After the satisfying resolution of my [question](https://mathoverflow.net/questions/437322/) on the Kondo-Brumer quintic, I decided to revisit my [old post](https://mathoverflow.net/questions/155087/) 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"](http://library.msri.org/books/Book45/files/book45.pdf) (*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?



Solving solvable septics using only cubics?
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.


