Entry 192

From Entry 191,

$$G_{11} = 2^{-1/4}x, \quad x^3 - 2x^2 + 2 x - 2=0\qquad$$

so \(x\) is the real root of a cubic \(x^3-2ax^2+2bx-2=0\) where \(a=b=1\). The discriminant \(d=11\) has class number \(n=h(-d)=1\). For \(G_d\), one can observe that if \(d\equiv 3\,\text{mod}\,8\), then \((a,b)\) are algebraic numbers at most of degree \(n\). Thus if \(n=3\), then \((a,b)\) are roots of cubics. 

There are 16 fundamental discriminants \(d\) with class number \(n=3\) and the largest is \(d=907\). The smallest two \(d = 23,31\) have different form \(d\equiv 7\,\text{mod}\,8\), and will be discussed in another entry while the rest are \(d\equiv 3\,\text{mod}\,8\),

$$\begin{align}G_{11} &= 2^{-1/4}x, \quad x^3 - 2rx^2 + 2 x - 2 = 0,\quad r = 1\\ G_{331} &= 2^{-1/4}x, \quad x^3 - 2rx^2 + 2 x - 2 = 0,\quad r^3-7r^2+9r-4=0\end{align}$$

and

$$\begin{align}G_{43} &= 2^{-1/4}x, \quad x^3 - 2x^2 + 2r x - 2 = 0,\quad r = 0\\ G_{83} &= 2^{-1/4}x, \quad x^3 - 2x^2 + 2r x - 2 = 0,\quad r^3-r^2-3r+4=0\end{align}$$

and

$$\begin{align}G_{67} &= 2^{-1/4}x, \quad x^3 - 2r x^2 - 2r x - 2 = 0,\quad r = 1\\ G_{211} &= 2^{-1/4}x, \quad x^3 - 2rx^2 - 2r x - 2 = 0,\quad r^3-3r^2+r-2=0\\ G_{283} &= 2^{-1/4}x, \quad x^3 - 2rx^2 - 2r x - 2 = 0,\quad r^3-4r^2-1=0\end{align}$$

and

$$\begin{align}\quad G_{163} &= 2^{-1/4}x, \quad x^3 - 2r x^2 +4 x - 2 = 0,\quad r = 3\\ G_{907} &= 2^{-1/4}x, \quad x^3 - 2rx^2 +4 x - 2 = 0,\quad r^3-29r^2+85r-66=0\end{align}$$

P.S. These are the simplest cubic "templates" and it seems interesting the largest discriminants for class numbers \(1\) and \(3\) have a \(G_d\) that share the same template. For class number \(5\), they get more complicated.