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 roots of,

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

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?

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?