principles, I do not fully understand why its solvability "carries over" to any $m$ (not just the original eta quotient), nor why $P(y)$ now relates eta quotients in two ways,

$$\text{If}\; m = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,\; \text{then}\; y = \left(\frac{\eta(\tau/7)}{\eta(7\tau)}\right)+1$$

$$\text{If}\; m = \left(\frac{\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4,\; \text{then}\; y = \left(\frac{7\eta(49\tau)}{\eta(\tau)}\right)+1$$

I just figure Nature is very economical with her polynomials.

Answer 3

Did you notice that the polynomial is essentially the same as the (unique) eta product identity for prime squared level for $p=7$? So there should be a connection, and equivalent ones for each prime (square).