Math 254A, UC Berkeley, Fall 2001 (Kedlaya) Lectures 28-29: Gauss Sums and Class Numbers November 7-9, 2001

Reference. Lang, Algebraic Number Theory, Chapter IV, or Washington, Introduction to Cyclotomic Fields, Chapter 6. (Washington gives slightly more general results than we will.)

Outline of lectures.

Throughout, let ζ_{n} denote a primitive n-th root of unity, let p be a prime, let m be an integer not divisible by p, let l be the order of p mod m, let q = p^{l}, and let 𝔽_{q} be the finite field of q elements.

1. By a character we will mean a homomorphism χ : 𝔽_{q}^{∗} → ℚ(ζ_{p(q−1)})^{∗}, which we extend to 𝔽_{q} by setting χ(0) = 0. For any character χ, define the Gauss sum

G(χ) = ∑_{a∈𝔽_{q}^{∗}} χ(a)ζ_{p}^{Trace(a)}

where the trace is from 𝔽_{q} to 𝔽_{a}. For example, if q = p, we could take χ to be the quadratic character χ(a) = (\frac{a}{p}) and get back the usual Gauss sum. Note that if χ has order dividing m, then G(χ) takes values in ℚ(ζ_{pm}).

2. Work out some basic properties of the Gauss sums:

(a) G(\overline{χ}) = χ(−1)\overline{G(χ)};

(b) for χ ≠ 1, g(χ)g(\overline{χ}) = χ(−1)q;

(c) for χ ≠ 1, g(χ)\overline{g(χ)} = q;

(d) G(χ^{p}) = G(χ).

3. Given characters χ_{1} and χ_{2}, define the Jacobi sum

J(χ_{1}, χ_{2}) = − ∑_{a∈𝔽_{q}} χ_{1}(a)χ_{2}(1 − a).

Work out some basic properties of the Jacobi sums:

(a) J(1, 1) = 2 − q;

(b) J(1, χ) = J(χ, 1) = 1 if χ ≠ 1;

(c) J(χ, \overline{χ}) = χ(−1) if χ ≠ 1;

(d) G(χ_{1})G(χ_{2}) = J(χ_{1}, χ_{2})G(χ_{1}χ_{2}) if χ_{1}χ_{2} ≠ 1.

4. Note that if χ_{1} and χ_{2} have order dividing m that G(χ_{1})G(χ_{2})/G(χ_{1}χ_{2}) is an algebraic integer in ℚ(ζ_{m}) (not just in ℚ(ζ_{pm})).

5. Choose a prime 𝔭 of ℚ(ζ_{pm}) above (p) and a prime 𝔮 of ℚ(ζ_{p(q−1)}) above 𝔭, choose an isomorphism of the residue field of 𝔮 with 𝔽_{q}, and let χ_{𝔮} be the character such that χ_{𝔮}(a) ≡ a^{−1} (mod 𝔮).

6. For 1 ≤ ν < q − 1, let s(ν) be the sum of the base p digits of ν. For future reference, observe that

s(ν) = (p − 1) ∑_{i=0}^{l−1} \{\frac{p^{i}ν}{q − 1}\} ,

where the braces denote fractional part.

7. Show that v_{𝔮}(G(χ_{𝔮}^{ν})) = s(ν), by induction on ν, by checking that v_{𝔮}(χ_{𝔮}) = 1 and that v_{𝔮}(J(χ_{𝔮}^{ν−1}, χ_{𝔮})) = 0 if ν is not divisible by p. This gives us an explicit formula for the prime factorization of the ideal (G(χ)) for any character χ.

8. Let G be the Galois group of ℚ(ζ_{m}) and let ℤ[G] and ℚ[G] be the group algebras of G over ℤ and ℚ, respectively. (That is, these algebras are free with generators [g] for each g ∈ G, multiplied by the rule [g_{1}][g_{2}] = [g_{1}g_{2}].) Define 𝔞^{β} for 𝔞 a fractional ideal of ℚ(ζ_{m}) and β ∈ ℤ[G] as follows: if β = ∑ z_{i}[g_{i}], then 𝔞^{β} = ∏ g_{i}(𝔞)^{z_{i}}. Define the Stickelberger element of ℚ[G] by

θ = ∑_{a∈(ℤ/mℤ)^{∗}} \{\frac{a}{m}\} σ_{a}^{−1},

where σ_{a} denotes the automorphism ζ_{m} ↦ ζ_{m}^{a}.

9. Check that

(G(χ_{𝔮}^{(q−1)/m})) = ∏_{a∈(ℤ/mℤ)^{∗}} σ_{a}^{−1}(𝔭)^{s(a(q−1)/m)},

where a runs over (ℤ/mℤ)^{∗}. Deduce that (G(χ_{𝔮}^{(q−1)/m})^{m}) = 𝔭^{mθ}. Then note that the generator of the ideal on the left actually lies in ℚ(ζ_{m}). (It generates an extension of ℚ(ζ_{m}) which is only ramified above primes dividing m, whereas ℚ(ζ_{pm})/ℚ(ζ_{m}) is totally ramified above primes dividing p.)

10. Prove Stickelberger’s theorem: if β ∈ ℤ[G] is such that βθ ∈ ℤ[G] as well, then 𝔞^{βθ} is principal for any fractional ideal 𝔞 of ℚ(ζ_{m}). It suffices to check this for 𝔞 prime and coprime to m (since such ideals generate the full class group; see a prior homework). In fact, if 𝔞 = 𝔭 ∩ ℚ(ζ_{m}), we have now shown that 𝔞^{βθ} is generated by G(χ_{𝔮}^{(q−1)/m})^{βθ}. (The more general statement of the theorem is that the same thing happens in any subfield of ℚ(ζ_{m}).)