Entry 147

This is the sextic overview. Using the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145, if $\tau$ are complex quadratics such that $j_n(\tau)$ is a radical, then the following sextics have a solvable Galois group, hence solvable in radicals

$$\begin{align}\qquad j_1 &=\frac{(x^2 + 10x + 5)^3}x\\ \color{red}{j_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\ \color{red}{{j_3}^2} &= \frac{j_3\,(2 x^6 + 29x^5 + 85x^4 + 50x^3) + 5^4 x^6}{(2x - 1)}\\ j_4 &=\frac{(x + 1)^5 (x + 5)}x \end{align}$$

with the ones in red by Joachim König. (It would be nice if the one for $j_3$ can be simplified.) They have discriminants,

$$\begin{align}D_1 &= 5^5\,(j_1-1728)^2\,{j_1}^4\\ D_2 &= 5^5\,(j_2-256)^3\,{j_2}^3 \\ D_3 &= 5^5(j_3-108)^3\,{j_3}^{11}\\ D_4 &= 5^5\,(j_4-64)^2\,{j_4}^4\qquad\end{align}$$

Examples. Let $\tau = \tfrac{1+\sqrt{-163}}2$ so $j_1(\tau) = -640320^3$. Let $\tau = \tfrac{\sqrt{-58}}2$ so $j_2(\tau) = 396^4$. Then

$$\frac{(x^2 + 10x + 5)^3}x = -640320^3\\ \frac{(x+1)^4(x^2+6x+25)}x=396^4$$ are sextics both solvable in radicals. There are infinitely many $\tau$ but special ones such that $j_n(\tau)$ are integers can be found in Entry 145.