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.

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