Math 254A, UC Berkeley, Fall 2001 (Kedlaya) Problem Set 10: Cyclotomic Fields Due in class Wednesday, November 14

Please submit nine of the following problems.

Leftover Problems about Ramification/Splitting.

1. Prove that ℚ(\sqrt{−5}) has exactly one quadratic extension which is everywhere unramified, and find it. (This has something to do with class numbers; we’ll get back to this next term.)

2. Let f be a monic polynomial over ℤ of degree greater than 1. Suppose that f has a root modulo p for all primes p. Prove that f cannot be irreducible. (Hint: if G is a finite group and H is a proper subgroup, the union of the conjugates of H cannot be all of G.)

Cyclotomic Fields.

1. Prove that the sign of the discriminant of the number field K with signature (r, s) equals (−1)^{s}. (You’ll need this for the next problem.)

2. Calculate the discriminant of ℚ(ζ_{n}) for n arbitrary. (Hint: choose an integral basis by multiplying together integral bases of ℚ(ζ_{p^{m}}) for each prime power p^{m} dividing n. Then reduce to the computation we did in class.)

3. Neukirch exercise I.10.1: Prove that for every natural number n, there are infinitely many primes congruent to 1 modulo n, without invoking Dirichlet’s theorem on primes in arithmetic progressions. (Hint: let ϕ_{n} be the minimal polynomial of ζ_{n}. Prove that every prime divisor of ϕ_{n}(m), for m ∈ ℤ, is congruent to 1 modulo n.)

4. Neukirch exercise I.10.2: Prove that for every finite abelian group A, there exists a Galois extension L/ℚ with Galois group A.

Interlude on Fermat’s Last Theorem.

Throughout this section, let p ≥ 5 be a regular prime, that is, a prime that does not divide the class number of ℚ(ζ_{p}). Our goal is to prove part of Fermat’s Last Theorem for the exponent p. Namely, if x, y, z are rational integers, none divisible by p, and p is regular as assumed above, we will prove that x^{p} + y^{p} + z^{p} ≠ 0. As usual, we may assume x, y, z are pairwise coprime (if x^{p} + y^{p} + z^{p} = 0 and two of x, y, z have a common prime factor, so does the third).

1. Let x and y be relatively prime rational integers such that x, y and x^{p} + y^{p} are not divisible by p. Prove that the ideals (x + ζ^{i}y) are pairwise coprime for i = 0, …, p − 1.

2. Suppose x^{p} + y^{p} + z^{p} = 0 and that p is regular. Prove that for i = 0, …, p − 1, there exists a unit ϵ_{i} and an integer α_{i} ∈ ℤ[ζ_{p}] such that x + ζ^{i}y = ϵ_{i}α_{i}^{p}. (Hint: the assumption that p is regular means that the p-th power map is a bijection on the class group.)

3. Let λ = 1 − ζ. Show that Norm(λ) = p, so that every element of ℤ[ζ_{p}] is congruent to a rational integer modulo λ. Then show that if a ≡ b (mod λ) for a, b ∈ ℤ[ζ_{p}], then a^{p} ≡ b^{p} (mod p). (Hint: use the fact that as ideals, (λ)^{p−1} = (p).)

4. Suppose x + ζy = ϵα^{p} for some unit ϵ and some integer α in ℤ[ζ_{p}]. Prove that there exists a rational integer g such that xζ^{g} + yζ^{g+1} ≡ xζ^{−g} + yζ^{−g−1} (mod p). (Hint: write ϵ as a power of ζ times an element of ℤ[ζ_{p} + ζ_{p}^{−1}], i.e., a real number; we saw this on a previous homework. Also note that α^{p} is congruent to a rational integer modulo p.)

5. Assume the conditions of the previous problem, and also assume x and y are not divisible by p. Prove that if xζ^{g} + yζ^{g+1} ≡ xζ^{−g} + yζ^{−g−1} (mod p) for some rational integer g, then 2g ≡ −1 (mod p). (Hint: first rule out the possibilities g ≡ 0 (mod p) and g ≡ −1 (mod p). Then use the fact that ζ, ζ^{2}, …, ζ^{p−1} form an integral basis to conclude that g, g + 1, −g, −g − 1 cannot be distinct modulo p.)

6. Suppose that p ≥ 5 is regular and that x^{p} + y^{p} + z^{p} = 0 for pairwise coprime rational integers x, y, z, none divisible by p. Conclude from the above that x ≡ y, y ≡ z, z ≡ x (mod p) and thus obtain a contradiction.