Algebraic number theory M 3/4 P 15

Answer Sheet 3

1. 4 marks. Let S be our finite set. Then the desired quadratic field exists provided S is not empty! Indeed, if d ≠ −1, there is always a prime p|d, and this p must be ramified in Q(\sqrt{d}). On the other hand, Q(\sqrt{−1}) is ramified at 2.

Suppose S is not empty. Let d be the product of all the primes from S. Then K = Q(\sqrt{±d}), for an appropriate sign, is the quadratic field we are looking for. Indeed, if 2 ∈ S, then the sign is not relevant. If 2 ∉ S, we can choose the sign so that ±d is congruent to 1 modulo 4. Then 2 is unramified, hence K is ramified exactly at the primes from S.

2. 1 mark each. (a) I = (2+\sqrt{−5}) is principal, hence I^{−1} = \frac{1}{2+\sqrt{−5}}A(−5) =

\frac{2−\sqrt{−5}}{9}A(−5). (b) We have I^{2} = (4, −6, 2\sqrt{−6}) = (2), hence I^{−1} = \frac{1}{2}I = (1, \frac{1}{2} \sqrt{−6}).

(c) We have I.\overline{I} = (5, \sqrt{−6} − 2).(5, \sqrt{−6} + 2) = (5), hence I^{−1} = (1, \frac{1}{5}(\sqrt{−6} + 2)).

3. (a) 1 mark. We observe that 5 is split in Q(\sqrt{−11}), and then (5) = (5, \sqrt{−11} − 2)(5, \sqrt{−11} + 2) is the factorization of (5) into prime ideals.

(b) 2 marks. We have (6) = (2)(3). The theory tells us that 2 is ramified (since -5 is congruent to 3 modulo 4), and that 3 is split. Hence the desired factorization is (6) = (2, \sqrt{−5} + 1)^{2}(3, \sqrt{−5} − 1)(3, \sqrt{−5} + 1).

(c) 3 marks. We know how to factorize ideals generated by n ∈ Z, so let us try to reduce our problem to this case. We have (1+\sqrt{−7})(1−\sqrt{−7}) = (8) = (2)^{3}. Now −7 is congruent to 1 modulo 8, hence 2 is split. More precisely,

(2) factors into prime ideals as (2) = (2, δ)(2, 1 − δ), where δ = \frac{1}{2} + \frac{1}{2} \sqrt{−7}. Note by the way, that N(δ) = N(1 − δ) = 2. Hence (2, δ) = (δ) and (2, 1 − δ) = (1 − δ).

On the other hand, we remark that (1 + \sqrt{−7}) = (2δ) = (2)(δ). By the previous calculations we conclude that the desired factorization into prime ideals is (1 + \sqrt{−7}) = (δ)^{2}(1 − δ).

4. An odd prime divisor p|d is clearly ramified in A(d). The norm of (p, \sqrt{d}) is ||(p, \sqrt{d})|| = p since this is the only prime ideal lying over p.

Suppose (p, \sqrt{d}) is principal, say, (p, \sqrt{d}) = (z). Then 2z = a + b\sqrt{d} for some a and b in Z. We have a^{2} + |d|b^{2} = 4p. Since d is divisible by another prime apart from p, we have |d| ≥ 2p, hence b = 0 or b = 1 are the only possibilities. The first case is impossible because 4p is not a square. If b = 1, then a^{2} must equal p (when |d| = 3p) or 2p (when |d| = 2p), which is also impossible. Finally, |d| ≠ 4p since d is square-free. Therefore (p, \sqrt{d}) is not principal.

5. Let z ∈ A(d)^{∗} be a unit. As before we write 2z = a + b\sqrt{d} for some a and b in Z. We now show that the equation a^{2} −db^{2} = −4 has no solutions in Z. Let p|d be congruent to 3 modulo 4. Then −4 is a square modulo p. This is equivalent to −1 being a square modulo p, which for odd p is equivalent to the condition that p is congruent to 1 modulo 4. This is a contradiction.