Riemann-Roch Theory for Function Fields

Andrei Jorza

March 15, 2005

1 Motivation

Let K be a global field (i.e., K is a finite extension of ℚ or of 𝔽_{q}(x)).

Definition 1.1. The ζ-function is

ζ_{K}(s) = ∑_{I} \frac{1}{NI^{s}} ,

where I runs through all the ideals of O_{K}.

Proposition 1.2. We have

ζ_{K}(s) = ∏_{𝔭∈Spec O_{K}} \frac{1}{1 − N𝔭^{−s}} .

Proof. There are sensible convergence issues here, but we will not worry about these. Since O_{K} is a Dedekind domain, with unique factorization, every ideal I = 𝔭_{1}^{n_{1}} ⋯ 𝔭_{k}^{n_{k}} so

ζ_{K}(s) = ∑(N𝔭_{1}^{n_{1}} ⋯ 𝔭_{k}^{n_{k}})^{−s} = ∏_{𝔭_{i}} ∑_{n_{i}=0}^{∞}(N𝔭_{i})^{−n_{i}s} = ∏_{𝔭_{i}} \frac{1}{1 − N𝔭_{i}^{−s}} .

The ζ_{K}(s) clearly converges for Res > 1 and moreover it has an analytic continuation to ℂ \ {1}.

Theorem 1.3 (Dirichlet’s Class Number Formula). If K is a number field then the residue of ζ_{K} at 1 is

vol(𝔸_{K}^{×}/K^{×}) = \frac{2^{r_{1}}(2π)^{r_{2}}h_{K}R_{K}}{w_{K} \sqrt{|D_{K}|}} .

This can be rewritten as

ζ_{K}^{r_{1}+r_{2}−1}(0) = −\frac{h_{K}R_{K}}{w_{K}} ,

which looks exactly like the Birch And Swinnerton-Dyer conjecture since the rank of 𝔾_{m}(O_{K}) is r_{1} + r_{2} − 1.

One way to prove this is to relate ζ_{K} to ζ-functions associated to characters 𝔸_{K}^{×}/K^{×} → ℂ^{×} and then use harmonic analysis. We will not prove this theorem here but we’ll see more of this analogy later when we study elliptic curves.

2 Riemann-Roch on Function Fields

2.1 Divisors on Adèles

We would like to obtain similar formulas for function fields.

Let K be a finite extension of 𝔽_{q}(x). Recall the topological rings K^{×} ↪ 𝔸_{K}^{1} ↪ 𝔸_{K}^{×} ↪ 𝔸_{K}. For each finite place v of K (all places are finite!) recall that we have K_{v}, O_{v}, ℘_{v}, k_{v}/𝔽_{q}, q_{v} = q^{d_{v}}.

Definition 2.1. Div(K) is the free abelian group generated by the (finite) places v of K, i.e. Div(K) = ⨁_{v} vℤ. The map Div(K) → ℤ given by deg : ∑ n_{v}v ↦ ∑ n_{v}d_{v} is a homomorphism with kernel Div^{0}(K).

There is an obvious map 𝔸_{K}^{×} → Div(K) given by div : 𝔞 = (a_{v}) ↦ ∑ v(a_{v})v, a homomorphism. Then we clearly have |𝔞|_{𝔸} = q^{− deg 𝔞} so this gives

Lemma 2.2. The map div is a surjection from 𝔸_{K}^{1} to Div^{0}(K) with kernel ∏ O_{v}^{×}.

Let P (K) = div(K^{×}) be the principal divisors. Then write Pic(K) = Div(K), Pic^{0}(K) = Div^{0}(K)/P (K).

Proposition 2.3. There is an isomorphism Cl(K) ≅ Pic^{0}(K) which proves that Cl(K) is finite.

Proof. There is an isomorphism between the group of fractional ideals and 𝔸_{K}^{1}. Moreover, the group 𝔸_{K}^{1}/K^{×} is compact and ∏ O_{v}^{×} is open so Pic^{0}(K) ≅ 𝔸_{K}^{1}/K^{×} ∏ O_{v}^{×} is finite.

2.2 Invertible sheaves associated with divisors

Definition 2.4. For 𝔞 = ∑ a_{v}v ∈ Div(K) let U(𝔞) = {b = (b_{v}) ∈ 𝔸_{K}^{×}||b_{v}|_{v} ≤ q_{v}^{−a_{v}}} = ∏{b ∈ K_{v}^{×}||b|_{v} ≤ q_{v}^{−a_{v}}} which is compact by Tychonov since each factor is compact. Let L(𝔞) = (U(𝔞) ∩ K^{×}) ∪ {0}. This is compact in K^{×} which is discrete, so L(𝔞) is a finite K ∩ ∏ O_{v} module. Since K ∩ ∏ K_{v} = 𝔽_{q} we get |L(𝔞)| = q^{ℓ(𝔞)} where ℓ(𝔞) = dim_{𝔽_{q}} L(𝔞).

Lemma 2.5. If deg 𝔞 < 0 then L(𝔞) = 0. If deg 𝔞 = 0 but 𝔞 ≠ 0 in Pic(K) then L(𝔞) = 0.

Proof. L(𝔞) consists of elements x ∈ K^{×} such that |x|_{𝔸} = q^{− deg 𝔞} > 1 which cannot be unless 𝔞 = 0.

If deg 𝔞 = 0 then the above proof shows that the only possible nonzero element in L(𝔞) must be x ∈ K^{×} such that x_{v} = −a_{v} which means that 𝔞 = divx = 0 in Pic(K) contradicting the hypothesis.

Lemma 2.6. Prove that L(𝔞) can be identified with divisors 𝔟 ∈ Div(K) such that 𝔟 ≥ 0 and 𝔟 = 𝔞 ∈ Pic(K).

Proof. L(𝔞) = K ∩ {𝔟 ∈ 𝔸_{K}|v(𝔟_{v}) + v(a_{v}) ≥ 0}. So the divisor 𝔟 + 𝔞 is nonnegative and is clearly linearly equivalent to 𝔞 since 𝔟 ∈ K.

Remark 2.7. We have ℓ(0) = 1 which corresponds to the fact that L(0) = K^{×} ∩ ∏ O_{v} = 𝔽_{q}.

Lemma 2.8. If I is an ideal of O_{K} then NI = q^{deg(div∘ideal(I))}, where ideal : I = ∏ 𝔭_{v}^{n_{v}} ↦ (π_{v}^{n_{v}}) for uniformizers π_{v} ∈ K_{v}.

Proof. Assume that I = ∏ 𝔭_{v}^{n_{v}} then div∘idealI = ∑ n_{v}v while NI = q^{∑ n_{v}d_{v}} = q^{deg(div∘ideal(I))}.

2.3 The canonical divisor

Let χ : 𝔸_{K}/K → ℂ^{×} be a nontrivial character, which corresponds to a collection of local characters χ_{v} : K_{v} → ℂ^{×} that are trivial on O_{v} for almost all v. Let ord(χ_{v}) be the smallest integer n_{v} such that χ_{v} is trivial on ℘_{v}^{n_{v}}. Then ord(χ) = ∑ ord(χ_{v})v ∈ Div(K).

Proposition 2.9. Let χ′ be another nontrivial character of 𝔸_{K}/K. Then ord(χ) = ord(χ′) in Pic(K).

Proof. Characters are defined up to scalar action so for each v there exists a b_{v} ∈ K_{v}^{×} such that χ′_{v}(x) = χ_{v}(b_{v}x) which means that χ′(x) = χ(bx) where b = (b_{v})_{v} ∈ 𝔸_{K}. But χ and χ′ are trivial on K by definition so χ(b) = 1 which implies that ord(χ′) = ord(χ(b)) = ord(χ)+div(b) = ord(χ) since b ∈ K.

Definition 2.10. There exists a unique divisor 𝔠 = ord(χ) for some χ nontrivial character of 𝔸_{K}/K. This is called the canonical divisor and ℓ(𝔠) = g is called the genus of K. (For number fields there are analogous notions genus of K coming from Arakelov geometry.)

2.4 Topological duality

For a topological group G we define the topological dual Ĝ = Hom_{continuous}(G, ℂ^{×}). Fix χ a nontrivial character of 𝔸_{K}/K for which 𝔠 = ord(χ). The map a ↦ χ(a−) identifies 𝔸_{K} to its topological dual.

Lemma 2.11. If H ⊂ G is an open topological subgroup and H^{⊥} = {f ∈ Ĝ|f(H) = 1} then { \atop G/H} ≅ H^{⊥}. Moreover, G is compact if Ĝ is discrete and G is discrete if Ĝ is compact. Via topological duals the exact sequence

1 → H → G → G/H → 1

becomes

1 → H^{⊥} → Ĝ → Ĥ → 1.

If G/H is compact and H is discrete in G then the measure on G is the composite of the discrete measure on H and the Haar measure on G/H.

Remark 2.12. This shows that { \atop 𝔸_{K}/K} = K is discrete.

Lemma 2.13. For 𝔞 ∈ Div(K) we write U = U(𝔞) and U′(𝔠 − 𝔞). Then 𝔸_{K}/(K + U) is the topological dual of K ∩ U′.

Proof. It is enough to show that (K + U)^{⊥} ≅ K ∩ U′. First, (K + U)^{⊥} ⊂ K^{⊥} = K. Now under the indentification 𝔸_{K} ≅ 𝔸̂_{K} the set (K + U)^{⊥} consists of x ∈ 𝔸_{K} such that χ(x(K + U)) = 1. This happens if and only if χ(x(k +b)) = 1 for all k ∈ K and b such that |b|_{v} ≤ q_{v}^{−a_{v}}. Therefore we want χ_{v}(x_{v}(k_{v} + b_{v})) = 1 which happens for all x_{v}(k_{v} + b_{v}) ∈ ℘_{v}^{ord(χ_{v})}. This happens when x_{v} ∈ ℘_{v}^{ord(χ_{v})+a_{v}} so x ∈ U′. Therefore, (K + U)^{⊥} = K ∩ U′.

2.5 Riemann-Roch

Lemma 2.14. Let μ be a Haar measure on 𝔸_{K} induced from the discrete measure on K and the normalized Haar measure on 𝔸_{K}/K. Then μ(U) = q^{− deg 𝔞} = q^{ℓ(𝔞)}μ((K + U)/U). Moreover, μ(𝔸_{K}/K) = q^{ℓ(𝔠−𝔞)}μ((K + U)/U).

Proof. We have an exact sequence 1 → K → 𝔸_{K} → 𝔸_{K}/K → 1 in which we have an exact sequence 1 → K ∩ U → U → (K + U)/K → 1. Therefore μ(U) = q^{ℓ(𝔞)}μ((K + U)/U) since μ(K ∩ U) = μ(L(𝔞)). The fact that μ(U) = q^{− deg 𝔞} is immediate from definition.

The last equality follows from the exact sequence 1 → (K+U)/K → 𝔸_{K}/K → (𝔸_{K}/K)/((K+ U)/K) = 𝔸_{K}/(K + U) → 1 because μ(K ∩ U′) = q^{ℓ(𝔠−𝔞)}.

Theorem 2.15 (Riemann-Roch). For each 𝔞 ∈ Div(K) we have

ℓ(𝔞) = ℓ(𝔠 − 𝔞) + deg 𝔞 − g + 1.

Proof. From Lemma 2.14 we have

μ(𝔸_{K}/K) = q^{ℓ(𝔠−𝔞)}μ((K + U)/U) = q^{ℓ(𝔠−𝔞)}q^{− deg 𝔞}/q^{ℓ(𝔞)}
= q^{ℓ(𝔠−𝔞)−deg 𝔞−ℓ(𝔞)}

The theorem follows if say q^{g−1} = μ(𝔸_{K}/K). Proposition 2.16. We have deg 𝔠 = 2g − 2 and g = ℓ(𝔠). Proof. Add ℓ(𝔞) = ℓ(𝔠 − 𝔞) + deg 𝔞 − g + 1 and ℓ(𝔠 − 𝔞) = ℓ(a) + deg(𝔠 − 𝔞) − g + 1 and get deg 𝔠 = 2g−2. So if 𝔞 = 0 in the formula we get ℓ(0) = ℓ(𝔠)+0−g+1 so ℓ(𝔠) = g−1+1 = g. Proposition 2.17. If deg 𝔞 > 2g − 2 then ℓ(𝔞) = deg 𝔞 − g + 1. Proof. Then deg(𝔠 − 𝔞) < 0 so by Lemma 2.5 we have ℓ(𝔠 − 𝔞) = 0.

3 Class number formula for function fields

Let’s get back to our analogy between the case of number and function fields. Let K be a finite extension of 𝔽_{q}(x). Recall that ζ_{K}(s) = ∑_{I}(NI)^{−s} and we have a homomorphism div : I → Div(K) that takes integral ideals to nonnegative divisors. Lemma 3.1. There exists u ∈ Div(K) such that deg u = 1. Proof. If u = ∑ n_{v}v then deg u = ∑ n_{v}d_{v} so it is enough to prove that the d_{v} have no common factor. I am not going to prove this, but you can think about what happens if K = 𝔽_{q}(x) (the analogous case of K = ℚ for number fields) and generalize. (See Weil, Basic Number Theory, pp 126 if impatient.) Lemma 3.2. We have

ζ_{K}(s) = ∑_{𝔞∈Div_{≥0}(K)}q^{deg 𝔞} = ∑_{k=0}^{∞} ∑_{deg 𝔞=k,𝔞≥0}q^{−ks}
= ∑_{𝔞_{i}∈Pic^{0}(K)} ∑_{k=0}^{∞} ∑_{𝔞≥0,𝔞=𝔞_{i}+ku} q^{−ks}
= (q − 1)^{−1} ∑_{𝔞_{i}∈Pic^{0}(K)} ∑_{k=0}^{∞} q^{ℓ(𝔞_{i}+ku)−ks} − ∑_{𝔞_{i}∈Pic^{0}(K)}1/(1 − q^{−s})

Proof. The first equality follows from Lemma 2.8. From Lemma 2.6 we get that the number of 𝔞 ≥ 0 such that 𝔞 = 𝔞_{i} + ku is equal to (q^{ℓ(𝔞_{i}+ku)} − 1)/(q − 1) (the 0 divisor corresponds to no ideal and any ideal defines a divisor up to a unit of 𝔽_{q}^{×}) so the last equality follows. (Here I used that 𝔞 − ku ∈ Pic^{0}(K) must equal one of the 𝔞_{i}-s, up to scalars, which do not count.)

Lemma 3.3. We have ∑_{k=0}^{∞} q^{ℓ(𝔞_{i}+ku)−ks} = ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks} + q^{g−s(2g−1)}/(1 − q^{1−s}).

Proof. Recall that for k > 2g − 2 we have ℓ(𝔞_{i} + ku) = deg(𝔞_{i} + ku) − g + 1 = k − g + 1 so ∑_{k=0}^{∞} q^{ℓ(𝔞_{i}+ku)−ks} = ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks}+∑_{k>2g−2} q^{k(1−s)−g+1} which is equal to ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks}+ ∑_{k>2g−2} q^{(k−2g+1)(1−s)+g−s(2g−1)} = ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks} + q^{g−s(2g−1))}/(1 − q^{1−s}).

Proposition 3.4. There exists a polynomial P of degree 2g such that

ζ_{K}(s) = \frac{P (q^{−s})}{(1 − q^{−s})(1 − q^{1−s})}.

Proof. By Lemma 3.3 and the fact that q^{s−s(2g−1)}/(1 − q^{1−s}) and −1/(1 − q^{−s}) have the above property, it is enough to show that for each i the sum ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks} has the required property.

But (1 − q^{−s})(1 − q^{1−s}) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks} has degree 2g in q^{−s} since only the term corresponding to k = 2g − 2 counts. The conclusion then follows.

The really interesting facts that are analogous to the analytic class number formula in the case of number fields occur when we apply the Riemann-Roch theory.

Theorem 3.5 (Class number formula). We have P (z) = q^{g}z^{2g}P (1/qz).

Proof. Clearly we can get rid of the (q − 1)^{−1} factor for the first part of the problem. Define P_{i}(q^{−s}) = (1 − q^{−s})(1 − q^{1−s}) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)−ks} + (1 − q^{−s})q^{g−s(2g−1)} − (1 − q^{1−s}) so P_{i}(z) = (1 − z)(1 − qz) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)}z^{k} + (1 − z)q^{g}z^{2g−1} − (1 − qz).

We need to show that ∑_{i} P_{i}(z) = q^{g}z^{2g}(∑_{i} P_{i}(1/qz)). This is equivalent to ∑_{i}((1 − z)(1 − qz) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)}z^{k}+(1−z)q^{g}z^{2g−1}−(1−qz)) = ∑_{i}(q^{g}z^{2g}((1−1/qz)(1−1/z) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)}q^{−k}z^{−k}+ (1 − 1/qz)q^{−(g−1)}z^{−(2g−1)}) + (1 − z)q^{g}z^{2g−1})

But the RHS is equal to

∑_{i}((1 − z)(1 − qz) ∑_{k=0}^{2g−2} q^{ℓ(a_{i}+ku)−k+g−1}z^{2g−2−k} + (qz − 1) + (1 − z)q^{g}z^{2g−1}).

But the Riemann-Roch formula gives that ℓ(a_{i} + ku) − k + g − 1 = ℓ(𝔠 − 𝔞_{i} + ku). But 𝔠 = 𝔞_{1} + (2g − 2)u for our choice of 𝔞_{1}.

So we need to show that ∑_{i}((1−z)(1−qz) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{i}+ku)}z^{k} +(1−z)q^{g}z^{2(g−1)} +(qz −1)) = ∑_{i}((1 − z)(1 − qz) ∑_{k=0}^{2g−2} q^{ℓ(𝔞_{1}−𝔞_{i}+(2g−2−k)u)}z^{2g−2−k} + (qz − 1) + (1 − z)q^{g}z^{2(g−1)}) which is obvious.

Proposition 3.6. We have P (0) = 1 and P (1) = h_{K}. Proof. We have P (z) = (q − 1)^{−1} ∑_{𝔞_{i}∈Pic^{0}(K)} P_{i}(z). Since P_{i}(1) = q − 1 we have P (1) = (q − 1)^{−1}h_{K}(q − 1) = h_{K}.

Also note that lim_{∞} ζ_{K}(−s) = 1 by definition. So P (0) = 1. Proposition 3.7 (Class number formula). The residue at 0 of ζ_{K} is

Res_{0}ζ_{K}(s) = \frac{h_{K}}{(1 − q) log q} .

Proof. The residue at 0 is

lim_{s→0} \frac{P (q^{−s})}{1 − q^{1−s}} \frac{s}{1 − q^{−s}} = \frac{h_{K}}{1 − q} \frac{1}{log q} .