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.

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