International Journal of Algebra, Vol. 7, 2013, no. 4, 157 - 166 HIKARI Ltd, www.m-hikari.com
Kummer Type Extensions in Function Fields
Marco S´anchez-Mirafuentes
Universidad Aut´onoma Metropolitana Unidad Iztapalapa Divisi´on de Ciencias B´asicas e Ingenier´ia Departamento de Matem´aticas Av. San Rafael Atlixco 186, Col Vicentina Del. Iztapalapa, C.P. 09240 M´exico, D.F.
Gabriel Villa Salvador
CINVESTAV IPN Departamento de Control Autom´atico Apartado Postal 14-740 M´exico, D.F. C.P 07000
Copyright c 2013 Marco S´anchez-Mirafuentes and Gabriel Villa Salvador. This is an open access article distributed under the Creative Commons Attribution License, which per- mits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
We present a generalization of Kummer extensions in algebraic func- tion fields with finite field of constants Fq, using the action of Carlitz- Hayes. This generalization of Kummer type extensions is due to Wen- Chen Chi and Anly Li and due also to Fred Schultheis. The main results of this article are Proposition 3.2 and Theorem 3.4. They provide a partial analogue of a theorem of Kummer, which establishes a bijection between Kummer extensions L/K of exponent n and subgroups of K∗ containing (K∗)n.
Mathematics Subject Classification: 11R60, 11R99, 12F05, 14H05
Keywords: Carlitz Module, Kummer extensions, Dual module, Bilinear map 158 M. S´anchez-Mirafuentes and G. Villa-Salvador
1 Introduction
An n-Kummer extension, is a field extension L/K such that n is relatively prime to char(K), the characteristic of K, μn ⊆ K, where μn is the group of n-th roots of unity, and √ L = K( n Δ),
∗ ∗n where√ Δ is a subgroup of K containing the group√ K of n-th powers, and K( n Δ) is the field generated by all the roots n a, with a ∈ Δ. This is the origin of the theory of Kummer which is of significance in class field theory, for example see [5] Chapter 1. In this setting we have the followings facts.
Let K be any field containing the group μn of n-th roots of unity, where n is a natural number prime to the characteristic of K. Then (1) An n-Kummer extension L/K is a Galois extension and Gal(L/K)is an abelian group of exponent n. √ (2) If L/K is an abelian extension of exponent n, then L = K( n Δ), where Δ ⊆ K∗ and (K∗)n ⊆ Δ. We will present a generalization of Kummer extensions via the Carlitz- s Hayes action. In what follows, p denotes a prime number, q = p , s ∈ N, RT denotes the ring Fq[T ], k denotes the field of rational functions Fq(T ) and k denotes an algebraic closure of k.
In Section 2 we present some facts over RT -modules that are used in Section 3. The main part of this work is precisely Section 3. We will discuss extensions L/K such that k ⊆ K ⊆ L ⊆ k. Next we give a brief outline of the Carlitz Hayes action. For details see [2] and [7].
q Let ϕ be the Frobenius automorphism ϕ : k → k, ϕ(u)=u , and let μT be the homomorphism μT : k → k, μT (u)=Tu. We have an action of RT in M k given as follows: if M ∈ RT and u ∈ k, then u := M(ϕ + μT )(u). It can be shown that with this definition, k becomes an RT -module, see [7] Chapter M 12. We have that z is a separable polynomial in z of degree d = deg(M). M M d qi Moreover, the polynomial z can be written as z = i=0(M,i)z , where (M,i) ∈ RT ,(M,0) = M and (M,d) is the leader coefficient of M, see [7] Chapter 12.
On the other hand, assuming that M ∈ RT \{0}, the M-torsion set of k, M denoted by ΛM ,isΛM := {u ∈ k | u =0}. We also call ΛM to be the set of M-roots of Carlitz. Note that if a ∈ k then set of all roots of the polynomial M M z − a is {α + λ | λ ∈ ΛM }, where α is any fixed root of z − a in k. Proposition 3.2 and Theorem 3.4 are analogous to facts (1) and (2) listed above, except that we only consider finite extensions. Kummer type extensions in function fields 159
2 Module theory
In this section, unless otherwise indicated, all modules and homomorphisms considered are RT -modules and RT -homomorphisms, respectively.
Let A be a module, and a ∈ A. Let ϕa : RT → A be the homomorphism defined by ϕa(M)=Ma. Definition 2.1. We say that A is a cyclic module, if there exists a ∈ A, such that ϕa is a surjective homomorphism. Note that the above definition is equivalent to say that there exists a ∈ A such that A =(a).
Let a ∈ A and consider the kernel of the homomorphism ϕa, ker(ϕa). If ker(ϕa) = {0} there exists a nonzero polynomial M that we may assume to be monic, such that ker(ϕa)=(M). Definition 2.2. Let a ∈ A. We say that a has infinite order if the kernel of ϕa is zero. We say that a has finite order M, where M is a monic polynomial, if ker(ϕa) is nonzero and ker(ϕa)=(M). Now if A is a module, an exponent of A is a nonzero M ∈ RT , such that Ma = 0 for each a ∈ A. Remark 2.3. If A is a finite cyclic module, there exists a ∈ A such that ϕa is a surjective homomorphism. Therefore there exists M ∈ RT \{0} such that
(i) ker(ϕa)=(M) and ∼ (ii) RT /(M) = A. As before, we may replace M by a monic polynomial and we say that A has order M. The proof of the following lemma is straightforward.
Lemma 2.4. Let A be a cyclic module of order N, with N =0 .LetN1 be a monic divisor of N. Then there exists a submodule of A, of order N1. Remark 2.5. With the conditions of Lemma 2.4, by Remark 2.3 we have Na =0.
On the other hand, if B is a cyclic submodule of A of order N1 then, again by Remark 2.3, there exists b ∈ B, such that ϕb is surjective and ker(ϕb)= (N1). Since b ∈ A, there exists N2 ∈ RT , such that b = N2a.
Now, because Nb = N(N2a)=N2(Na) = 0 we have N ∈ ker(ϕb)= (N1). Therefore N1 is a divisor of N. Since all cyclic modules of order M are isomorphic to RT /(M) we have there is only one cyclic module of order M,it follows that for each monic divisor M of N, there is a unique cyclic submodule of A of order N. 160 M. S´anchez-Mirafuentes and G. Villa-Salvador
Analogously to the case of cyclic groups Cm, we denote by CM the module RT /(M) which is cyclic of order M. Definition 2.6. Let A be a module. We denote by A or by HomRT (A, CM ), the group of homomorphisms from A into CM , A is the dual module of A.
Let f : A → B be a homomorphism, such that A and B have exponent M. Then we have a homomorphism f : B → A defined by f(ψ)=ψ ◦ f. Note that this is a contravariant functor, namely () : RT -modules of exponent M → RT -modules such that if f : A → B and g : B → C are homomorphisms, then (1) g ◦ f = f◦ g and (2) 1=1. Lemma 2.7. If A is a finite module, with exponent M such that A = B×D, then A is isomorphic to B × D.
Proof. The natural projections π1 : B × D → B and π2 : B × D → D induce homomorphisms π1 : B → B × D and π2 : D → B × D. Therefore we may define θ : B × D → B × D given by θ(ψ1,ψ2)=π1(ψ1)+π2(ψ2), where (ψ1,ψ2) ∈ B × D. We have θ is a homomorphism. Moreover if ψ ∈ B × D, since ψ is a homomorphism, we get ψ(x, y)=ψ(x, 0) + ψ(0,y) for all (x, y) ∈ B × D. Now, if we define ψ1 : B → CM by ψ1(x)=ψ(x, 0) and ψ2 : D → CM by ψ2(y)=ψ(0,y) then ψ1 and ψ2 are homomorphisms. Thus, we obtain the induced map δ : B × D → B × D given by δ(ψ)=(ψ1,ψ2) which is a module homomorphism whose inverse is θ. This completes the proof. Proposition 2.8. A finite module A with exponent M is isomorphic to its dual. In other words ∼ A = A ∼ Proof. By Theorem 4.7, Chapter 5 of [3], we can write A = P AP . The above sum is over all irreducible polynomials P and AP denotes the set of elements of A having order a power of P . ∼ Now by Theorem 4.9 of Chapter 5 of [3], each AP can be written as AP = CP α1 ⊕ ...⊕ CP αk , where α1 ≥ ...≥ αk ≥ 1, and each CP αi is a cyclic module αi αi whose generator has order P , so each CP αi has order P . Note that each AP and each CP αi have exponent M. Kummer type extensions in function fields 161
By the above observation and by Lemma 2.7, we may assume that A is cyclic generated by a of order P α, where α ∈ N and P is irreducible. Hence α the function ϕa is an epimorphism and (P )=ker(ϕa). Since M is an exponent of A, we have P α | M. Now from Lemma 2.4 and α Remark 2.3, CM has a unique cyclic submodule of order P . We denote such module by CP α . The homomorphism ϕa : RT → A induces an isomorphism, α denoted again by ϕa : RT /(P ) → A.
The inverse of the isomorphism ϕa will be denoted by ψ. Let y = ψ(a). Then y is a generator of CP α . The composition ψ with the natural inclusion CP α → CM , gives an element of A, also denoted by ψ. Let ϕ ∈ A. Note that Im(ϕ) ⊆ CM is a cyclic submodule of order N. Thus, there exists w ∈ Im(ϕ) such that w = ϕ(aw) where aw ∈ A generates Im(ϕ) and its order is N. α α On the other hand P ∈ (N). Therefore P = ND, for some D ∈ RT . γ γ Consequently N = P for some γ ≤ α, i.e., w has order P and as aw generates Im(ϕ), we have Im(ϕ) ⊆ CP α . Now ϕ is determined by its action on a, where a ∈ A is a generator of A. Therefore ϕ(a)=Ny.NowifψN = Nψ then ψN (a)=Nψ(a)=Ny = ϕ(a), that is, ϕ = ψN ∈ (ψ). P α ∼ So A =(ψ) and the order of A is qdeg( ). Therefore A = A. Definition 2.9. Let A and B be modules. A bilinear map of A × B into a module C, denoted by (a, b) →
Let A and B be the left and right kernels of the bilinear map respectively. ∈ → An element b B gives an element in HomRT (A, C) given by a
Similarly we obtain the exact sequence → → 0 A/A HomRT (B/B ,C). (2) Proposition 2.10. Let A × A → CM be a bilinear map of modules into a cyclic module CM of finite order M.LetB, B be their respective left and right kernels. Suppose that A /B is finite with exponent M and that A/B have exponent M. Then A/B is finite and A /B is isomorphic to the dual module of A/B. Proof. From the exact sequences (1) and (2) it follows that the following se- quences are exact → → 0 A /B HomRT (A/B, CM ) (3) and → → 0 A/B HomRT (A /B ,CM ). (4)
From (4) we obtain that A/B can be viewed as a submodule of A /B . From this follows the finiteness of A/B. On the other hand we obtain from the exact sequences (3) and (4) and Proposition 2.8
card(A/B) ≤ card((A /B )) = card(A /B ). (5) and card(A /B ) ≤ card(( A/B)) = card(A/B). (6) the second equality follows from Proposition 2.8. Now from (3) we have obtain injective map φ : A /B → A/B . Then by (5) and (6), φ is surjective, and this finishes the proof.
3 Kummer theory
In this section we give a generalization of Kummer extensions, a little different from those given by Chi and Li [1] and Schultheis [6]. In what follows we will assume that the extensions considered are subextensions of k/k. Let M ∈ RT be a non-constant polynomial and ϕ : K → K defined by ϕ(u)=uM , where K = k(ΛM ). Then ϕ is a homomorphism. Moreover consider a submodule B of K under the action of Carlitz-Hayes containing KM = ϕ(K). √ Let b ∈ K we shall use the symbol M b to denote any such element β, which will be called an M-root of b. Since the M-roots of Carlitz are in K,we observe that field K(β) is the√ same no matter which M-root β of b we select. We denote this field by K( M b). For our next proposition we need the definition give in [1]. Kummer type extensions in function fields 163
Definition 3.1. An abelian extension L/K with Galois group G, is said to be an RT -abelian extension if its Galois group has an RT -module structure. An RT -abelian extension L/K is said to be of exponent M if G is an M-torsion RT -module, i.e., M · σ = 1 for all σ ∈ G.
M Proposition 3.2. (1) Let B be an RT -submodule√ of K containing K M and let KB be the composite of all fields K( b) with b ∈ B. Then KB/K is a Galois abelian extension.
(2) Assume that KB/K is an RT -abelian extension of exponent M with Ga- lois group G. Then there is a bilinear map
G × B → ΛM given by (σ, a) →<σ,a>
where <σ,a>= σ(α)−α and αM = a. The left kernel is 1 and the right kernel is KM . M The extension KB/K es finite if only if (B : K ) is finite. In this case we have M ∼ B/K = G In particular we have the equality
M [KB : K]=(B : K ).
Proof. (1) Let b ∈ B and let β be an M-root of b. The polynomial zM −b splits into linear factors in KB for all b ∈ B.ThusKB/K is a Galois extension.
Now let G = Gal(KB/K), σ ∈ G, b ∈ B and β a root of the polynomial M Mσ z − b. Then σ(β)=β + λ for some Mσ ∈ RT , where λ is a generator of ΛM . Therefore the map σ → Mσ is a injective homomorphism of G into ΛM .
(2) We define G × B → ΛM by (σ, b) →<σ,b>, where <σ,b>= σ(β) − β and βM = b. This definition is independent of the choice of the M-root of b. We have <σ,a+ b>=<σ,a>+ <σ,b>for all a, b ∈ B and since the definition of <σ,b>is independent of the choice of the M-root of b, it follows <σ· τ,b >=<σ,b>+ <τ,b>. Let σ ∈ G. Suppose <σ,a>= 0 for all a ∈ B. Then for every α of M KB such that α = a ∈ B we have σ(α)=α. Hence σ induces the identity on KB and the left kernel is 1. On the other hand, let a ∈ B and suppose <σ,a>= 0 for all σ ∈ G. Then σ(α)=α for all σ ∈ G. Therefore α ∈ K and a = αM ∈ KM . So the right kernel is KM . Now suppose that B/KM is finite. Since the right kernel is KM , from Proposition 2.10 we obtain that G = G/1 is finite. In particular KB/K is 164 M. S´anchez-Mirafuentes and G. Villa-Salvador
finite. On the other hand if KB/K is finite then, from Proposition 2.10, we have that the sequence → M → 0 B/K HomRT (G/1, ΛM )
M is exact. Since HomRT (G/1, ΛM ) is finite, it follows that B/K is finite. Finally, since by Proposition 2.10 B/KM is isomorphic to the dual module M ∼ M of G, we have that B/K = G,so[KB : K]=(B : K ). For our next result we need definition given in [1]
Definition 3.3. An RT -abelian extension L/K is said to be RT -cyclic if ∼ its Galois group is a cyclic RT -module. In this case, if Gal(L/K) = RT /(M), where M is a monic polynomial, we say that the RT -cyclic extension L/K is of order M.
In the following theorem let M denote the set of RT -submodules which M contain K and let F denote the set of RT -abelian extensions of K with exponent M.
Theorem 3.4. With the notation of Proposition 3.2, the function ϕ : M → F given by ϕ(B)=KB is injective. Also if L/K is a finite RT -abelian extension finite of exponent M, then there is a submodule B of K containing KM , such that L = KB.
Proof. To show that the function ϕ is injective, it is enough to prove that if ⊆ ⊆ KB1 KB2 then B1 B2, since from the equality ϕ(B1)=ϕ(B2√), it follows ⊆ ⊆ ∈ M ⊆ that KB1√ KB2 and that KB2 KB1 . Let b B1. Then K( b) KB2 M and K( b) is contained in a finitely generated subextension of KB2 .Thus M we may assume, without loss of generality, that B2/K is finitely generated, hence finite. M Now let β be such that β = b. Let B3 be the submodule of K generated ⊆ by B2 and b. We will show that KB2 = KB3 . We have KB2 KB3 .To ⊇ ∈ show KB2 KB3 , let α be any M-root of c B3. Therefore c is of the form N s Ni ∈ M N s Ni MN s MNi b + i=1 bi , with bi B2. Then α =b + i=1 bi = β + i=1 βi M N s Ni A with βi = bi, i =1,... ,s, i.e., α = β + i=1 βi +λ , where λ is a generator of ΛM . Therefore K(α) ⊆ KB2 . It follows that KB3 ⊆ KB2 . M M Thus by Proposition 3.2 (2) (B2 : K )=(B3 : K ). Hence b ∈ B2,so that B1 ⊆ B2. On the other hand, let K be a finite RT -abelian extension of K of exponent M. Let G = Gal(K /K). Then, by Theorem 4.7 and Theorem 4.9 Chapter 4 of [3], G is a finite direct sum of RT -submodules of exponent M. By Galois theory we may assume that the extension K /K is cyclic of exponent M.Now, Kummer type extensions in function fields 165
by Proposition 2.6 of [1], we have that every RT -cyclic extension of exponent M is obtained by attaching an M-root of an element of K. M Therefore there exits sets {bj}⊆K, {αj}⊆K such that αj = bj and M K = K({αj}). Let B be the submodule of K generated by {bj} and K . Then M K ⊆ KB. On the other hand, consider an M-root of c ∈ B,sayα. So, α = c. s Nj M ∈ s Nj Other hand c = j=1 bj + a , a K. Then, we have α = j=1 αj + a therefore K(α) ⊆ K . It follows that KB ⊆ K and ϕ(B)=K . This completes the proof.
Proposition 3.5. Let L/K be a finite RT -abelian extension. Assume ΛN ⊆ K, with N ∈ RT .Let √ W = {a = a + KN ∈ K/KN | N a ∈ L}. ∼ Then W = Hom(G, ΛN ), where G = Gal(L/K).
Proof. Let a ∈ W . We define a function ϕa : G → ΛN , given by ϕa(σ)= N σ(α) − α, where α is an N-root of a, i.e., α = a. The definition of ϕa is independent of the root used. Note that
ϕa(σ ◦ τ)=σ(τ(α)) − α = σ(τ(α) − α + α) − α = σ(τ(α) − α)+σ(α) − α = τ(α) − α + σ(α) − α.
Thus ϕa is a homomorphism of abelian groups. Therefore it is possible to define f : W → Hom(G, ΛN ) given by f(α)=ϕα.
We have that f is a homomorphism of abelian groups. Now if f(a)=ϕa =0 then σ(α) − α = 0, for all σ ∈ G. In this way we obtain that α ∈ K. Since a = αN then a ∈ KN . In this way a =0,sof is injective.
Now let ϕ : G → ΛN be a homomorphism of abelian groups. Then
ϕ(σ ◦ τ)=ϕ(σ)+ϕ(τ)=ϕ(σ)+σ(ϕ(τ)), that is, ϕ is a crossed homomorphism. By the additive Hilbert Theorem 90, there exists α ∈ L such that ϕ(σ)=σ(α) − α. Therefore (σ(α) − α)N = σ(αN ) − αN =0.Thusa = αN ∈ K, and this shows that the function f is surjective.
References
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Received: January, 2013