Carlitz Modules and the Function Field Analogue of Kronecker-Weber

Li-Chung Chen

May 6, 2002

1 Introduction

Hilbert’s 9th problem asks for an explicit class field theory for all number fields. This has proved difficult, and the only known cases are for ℚ (cyclotomic theory) and for imaginary quadratic number fields (the theory of complex multiplication). However, the function field analogue has been solved, as independent work by L. Carlitz, D. Hayes, and V. Drinfeld has provided explicit construction of all abelian extensions of any global function field of positive characteristic.

This paper focuses on Carlitz’s function field analogues to cyclotomic number fields. Using Carlitz modules, we will construct explicit abelian extensions of 𝔽(T), where 𝔽 is a finite field, and we will state Hayes’ result on the maximal abelian extensions of 𝔽(T). Then we will briefly discuss Drinfeld’s generalization for general global function fields.

2 Drinfeld Modules Over 𝔽(T)

Let 𝔽 be a finite field with q = p^{s} elements, and set A = 𝔽[T] and k = 𝔽(T). For a commutative k-algebra B (which we will later take to be the algebraic closure of k, k̄), let τ : B → B be given by τ(x) = x^{q}. Because k has characeristic p, (x + y)^{p} = x^{p} + y^{p} holds in B. Repeated application of this equality shows that τ is additive. In fact, a similar argument shows that any polynomial of the form a_{0}x + a_{1}x^{q} + … + a_{r}x^{q^{r}} acts additively on B. The set of these polynomials forms a ring (under composition) that is identified with the “twisted” polynomial ring k⟨τ⟩, in which we have τa = a^{q}τ for all a ∈ k. In other words, we will consider elements of k⟨τ⟩ to be polynomials that act on B, with τ ↔ x^{q} and [1 ∈ k] ↔ x.

Now we can define a Drinfeld module over k = 𝔽(T):

Definition 2.1 A Drinfeld module for A defined over k is an 𝔽-algebra homomorphism ρ : A → k⟨τ⟩ such that

1. the constant term of ρ_{a} is a for all a ∈ A, and

2. ρ_{a} ∉ k for at least one a ∈ A.

This is called a module because any commutative k-algebra B can be made into an A-module B_{ρ} by defining

a · u = ρ_{a}(u) for all a ∈ A, u ∈ B.

In this special case where A = 𝔽[T], giving ρ is the same as specifying ρ_{T}. The first condition implies that the constant term of ρ_{T} is T, and the converse clearly holds. Now if ρ_{T} ∈ k, then ρ would map into k. Hence the second condition is equivalent to ρ_{T} ∉ k. Therefore, a Drinfeld module for A over k is given by specifying a nonconstant ρ_{T} with constant term T. We say that the rank of the Drinfeld module ρ is deg(ρ_{T}).

Consider the Drinfeld module given by ρ_{T} = T + c_{1}τ + c_{2}τ^{2} + … + c_{r}τ^{r}, where c_{i} ∈ k and c_{r} ≠ 0. This induces an A-module structure k̄_{ρ} on k̄ as described above. For nonzero a ∈ A, define the torsion submodule

Λ_{ρ}[a] = {λ ∈ k̄ : ρ_{a}(λ) = 0} .

It is clear by induction that ρ_{T^{n}} is a degree-rn polynomial in τ with constant term T^{n}. Hence for a ∈ A, ρ_{a} has degree r deg(a) in τ (and degree q^{r deg(a)} in x) with constant term a. But 𝔽⟨τ⟩ corresponds to polynomials in x whose monomials have degrees that are powers of q. Thus the derivative of ρ_{a}(x) is a and ρ_{a}(x) is separable. Since Λ_{ρ}[a] is just the set of roots of ρ_{a}(x), it has size q^{r deg(a)}. This information is sufficient to determine Λ_{ρ}[a]:

Theorem 2.2 Let ρ be a Drinfeld module of rank r. Then for any nonzero a ∈ A, Λ_{ρ}[a] ≅ (A/aA)^{r} as A-modules (and hence as (A/aA)-modules because both sides are annihilated by a).

Proof

First consider the case that a = P^{e} is a prime power. Since Λ_{ρ}[a] is finite, by the structure theorem of modules over a PID, we obtain an A-module isomorphism

Λ_{ρ}[P^{e}] ≅ A/Q_{1}^{f_{1}}A ⊕ A/Q_{2}^{f_{2}}A ⊕ … ⊕ A/Q_{r}^{f_{r}}A

for some irreducible polynomials Q_{i}. But P^{e} annihilates Λ_{ρ}[P^{e}], so it annihilates each A/Q_{i}^{f_{i}}A and forces Q_{i}/P ∈ 𝔽^{∗} (so we may assume Q_{i} = P) and f_{i} ≤ e for all i.

Now note that

Λ_{ρ}[P] = {λ ∈ Λ_{ρ}[P^{e}] : ρ_{P} (λ) = 0}

because any element annihilated by P is also annihilated by P^{e}. The left hand side has q^{r deg(P)} elements as determined earlier, and the right hand side has |(A/P A)^{t}| = q^{t deg(P)} elements. Hence t = r. This implies that Λ_{ρ}[P^{e}] has q^{f_{1}+...+f_{r}} elements. But we already know that Λ_{ρ}[P^{e}] has q^{r deg(P^{e})} = q^{re deg(P)} elements and that each f_{i} ≤ r. Hence f_{i} = r for all i and we are done with the prime power case.

For the general case, let the prime decomposition of a be a = αP_{1}^{e_{1}}P_{2}^{e_{2}} … P_{t}^{e_{t}} (where α ∈ 𝔽^{∗}). We claim that Λ_{ρ}[a] = ⊕_{i=1}^{t}Λ_{ρ}[P_{i}^{e_{i}}]. Clearly Λ_{ρ}[P_{i}^{e_{i}}] ⊂ Λ_{ρ}[a] for all i. Let a_{i} = \frac{a}{αP_{i}^{e_{i}}}. Because the a_{i}’s have greatest common divisor 1, there exist b_{i} ∈ A such that ∑_{i=1}^{t} b_{i}a_{i} = 1. Then for any m ∈ Λ_{ρ}[a], m = ∑_{i=1}^{t} b_{i}(a_{i}m). But a_{i}m ∈ Λ_{ρ}[P_{i}^{e_{i}}] because P_{i}^{a_{i}}(a_{i}m) = α^{−1}am = 0. Thus Λ_{ρ}[a] = ∑_{i=1}^{t} Λ_{ρ}[P_{i}^{e_{i}}]. Now if we have c_{i} ∈ Λ_{ρ}[P_{i}^{e_{i}}] such that ∑_{i=1}^{t} c_{i} = 0, then 0 = a_{j} ∑_{i=1}^{t} c_{i} = a_{j}c_{j} because P_{i}^{e_{i}} divides a_{j} whenever i ≠ j. But a_{j} is a unit in A/P_{j}^{e_{j}}A and Λ_{ρ}[p_{j}^{e_{j}}] is an (A/P_{j}^{e_{j}}A)-module. Thus c_{j} = 0 for all j and Λ_{ρ}[a] = ⊕_{i=1}^{t}Λ_{ρ}[P_{i}^{e_{i}}].

Hence by the prime power case, we have

Λ_{ρ}[a] = ⊕_{i=1}^{t}Λ_{ρ}[P_{i}^{e_{i}}] ≅ ⊕_{i=1}^{t}(A/P_{i}^{e_{i}}A)^{r} ≅ (A/aA)^{r}

(using the Chinese remainder theorem).

□

Let K_{ρ,a} = k(Λ_{ρ}[a]). It was noted above that Λ_{ρ}[a] is the set of roots of a separable polynomial. Hence K_{ρ,a}/k is a Galois extension. Because ρ_{a}(x) ∈ k[x], ρ_{a}(λ) = 0 iff ρ_{a}(σλ) = 0 for all σ ∈ Gal(K_{ρ,a}/k). Thus ρ maps Λ_{ρ}[a] into itself bijectively. But further, for any b ∈ A, λ ∈ Λ_{ρ}[a], and σ ∈ Gal(K_{ρ,a}/k), we have σ(ρ_{b}λ) = ρ_{b}(σλ) because ρ_{b}(x) ∈ k[x]. Thus σ acts as an A-module automorphism on Λ_{ρ}[a] and hence as an A/aA-module automorphism because a annihilates Λ_{ρ}[a]. Therefore, we get a homomorphism

Gal(K_{ρ,a}/k) → Aut_{A/aA}(Λ_{ρ}[a]).

Now if σ ∈ Gal(K_{ρ,a}/k) maps to the identity, then it fixes every element in Λ_{ρ}[a]. But these elements generate K_{ρ,a}, so σ must be the identity. Therefore the above map is injective, and we have established this important fact:

Proposition 2.3 If K_{ρ,a} = k(Λ_{ρ}[a]), then K_{ρ,a}/k is Galois and there is an injection

Gal(K_{ρ,a}/k) → GL_{r}(A/aA).

Corollary 2.4 If ρ has rank 1, then K_{ρ,a}/k is an abelian extension.

Proof

This follow immediately because GL_{1}(A/aA) = (A/aA)^{∗} is abelian and Gal(K_{ρ,a}/k) is a subgroup of it.

□

Therefore, a Drinfeld module (on 𝔽[t] over 𝔽(t)) given by ρ_{T} = T + cτ (where c ∈ k^{∗}) produces many examples of abelian extensions of 𝔽(t). We will be interested in the special case where ρ is given by ρ_{T} = T + τ. In this case we denote ρ as C and call it a Carlitz module.

3 Carlitz Modules and Their Induced Extensions

In this section, we will work solely with Carlitz modules and hence will abbreviate Λ for Λ_{C} = Λ_{ρ}, Λ_{a} for Λ_{C}[a], and K_{a} for K_{C,a}. Corollary 2.4 shows that Gal(K_{a}/k) is a subgroup of (A/aA)^{∗} for a ≠ 0. We will show that Gal(K_{a}/k) ≅ (A/aA)^{∗}, and along the way we will establish some properties of Carlitz modules. We will see that K_{a} is analogous to cyclotomic extensions of ℚ.

First some important remarks about the polynomial C_{a}(x):

Lemma 3.1 For nonzero a ∈ A of degree d, C_{a}(x) is of the form

[a, 0]x + [a, 1]x^{q} + … + [a, d]x^{q^{d}}

with [a, 0] = a, [a, d] ≠ 0, and each [a, i] ∈ A. If a is monic, then [a, d] = 1.

Proof

Reduce to the case a = T^{n} and induct on n. The base case works because C_{T} = T + τ is monic in A⟨τ⟩. Note that this lemma fails for any other rank-1 Drinfeld module.

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Recall that Λ_{a} ≅ A/aA as A-modules. Hence it has Φ(a) = |(A/aA)^{∗}| generators. If λ_{a} is a generator, then every λ ∈ Λ_{a} is of the form ϕ_{b}(λ_{a}) for some b ∈ A. But ϕ_{b}(x) is just a polynomial in k[x], so λ ∈ k(λ_{a}). Therefore, K_{a} = k(Λ_{a}) = k(λ_{a}).

Let O_{a} denote the integral closure of A in K_{a}. The integral closure of A in k is A because A is factorial and hence integrally closed. Thus k and A are analogous to ℚ and ℤ, and (K_{a}, O_{a}) is analogous to a rational cyclotomic field and its ring of integers.

Lemma 3.2 Suppose λ_{a} ∈ Λ_{a} is a generator and b ∈ A is relatively prime to a. Then C_{b}(λ_{a})/λ_{a} is a unit in O_{a}.

Proof

It is clear that if a = βa′ for a′ monic and β ∈ 𝔽^{∗}, then Λ_{a′} = Λ_{a} and the other corresponding items are equal. Thus we may assume that a is monic.

By Lemma 3.1, C_{a}(λ_{a}) = 0 implies λ_{a} ∈ O_{a}. Further, C_{b}(λ_{a})/λ_{a} is a polynomial in λ_{a} with coefficients in A, so it is also in O_{a}. Thus it suffices to check that λ_{a}/C_{b}(λ_{a}) is in O_{a}.

Because a and b are relatively prime, there exist f, g ∈ A such that fb = 1 + ga. Then C_{f}C_{b} = C_{1} + C_{g}C_{a} and C_{f}(C_{b}(λ_{a})) = λ_{a} + C_{g}(C_{a}(λ_{a})) = λ_{a}. Hence

λ_{a}/C_{b}(λ_{a}) = C_{f}(C_{b}(λ_{a}))/C_{b}(λ_{a}).

Now C_{b}(λ_{a}) is a root of C_{a}(x) and is hence in O_{a}. Thus similar to above, C_{f}(C_{b}(λ_{a}))/C_{b}(λ_{a}) is a polynomial in C_{b}(λ_{a}) with coefficients in A; thus λ_{a}/C_{b}(λ_{a}) ∈ O_{a}.

□

With the lemma, we can do the prime power case:

Proposition 3.3 Suppose P ∈ A is monic irreducible and e ≥ 1. Then K_{P^{e}} is totally ramified with ramification index Φ(P^{e}) at the prime P A, but is unramified at all other primes. Hence

[K_{P^{e}} : k] = Φ(P^{e}) and Gal(K_{P^{e}}/k) ≅ (A/P^{e}A)^{∗}.

Finally, the prime ideal above P A is (λ) = λO_{P^{e}}, where λ is any generator of Λ_{P^{e}}.

Proof

Let λ be a generator of Λ_{P^{e}}. It satisfies the monic polynomial C_{P^{e}}(x) ∈ A[x] and is thus in O_{P^{e}}. Because A = 𝔽[T] is a factorial integral domain, by Gauss’ lemma the minimal polynomial g(x) of λ is a monic polynomial in A[x]. Hence C_{P^{e}}(x) = f(x)g(x) where f(x) ∈ A[x].

By Lemma 3.1, the derivative of C_{P^{e}}(x) is P^{e}. Thus P^{e} = C′_{P^{e}}(λ) = f(λ)g′(λ) + f′(λ)g(λ) = f(λ)g′(λ). Because λ is a generator, we showed above that K_{P^{e}} = k(λ). Hence g′(λ) is contained in the different D of O_{P^{e}}/A. Now because f(x) ∈ A[x] and λ ∈ O_{P^{e}}, we see that f(λ) ∈ O_{P^{e}}. Hence for any prime ideal I ⊂ O_{P^{e}},

I ramifies over k ⟹ I|D
⟹ P^{e} = f(λ)g′(λ) ∈ g′(λ)O_{P^{e}} ⊂ D ⊂ I
⟹ P ⊂ I ∩ A
⟹ I is over P A.

Thus P A is the only possible prime ideal in A ramified in O_{P^{e}}.

Let d = deg(P). Because Λ_{P^{e}} ≅ A/P^{e}A, we see that λ is a generator of Λ_{P^{e}} iff C_{P^{e}}(λ) = 0 and C_{P^{e−1}}(λ) ≠ 0. Thus the Φ(P^{e}) generators are all the roots of

\frac{C_{P^{e}}(x)}{C_{P^{e−1}}(x)} = \frac{C_{P} (C_{P^{e−1}}(x))}{C_{P^{e−1}}(x)}
= P + [P, 1]C_{P^{e−1}}(x)^{q−1} + … + [P, d]C_{P^{e−1}}(x)^{q^{d}−1}

(using Lemma 3.1). Note that this polynomial has degree q^{deg(P^{e−1})}(q^{d}−1) = q^{d(e−1)}(q^{d}−1) = Φ(P^{e}). This polynomial and Lemma 3.2 imply that

P = ± ∏_{generator λ} λ = λ^{Φ(P^{e})} × unit.

Hence P O_{P^{e}} = (λ)^{Φ(P^{e})}.

Thus if I ⊂ O_{P^{e}} is a prime above P A, then I | λ and e(I/P A) is a multiple of Φ(P^{e}). In particular, [K_{p^{e}} : k] ≥ e(I/P A) ≥ Φ(P^{e}). However, since K_{p^{e}} = k(λ) and λ is a root of the degree-Φ(P^{e}) polynomial above, we must have [K_{p^{e}} : k] ≤ Φ(P^{e}). Therefore, we must have equalities and [K_{p^{e}} : k] = e(I/P A) = Φ(P^{e}). Since K_{p^{e}}/k is Galois, we have efr = [K_{p^{e}} : k] and hence f = r = 1, i.e. P A is totally ramified and there is only one prime above P A. If (λ) factors, then there would either be more than one prime above P A or a larger ramification index. Hence (λ) is prime and is the only prime above P A.

Lastly, we already know that Gal(K_{P^{e}}/K) is a subgroup of the group (A/P^{e}A)^{∗} of order Φ(P^{e}). Because Gal(K_{P^{e}}/K) has order Φ(P^{e}), it must be the entire group.

□

The general case now follows fairly directly:

Theorem 3.4 Suppose a = αP_{1}^{e_{1}}P_{2}^{e_{2}} … P_{t}^{e_{t}} is the prime decomposition of a, where each P_{i} is monic. Then K_{a} is the compositum of the fields K_{P_{i}}^{e_{i}}. The only ideals in A that ramify in O_{a} are the P_{i}A’s. Furthermore, [K_{a} : k] = Φ(a), from which it follows that

Gal(K_{a}/k) ≅ (A/aA)^{∗}.

Proof

Set a_{i} = \frac{m}{αP_{i}^{e_{i}}}. Let λ_{a} be a generator of Λ_{a} ≅ A/aA. Then λ_{P_{i}}^{e_{i}} = C_{a_{i}}(λ_{a}) is a generator of Λ_{P_{i}}^{e_{i}} ≅ a_{i}(A/aA). Thus K_{P_{i}}^{e_{i}} = k(λ_{P_{i}}^{e_{i}}) = k(C_{a_{i}}(λ_{a})) ⊂ k(λ_{a}) = K_{a}, and we have ∏_{i=1}^{t} K_{P_{i}}^{e_{i}} ⊂ K_{a}.

Conversely, because the a_{i}’s have greatest common divisor 1, there exist polynomials b_{i} ∈ A such that ∑_{i=1}^{t} b_{i}a_{i} = 1. Thus

λ_{a} = ∑_{i=1}^{t} C_{b_{i}}(C_{a_{i}}(λ_{a})) = ∑_{i=1}^{t} C_{b_{i}}(λ_{P_{i}}^{e_{i}}) ∈ ∏_{i=1}^{t} K_{P_{i}}^{e_{i}}.

Since λ_{a} generates K_{a}, we have ∏_{i=1}^{t} K_{P_{i}}^{e_{i}} = K_{a}.

Next, if P A ⊂ A is a prime different from each P_{i}A, then by Proposition 3.3 it is unramified in every K_{P_{i}}^{e_{i}}. Hence it is also unramified in the compositum K_{a}. Conversely, each P_{i}A is ramified in K_{a} because it is already ramified in K_{P_{i}}^{e_{i}}.

Now we prove that [K_{a} : k] = Φ(a) by inducting on t. The base case t = 1 is exactly Proposition 3.3. If the result holds for t − 1, then [K_{a_{t}} : k] = Φ(a_{t}). But by the above and by Proposition 3.3, P_{t}A is unramified in K_{a_{t}} and totally ramified in K_{P_{t}}^{e_{t}}. Hence K_{a_{t}} and K_{P_{t}}^{e_{t}} are linearly disjoint over K, and

[K_{a} : k] = [K_{a_{t}} : k][K_{P_{t}}^{e_{t}} : k] = Φ(a_{t})Φ(P_{t}^{e_{t}}) = Φ(a).

Induction is complete. So once again, because Gal(K_{a}/k) is a subgroup of (A/aA)^{∗} of the correct size, it must be the entire group.

□

4 The Maximal Abelian Extension of 𝔽(t)

From the preceding discussion, the compositum k(Λ) of all the fields K_{a} (for nonzero monic a ∈ A) is an abelian extension of k. Further, it is clear that k(Λ) = \underrightarrow{lim}_{a≠0} K_{a} and that Gal(k(Λ)/k) ≅ \underleftarrow{lim}_{a≠0}(A/aA)^{∗}, which is analogous to Gal(ℚ(λ_{∞})/ℚ) ≅ \underleftarrow{lim}_{m}(ℤ/mℤ)^{∗} = ℤ̂^{∗} for number fields.

To construct more abelian extensions, we substitute 1/T for T everywhere above and obtain new but isomorphic abelian extensions. In particular, K_{T^{n+1}} becomes what we call k(Λ_{T^{−(n+1)}}), whose Galois group over k is still of order Φ(T^{n+1}) = q^{n}(q−1). Since (q^{n}, q−1) = 1, the Galois group decomposes uniquely into the product of groups of order q^{n} and q−1. Let L_{n} be the fixed field of the order-(q − 1) subgroup. Then L_{n}/k is still an abelian extension. Let L_{∞} = ∏_{n=1}^{∞} L_{n}.

However, we have not yet obtained all the abelian extensions of k. It can be shown that the constant fields (i.e. the intersection with the algebraic closure 𝔽̄) of k(Λ) and L_{∞} are both 𝔽. In other words, k(Λ) and L_{∞} are what are called geometric extensions of k. By extending the constant field, we obtain another abelian extension:

Proposition 4.1 Gal(𝔽̄(T)/𝔽(T)) ≅ ℤ̂.

Proof

We claim that Gal(𝔽̄(T)/𝔽(T)) ≅ Gal(𝔽̄/𝔽). Any element σ ∈ Gal(𝔽̄/𝔽) extends to an element in Gal(𝔽̄(T)/𝔽(T)) by setting σ(T) = T. Conversely, if σ ∈ Gal(𝔽̄(T)/𝔽(T)) and x ∈ 𝔽̄, then x is algebraic over 𝔽 (which is fixed by σ), so σ(x) is also algebraic over 𝔽 and σ(x) ∈ 𝔽̄. Thus σ sends 𝔽̄ to itself, and σ is the extension of an element of Gal(𝔽̄/𝔽).

Now 𝔽 is a finite field of size q. Hence 𝔽̄ ≅ \underrightarrow{lim}_{d} 𝔽_{q^{d}} and Gal(𝔽̄(T)/𝔽(T)) ≅ Gal(𝔽̄/𝔽) ≅ \underleftarrow{lim}_{d} ℤ/dℤ = ℤ̂.

□

Now we can state the function-field analogue of the Kronecker-Weber theorem: Theorem 4.2 (Hayes) k(Λ), L_{∞}, and 𝔽̄(T) are linearly disjoint over k and generate the maximal abelian extension of k. The proof uses class field theory and can be seen in [2] or [3].

5 Drinfeld Modules

Finally, we examine Drinfeld modules, the generalization of Carlitz modules over function fields other than 𝔽(t), and suggest how they relate to abelian extensions. We begin with some preliminary setup and definitions.

Definition 5.1 A function field k over 𝔽 is an extension of 𝔽 of transcendence degree 1.

Definition 5.2 A prime of a function field k/𝔽 is a discrete valuation ring R with maximal ideal P such that 𝔽 ⊂ R ⊂ k and R has quotient field k. This prime is referred to as P, and the associated valuation is denoted ord_{P} (·).

Now let 𝔽 = 𝔽_{q} and let k/𝔽 be a function field with constant field 𝔽. Let ∞ be a fixed prime of k and let A be the ring of elements of k whose only poles are at ∞ (i.e. have nonnegative valuations at all other primes). Let L be a field containing 𝔽. As before, regard τ as the map raising an element to the qth power, and let L⟨τ⟩ be the twisted polynomial ring with relation τa = a^{q}τ for a ∈ L. Fix an 𝔽-algebra homomorphism δ : A → L. Then Definition 5.3 A Drinfeld A-module over L is an 𝔽-algebra homomorphism ρ : A → L⟨τ⟩ such that for all a ∈ A, the constant coefficient of ρ_{a} is δ(a). Also, we require that the image of ρ not be contained in L. The set of all such ρ is denoted Drin_{A}(L).

Similar to before, if Ω is an L-algebra, then a Drinfeld module makes Ω into an Amodule Ω_{ρ} by setting a · u = ρ_{a}(u) for a ∈ A, u ∈ Ω. This construction applies to M, a fixed algebraically closed field extension of L. Thus we can define the torsion submodule M_{ρ}[a] = {u ∈ M_{ρ} : ρ_{a}(u) = 0} as before.

Note that the above definition is consistent with our prior definition of Drinfeld modules over 𝔽(T). There k = 𝔽(T), ∞ is the prime at infinity, A = 𝔽[T], L = k, and δ is the inclusion. There is something to check here, namely that 𝔽[T] is the set of elements whose only poles are at ∞. Now by ∞ we mean the valuation where for polynomials f, g ∈ 𝔽[T], ord_{∞}(f/g) is the highest exponent in g minus the highest exponent in f. It can be shown that the other primes of k correspond to nonzero prime ideals P ⊂ 𝔽[T], with valuation ord_{P} (z) being the exponent of P in the prime factorization of z. Hence 𝔽[T] is indeed the set of elements regular at all the non-∞ primes.

In the case over 𝔽(T), giving a Drinfeld module is the same as setting ρ_{T} to be a nonconstant polynomial with constant T. The existence of Drinfeld modules in general is less clear, but it can be shown using analytic methods. The construction is similar to the construction of elliptic curves over the complex numbers ℂ by means of two-dimensional ℤ-lattices and the associated Weierstrass P-functions.

Now in general δ need not be injective. Let Q = ker(δ). Then Q is a prime ideal of A because A/Q is contained in the integral domain L. We say that Q is the A-characteristic of L. Then the following theorem defines the rank of a Drinfeld module:

Theorem 5.4 Let ρ ∈ Drin_{A}(L), let Q be the A-characteristic of L, and let M be an algebraically closed field containing L. If P ≠ Q is a nonzero prime ideal of A and e ≥ 1, then there exist a positive integer r independent of P and e such that

M_{ρ}[P^{e}] ≅ (A/P^{e})^{r}.

If Q ≠ (0) and e ≥ 1, then there exist an integer h, independent of e, such that

M_{ρ}[Q^{e}] ≅ (A/Q^{e})^{r−h}.

We say that r and h are the rank and height of ρ, respectively.

Corollary 5.5 This definition of rank coincides with the previous definition over 𝔽(T).

Proof

Use Theorem 2.2.

□

Proposition 5.6 Let r = rank(ρ). Then for each a ∈ A \ Q, there is an injection

Gal(L(M_{ρ}[a])/L) → GL_{r}(A/aA).

Proof

This is analogous to the proof in the case of 𝔽(T). Because a ∉ ker(δ), the constant term of ρ_{a} is nonzero. Hence for some polynomial f(x) = c_{0}x + c_{1}x^{q} + … + c_{t}x^{q^{t}} ∈ L[x] where c_{0} ≠ 0, we have ρ_{a}(u) = f(u) for all u ∈ M. Because M is algebraically closed, M_{ρ}[a] is just the set of roots of f, and L(M_{ρ}[a]) is f’s splitting field. Now since 𝔽 ⊂ L, the characteristic of L divides q and f′(x) = c_{0} ∈ L^{∗}. Thus f is separable and L(M_{ρ}[a])/L is Galois.

By arguments similar to before, each ρ ∈ Gal(L(M_{ρ}[a])/L) acts on M_{ρ}[a] ≅ (A/aA)^{r} as an A-module (hence (A/aA)-module) isomorphism. This defines a homomorphism Gal(L(M_{ρ}[a])/L) → GL_{r}(A/aA) that is injective because M_{ρ}[a] generates L(M_{ρ}[a]).

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Corollary 5.7 If ρ is a rank-1 Drinfeld A-module over L and a ∈ A is not in the Acharacteristic of L, then L(M_{ρ}[a])/L is an abelian extension.

Thus we see that adjoining torsion points of a Drinfeld module creates abelian extensions. This is the start of an explicit construction of the maximal abelian extension of a function field. For more details, see [1] and [4].

References

[1] V.G. Drinfeld. Elliptic Modules (Russian), Math. Sbornik 94 (1974), 597-627. English translation: Math. USSR, Sbornik 23 (1977), 159-170.

[2] D.R. Hayes. A brief introduction to Drinfeld modules, in The Arithmetic of Function Fields (eds. D. Goss et al), Walter de Gruyter & Co., New York-Berlin, 1992.

[3] D.R. Hayes. Explicit class field theory for rational function fields, Trans. Amer. Math. Soc. 189 (1974), 77-91.

[4] D.R. Hayes. Explicit class field theory in global function fields, Studies in Algebra and Number Theory, Adv. Math. Supplementary Studies 6 (1979), 173-217.

[5] M. Rosen. Number Theory in Function Fields. Springer-Verlag, New York, 2002.