## The class number of QQ(sqrt(3)) is 1

Compute the class number of $K = \mathbb{Q}(\sqrt{3})$.

Recall that $\mathbb{Z}[\sqrt{3}] = \mathcal{O}$ is the ring of integers in $K$, and that $\{1,\sqrt{3}\}$ is an integral basis for $K$.

We will now compute an upper bound on the constant $C$ which is shown to exist in Theorem 10.2. That is, we will find a number $C$ such that, for every ideal $A$, there exists an element $\alpha \in A$ such that $|N(\alpha)| \leq C \cdot N(A)$.

To this end, let $t = \lfloor \sqrt{N(A)} \rfloor$– that is, $t$ is the largest integer such that $t^2 \leq N(A)$. Consider the set of integers $\{b_0 + b_1\sqrt{3} \ |\ 0 \leq b_0,b_1 \leq t \}$. Since $(t+1)^2 > N(A)$ by our choice of $t$, there must exist two elements $\beta_1,\beta_2$ in this set which are congruent mod $A$. Say $\alpha = \beta_2 - \beta_1 = a_0 + a_1\sqrt{3}$.

Using the triangle inequality, we have $|N(\alpha)| = |a_0^2 - 3a_1^2| \leq |a_0|^2 + 3|a_1|^2 \leq t^2 + 3t^2 = 4t^2$ $\leq 4 \cdot N(A)$. Thus the constant $C$ is at most 4.

By our proof of Theorem 10.3, every class of ideals in $\mathcal{O}$ contains an ideal of norm at most 4. We will now compute all of the ideals of norm $k$ for $k \in \{1,2,3,4\}$.

$(k=1)$ There is only one ideal of norm 1; namely $D_1 = (1)$.

$(k=2)$ Suppose $A$ is an ideal of norm 2. By Theorem 9.19, $A$ is a divisor of $(2)$. We claim that $(2) = (2,1+\sqrt{3})^2$; indeed, $(2,1+\sqrt{3})^2 = (4,2+2\sqrt{3},-2) = (2)$. Now $4 = N((2)) = N((2,1+\sqrt{3}))^2$, so that $N((2,1+\sqrt{3})) = 2$. By Theorem 9.15, the ideal $(2,1+\sqrt{3})$ is prime, and so $(2) = (2,1+\sqrt{3})^2$ is the prime factorization of $(2)$. So $D_2 = (2,1+\sqrt{3})$ is the only ideal of norm 2 in $\mathcal{O}$.

$(k=3)$ Suppose $A$ is an ideal of norm 3. By Theorem 9.19, $A$ is a divisor of $(3)$. Note that $(3) = (\sqrt{3})^2$, and that $N((\sqrt{3})) = N(\sqrt{3}) = 3$. Thus $(3) = (\sqrt{3})^2$ is the prime factorization of $(3)$, and $D_3 = (\sqrt{3})$ is the only ideal of norm 3 in $\mathcal{O}$.

$(k=4)$ 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 prime ideal of norm 2 is $D_2$, so that $A = D_2^2 = (2)$. If $A$ is prime, then by Theorem 9.19, $A \supseteq (2)$. Since $(2)$ has norm 4, in fact $A = (2)$, which in this case is a contradiction. So $D_4 = (2)$ is the only ideal of norm 4.

It is clear that $D_1 \sim D_3 \sim D_4$. Moreover, note that $(1-\sqrt{3})D_2 = (2-2\sqrt{3}, -2) = (2) = (1)(2)$, so that $D_2 \sim D_4$.

Thus the class number of $\mathbb{Q}(\sqrt{3})$ is 1.