## The class number of QQ(sqrt(-6)) is 2

Compute the class number of $\mathbb{Q}(\sqrt{-6})$.

Recall that the ring of integers in this field is $\mathcal{O} = \mathbb{Z}[\sqrt{-6}]$, and that $\{1,\sqrt{-6}\}$ is an integral basis. We will now give an upper bound on the constant $C$ which is proven to exist in Theorem 10.2.

Let $A$ be an ideal in $\mathcal{O}$, and let $t = \lfloor \sqrt{N(A)} \rfloor$. That is, $t$ is the largest integer such that $t^2 \leq N(A)$. Now consider $\{b_0 + b_1\sqrt{-6} \ |\ 0 \leq b_0,b_1 \leq t\}$. Since $(t+1)^2 < N(A)$, there exist some $\beta_1,\beta_2$ in this set such that $\beta_2 - \beta_1 \in A$. Say $\alpha = \beta_2 - \beta_1 = a_0+a_1\sqrt{-6}$, where (clearly) $|a_0|,|a_1| \leq t$. By the triangle inequality, $|N(\alpha)| = |a_0^2 + 6a_1^2| \leq |a_1|^2 + 6|a_1|^2 \leq t^2 + 6t^2 = 7t^2$ $\leq 7 \cdot N(A)$. By our proof of Theorem 10.3, every class of ideals in $\mathcal{O}$ contains an ideal whose norm is at most 7. We will now construct all of the ideals having norm $k$ where $k \in \{1,2,3,4,5,6,7\}$.

$(k=1)$ There is only one ideal of norm 1, namely $A_1 = (1)$.

$(k=2)$ If $A$ is an ideal of norm 2, then $A$ divides $(2)$. Note that $(2,\sqrt{-6})^2 = (4,2\sqrt{-6},-6)$ $= (2)$. Now $4 = N((2)) = N((2,\sqrt{-6}))^2$, so that $N((2,\sqrt{-6})) = 2$. By Corollary 9.15, $(2,\sqrt{-6})$ is prime. Thus $A_2 = (2,\sqrt{-6})$ is the only ideal of norm 2.

$(k=3)$ If $A$ is an ideal of norm 3, then $A$ divides $(3)$. Note that $(3,\sqrt{-6})^2 = (9,3\sqrt{-6},-6) = (3)$. Now $9 = N((3)) = N((3,\sqrt{-6}))^2$, so that $N((3,\sqrt{-6})) = 3$. By Corollary 9.15, $(3,\sqrt{-6})$ is prime. Thus $A_3 = (3,\sqrt{-6})$ is the only ideal of norm 3.

$(k=4)$ Suppose $A$ is an ideal of norm 4. If $A$ is not prime, then it is a product of two prime ideals of norm 2; the only such ideal is $(2)$. If $A$ is prime, then by Theorem 9.19, $A$ divides $(2)$. But no prime ideals dividing $(2)$ have norm 4. So the only ideal of norm 4 is $A_4 = (2)$.

$(k=5)$ If $A$ is an ideal of norm 5, then $A$ divides $(5)$. Note that $(5) = (5,2+\sqrt{-6})(5,2-\sqrt{-6})$, since $5 = 5^2 - 2(2+\sqrt{-6})(2-\sqrt{-6})$. We claim that both of these factors are proper. Indeed, if $1 \in (5,2+\sqrt{-6})$, then $1 = 5(a_b\sqrt{-6}) + (2+\sqrt{-6})(h+k\sqrt{-6})$ for some $a,b,h,k \in \mathbb{Z}$. Comparing coefficients mod 5, we have $1 \equiv 0$, a contradiction. Thus $(5,2+\sqrt{-6})$ is proper. Similarly, $(5,2-\sqrt{-6})$ is proper. Now $25 = N((5)) = N((5,2+\sqrt{-6}))N((5,2-\sqrt{-5}))$, and neither factor is 1. So $N((5,2+\sqrt{-6})) = N((5,2-\sqrt{-6})) = 5$. Thus $A_5 = (5,2+\sqrt{-6})$ and $A_6 = (5,2-\sqrt{-6})$ are the only possible ideals of norm 5 in $\mathcal{O}$. Note that the discriminant of this field is not divisible by 5, so that these factors are distinct by Theorem 9.6. (Though this last step is not necessary.)

$(k=6)$ If $A$ is an ideal of norm 6, then $A$ divides $(6) = (2)(3)$. Thus the prime factors of $A$ come from the set $\{(2,\sqrt{-6}), (3,\sqrt{-6})\}$. The only such ideal is $A_7 = (\sqrt{-6})$.

$(k=7)$ If $A$ is an ideal of norm 7, then $A$ divides $(7)$. Note that $(7) = (1+\sqrt{-6})(1+\sqrt{-7})$. Moreover, by Corollary 9.11, $N((1+\sqrt{-6})) = N((1-\sqrt{-6})) = 7$, and by Corollary 9.15, these ideals are prime. So the only ideals of norm 7 are $A_8 = (1+\sqrt{-6})$ and $A_9 = (1-\sqrt{-6})$.

Certainly we have $A_1 \sim A_4 \sim A_7 \sim A_8 \sim A_9 \sim (1)$.

Note that $A_2A_3 = A_7 \sim (1)$ and that $A_2 = A_4 \sim (1)$. Thus $A_2 \sim A_3$.

We also have $A_5A_6 = (5)$ and $A_5^2 = (1-2\sqrt{-6})$, since $1-2\sqrt{-6} = 25 + 2(2+\sqrt{-6})^2 - 2 \cdot 5 \cdot (2+\sqrt{-6})$. So $A_5A_6 \sim A_5^2$, and thus $A_5 \sim A_6$.

We also that $A_5A_2 = (2+\sqrt{-6})$ since $5 \cdot 2 + 5 \cdot \sqrt{-6} - 2(2)(2+\sqrt{-6}) = 2+\sqrt{-6})$. So $A_5A_2 \sim A_5^2$, and thus $A_5 \sim A_2$. So $A_2 \sim A_3 \sim A_5 \sim A_6$.

Finally, we claim that $A_2$ is not principal. If $(2,\sqrt{-6}) = (\alpha)$, then $N(\alpha)$ divides both 4 and 6, and so is either 1 or 2. But no element of $\mathbb{Z}[\sqrt{-6}]$ has norm 2, and $A_2$ is proper; so $A_2$ is not principal, and $A_2 \not\sim (1)$.

Thus $\mathbb{Q}(\sqrt{-6})$ has two ideal classes.