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
- 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?
- 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?
Best Answer
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.