Computing the Class Number H(-11)

# Computing the Class Number H(-11)

 Theorem 1: The class number $H(-11) = 1$.
• Proof: Consider the binary quadratic form:
(1)
\begin{align} \quad f(x, y) = x^2 + xy + 3y^2 \end{align}
• The discriminant is $d = 1^2 - 4(1)(3) = -11$. Observe that $f(x, y)$ is a reduced binary quadratic form since $-a < b \leq a < c$, that is:
(2)
\begin{align} \quad -1 < 1 \leq 1 < 3 \end{align}
• To show that $H(-11) = 1$ we want to show that the only reduced binary quadratic form with discriminant $-11$ is $f(x, y)$. Observe that:
(3)
\begin{align} a < \sqrt{\frac{-d}{3}} =\sqrt{\frac{11}{3}} = \sqrt{3 + \frac{1}{3}} < 2 \end{align}
• Therefore $a = 0$ or $a = 1$. If $a = 0$ then $d = b^2 -4ac$ implies that $-11 = b^2$ which is a contradiction. So $a = 1$. Hence $-1 < b \leq 1 < c$ implies that $b = 0$ or $b = 1$. If $b = 0$ then $d = b^2 - 4ac$ implies that $-11 = -4ac$ which is a contradiction. So $b = 1$. Thus $-11 = b^2 - 4ac$ implies that $c = 3$.
• So $f(x, y) = x^2 + xy + 3y^2$ is the only reduced binary quadratic form with discriminant $-11$. So every binary quadratic form with discriminant $-11$ is equivalent to $f(x, y)$, i.e., $H(-11) = 1$. $\blacksquare$