Algebraic number theory M 3/4 P 15

Answer Sheet 2

This coursework turned out to be more difficult than I intended it to be. I apologize for this.

1. 1.5 mark each. (a) 2, 5; (b) 7; (c) 2; (d) 2, 11.

2. 3 marks. We have seen that 2 and 5 are ramified, so we do not consider these primes. By a result from the course a prime p is split in Q(\sqrt{10}) iff 1 = (\frac{10}{p}) = (\frac{2}{p})(\frac{5}{p}) = (−1)^{(p^{2}−1)/8}(\frac{p}{5}). (We have (\frac{5}{p}) = (\frac{p}{5}) by the quadratic reciprocity law since 5 is congruent to 1 modulo 4.) It is clear that (\frac{p}{5}) = 1 when p is congruent to ±1 modulo 5, and -1 if not. Hence p is split iff p is congruent to ±1, ±3, ±9, ±13 modulo 40. [Note that there are precisely 16 invertible residues modulo 40: |(Z/40)^{∗}| = ϕ(40) = ϕ(8)ϕ(5) = 4.4 = 16. Isn’t it remarkable that precisely half of them correspond to split cases and another half to inert ones? Can you explain it?]

3. (a) 3 marks. We note that p is odd and coprime with d, so that p can only be split or inert. If p is a norm, then p = z.\overline{z}, for some z ∈ A(d). The ideal I = (z) is prime^{1}. The ideal I contains p, hence is lying over p. Since the norm of z is p, and the norm of p is p^{2}, we have z ∉ (p). This means that there is a prime ideal I over p, which is distinct from (p). Therefore p is split. This is known to be equivalent to (\frac{d}{p}) = 1.

A more direct proof is this. We have p = N(z), for some z ∈ A(d). Then 2z can be written as a+\sqrt{d}b with integer a and b. Then 4p = a^{2} −db^{2}. Since p ≠ 2 and (p, d) = 1, it follows that a and b are not divisible by p. Reducing modulo p we conclude that d is a square modulo p.

THE CONVERSE IS TRUE PROVIDED d < 0. I APOLOGIZE FOR FORGETTING TO PUT THIS CONDITION. Indeed, if p is split, then (p) = (p, a + \sqrt{d})(p, a − \sqrt{d}). Let z be a generator of the first of these ideals, then it is easy to see that \overline{z} generates the second one. Hence (p) = (N(z)) as ideals in A(d). This implies that p = u.N(z), where u ∈ A(d)^{∗} is a unit. Since p and N(z) are in Z, u is 1 or −1. It is here that we need the condition that d is negative: then N(z) > 0, hence u = 1, and p is a norm. [The statement is not true in general for d > 1: take d = 3, then A(3) is known to be a Principal Ideal Domain, take p = 2, then −2 = 1^{2} − 3.1^{2} = N(1 + \sqrt{3}). If 2 = N(z) for some z ∈ Q(\sqrt{3}), then (1 + \sqrt{3})/z has norm -1. Let us show that it is not possible. Suppose that −1 is the norm of \frac{a}{c} + \frac{b}{c} \sqrt{3}, where a, b, c are not divisible by a common prime. We then have a^{2} − 3b^{2} + c^{2} = 0. Looking at this modulo 3, and using the fact that −1 is not a square modulo 3, we conclude that 3|a and 3|c. But then 3|b. Contradiction.]

1Indeed, z, as an element with prime norm, is irreducible. In a Euclidean ring the factorization into irreducibles is unique. Hence if ab = zt, then either a or b is divisible by z, hence I is prime. Actually this argument shows that in a Unique Factorization Domain every principal ideal generated by an irreducible is prime.

(b) 1 mark. This is clear from the standard formula for (\frac{−1}{p}).

(c) 2 marks. It is clear that a^{2} + 2b^{2} = N(a + b\sqrt{−2}). We compute (\frac{−2}{p}) = (\frac{−1}{p})(\frac{2}{p}) using the standard formula (\frac{2}{p}) = (−1)^{(p^{2}−1)/8}.

(d) 2 marks. Observe that a^{2} + ab + b^{2} = N(a + \frac{1+\sqrt{−3}}{2}b). Then (\frac{−3}{p}) = (\frac{−1}{p})(\frac{3}{p}). By the quadratic reciprocity law we have (\frac{3}{p}) = (−1)^{(p−1)/2}(\frac{p}{3}), hence the answer.

(e) 3 marks. Call a prime p “good” for A(d), d = −1, −2, −3, if it is a norm of an element of A(d), and “bad” otherwise. It is clear that any positive integer divisible exactly by an even power of all its bad prime divisors is a norm. Let us show that this condition is also necessary. Let n = a^{2} + 2b^{2}, and let p be a bad prime for d = −2, such that p divides n to an odd power. Then we can assume that a and b are coprime with p (if one is divisible by p, then also the other; now p^{2}|n, and we divide both sides by p^{2}). Reducing modulo p we conclude that −2 is a square modulo p. Contradiction. The cases of −1 and −3 are analogous.