Entry 148

This is the 7th-deg overview though only results by Klein for the j-function \(j = j_1(\tau)\) are known. In Klein's "On the Order-Seven Transformations of Elliptic Functions", he gave two elegant resolvents of degrees 7 and 8 in pages 306 and 313. Translated to more understandable notation, we have,

$$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$$

$$y^8+14y^6+63y^4+70y^2-7 = y\sqrt{j-1728}$$

If \(\tau\) are complex quadratics such that \(j= j_1(\tau)\) is a radical, then the two resolvents have a solvable Galois group, hence solvable in radicals

Example. Let \(\tau = \tfrac{1+\sqrt{-163}}2\), then \(j = -640320^3\) and $$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3  = -640320^3$$ is solvable in radicals. There are infinitely many such \(\tau\) and some can be found in Entry 145. Note also that $$\frac{(y^4 + 14y^3 + 63y^2 + 70y - 7)^2}y + 1728 = \frac{(y^2 + 5y + 1)^3 (y^2 + 13y + 49)}y$$ where the octic on the RHS will appear in the next entry.