Algebraic number theory M 3/4 P 15

Answer Sheet 4

1. 3 marks. We have λ(5) = \sqrt{5}/2 < 2. Hence every ideal class is represented by an integral ideal of norm 1, that is, A(−5) itself. Thus the class group is trivial.

4 marks. We have λ(6) = \sqrt{6} < 3, hence we must consider only ideals I ⊂ A(6) of norm 2. These are necessarily prime ideals lying over 2. But 2 is ramified, hence I = (2, \sqrt{6}). If it is principal, then the norm of the generator must be ±2. (Since ||zA(d)|| = |N(z)|.) Indeed, a^{2} −6b^{2} = −2 has a solution a = 2, b = 1. One checks immediately that we have (2, \sqrt{6}) = (2 + \sqrt{6}). Hence all the ideal classes of the class group C(6) are represented by principal ideals, thus C(6) is trivial.

2. We have λ(−163) < 9, hence we must look at the integer ideals whose prime factors are lying over 2, 3, 5 and 7. By computing the Legendre symbols one observes that 3, 5 and 7 are all inert in Q(\sqrt{−163}). Since −163 is congruent to 5 modulo 8, by the result from the lectures we know that 2 is also inert! Hence all prime ideals lying over 2, 3, 5 and 7 are principal, hence also every ideal of norm less than 9 is principal. The class number is 1. [It is a difficult result that if d < −163 and d is squarefree, then the class number is greater than 1.]

3. Let d = −p_{1} … p_{n}, where p_{1}, …, p_{n} are different prime numbers, n > 2. To fix ideas assume that 2|p. Consider the ideal J_{i} = (p_{i}, \sqrt{d}). This is a prime ideal lying over p_{i}, which is ramified in Q(\sqrt{d}). Hence J_{i}^{2} = p_{i}A(d), and J_{i}^{−1} = p_{i}^{−1}(p_{i}, \sqrt{d}). Thus we have J_{i}J_{k}^{−1} = p_{k}^{−1}J_{i}J_{k} = p_{k}^{−1}(p_{i}p_{k}, p_{i}\sqrt{d}, p_{k}\sqrt{d}, d) = p_{k}^{−1}(p_{i}p_{k}, \sqrt{d}), because (p_{i}, p_{k}) = 1. I claim that the ideals J_{i} = (p_{i}, \sqrt{d}) and (p_{i}p_{j}, \sqrt{d}) for i ≠ j are not principal. Let us assume this for a moment. Then the classes [J_{i}] in the class group of Q(\sqrt{d}) are all non-trivial since J_{i} are not principal. These classes are also pairwise distinct since J_{i}J_{k}^{−1} are not principal. This produces n different non-trivial elements in the class group, hence the result.

It remains to show that (p_{i}, \sqrt{d}) and (p_{i}p_{j}, \sqrt{d}) are not principal. The norms of these ideals are p_{i} and p_{i}p_{j}, respectively. But there are no elements with such norms in A(d), since the equations x^{2} + |d|y^{2} = p_{i} and x^{2} + |d|y^{2} = p_{i}p_{j} have no integer solutions. Indeed, |d| > p_{i}p_{j}, hence y = 0, and now it is clear that there are no solutions (recall that p_{i} ≠ p_{j}). This completes the proof.

4. We have λ(−21) < 7, so we look into prime ideals over 2, 3 and 5. Fortunately, 5 is inert (−21 is a square modulo 5), hence the only ideal over 5 is principal. It remains to consider 2 and 3. Since 2 and 3 are ramified, the corresponding prime ideals are P = (2, \sqrt{−21} − 1) and Q = (3, \sqrt{−21}), respectively. Using the same method as before, we show that there are no elements in A(−21) with norm 2 or 3, hence P and Q are not principal. In other words, [P] ≠ 1, [Q] ≠ 1. It is clear that P^{2} = (2) and Q^{2} = (3), hence [P]^{2} = 1 and [Q]^{2} = 1. Next we study the product P Q. The norm of P Q is 6, and the same usual method shows that there are no elements in A(−21) with norm 6. Hence P Q is not principal. Now every fractional ideal can be expresed in terms of powers of P and Q, and these two generate the group isomorphic to (Z/2)^{2}. Therefore the class group is isomorphic to (Z/2)^{2}.