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78 D R HAYES, with coefficients in RT whose set of roots Aw the M torsion points of A 0. generate a finite abelian extension field k AM of k The properties of these. extensions are quite similar to those of the cyclotomic extensions of Q. Carlitz arrives at his definition of the RT action in a remarkable way The. choice of T singles out an infinite prime of k namely the unique pole Px of T. In a previous paper 3 he had defined an analytic function ip u on kx the. completion of k at with properties closely resembling those of the function. exp on the real numbers This p u is defined by an everywhere convergent. power series with coefficients from k Carlitz then notices that for a given. M G RT p Mu uM u where wM is a uniquely determined additive poly. nomial over RT The properties of p u made it evident that the action. M u uM u gives the additive group of cac the structure of an rmodule. Carlitz was able to give a purely algebraic description of uM and therefore of the. RT action He noted also that the roots of wM are expressible in the form. i A M where is a transcendental element lying in the completion of the. algebraic closure of kx and A is in RT Thus the elements of AM are the values. of an analytic function at the rational points of the form A M For the details I. refer the reader to the original papers, Carlitz makes a careful study of the polynomials uM u and proves the analog. of the theorem which states that the cyclotomic polynomial is irreducible over Q. In 1 4 below I give an exposition of Carlitz results in the language of modern. algebraic number theory A discussion of how the prime splits in k hM is also. included This question does not arise naturally in Carlitz set up but it is crucial. for the application to the class field theory of k For what it is worth I also. calculate the different of k AM when M is a power of an irreducible and thereby. get a formula for the genus of k AM in that case, 5 6 and 7 are modeled after the usual explicit construction of the norm. residue symbol for cyclotomic extensions of Q see Chapter 7 of 1 or Chapter. 7 of 4 It turns out that the extensions k AM together with the constant field. extensions almost generate the maximal abelian extension of k What is lacking. is a piece containing the extensions where P is wildly ramified This piece is. constructed out of the theory which results from the choice of 1 7 as generator. instead of T, It is perhaps surprising that the results come by using endomorphisms of the. additive group of kiC Note however that the formal group law constructed from. itT Tq in the Lubin Tate theory is just X Y in the equicharacteristic case. This gives some hope that the additive group might be used in a similar way to. do explicit class field theory for an arbitrary function field in one variable over Fr. 1 The RT action As above k is the field of rational functions over the finite. field F of q elements We arbitrarily choose a generator T of k and put. R T Fq T the polynomial subring of k generated by T over q Most of the. results will be relative to this choice of T although this fact is suppressed more. or less in the notation, License or copyright restrictions may apply to redistribution see https www ams org journal terms of use.

EXPLICIT CLASS FIELD THEORY 79, Let A 0be the algebraic closure of tc The F algebra Endive40 of all Fq. endomorphisms of the additive group of fe 0contains the Frobenius automor. phism pdefined by p uq and the map ir defined by Lj u Tu Since RT. is a polynomial ring over F the substitution T t p tr yields a ring homomor. phism RT End 7cac which provides ktc with the structure of an j module If. we write uM for the action of M G RT on G fcac then we have. 1 1 M M tp ftT u, Note that for a G F a aw so that our RT action respects the E algebra. structure of tc,Proposition 1 1 Ifd deg M then, where each f is a polynomial in RT of degree d i q Further M and. sthe leadingcoefficientof M, Proof Since each element of RT is an E linear combination of powers of T it. suffices to verify the proposition for the special case M Td The endomor. phisms p and iT do not commute but rather obey the rule p ir if o p. Therefore one can write p nT d as a sum of terms of the form if o p Since. if o p u r5 we see that uM is indeed a polynomial in u of the form 1 2. For d there is a unique term pdand w u d For 0 there is a unique. term if and if Tdu For 0 d there is a unique term with maximum. 5 namely qp tf i p,This completes the proof, Put 0 for i 0 and deg M In calculating the polynomial f one.

can make use of the following easily established properties. In 2 Equation 1 6 Carlitz gives an explicit formula for these polynomials. Definition 1 2 Let Aw denote the set of M torsion points of 7cac i e the set of. zeros of the polynomial uM Since RT is commutative AMis an rsubmodule of. Proposition 13 As a polynomial in u over k uM is separable of degree qd where. d deg M The submodule AM is finite of order qd and is therefore a vector space. over Fqof dimension d, License or copyright restrictions may apply to redistribution see https www ams org journal terms of use. 80 D R HAYES, Proof From Proposition 1 1 we see that uM is of degree qd in u and that its. derivative with respect to u is just if M The proposition follows immediate. The structure of the Pj module AM will now be determined As one might. expect from the analogous cyclotomic theory Au turns out to be a cyclic Rf. Proposition 1 4 Let M a TJ P be a factorization of M into powers of monk. irreducibles Then,1 3 Aw 2 A,and the sum is direct. Proof This follows from the general theory of modules over principal ideal. domains In fact AP is the P primary submodule of Au and so 1 3 is the. canonical decomposition of AMinto its P primary components. Proposition 1 5 If M P where P is irreducible then AM is a cyclic Rr. Proof Let d deg P The proof goes by induction on n For n 1 A is a. vector space over RT P Since both Rt P and Ay contain qd elements AP is. 1 dimensional hence cyclic over RT P and hence cyclic over RT Now assume. the proposition true for n k k 1 The map m h ur from A A is. surjective since its domain kernel and range contain respectively qd k l qd and. qdk elements Since Apk is cyclic by the induction hypothesis one can therefore. choose X G Apic iso that X generates A This X will generate APk over RT To. prove it let i E Apk nbe given Then choose A E RT such that pr XPA Then. H Xa belongs to AP Now X G AP is not zero since X generates A. Therefore since AP is a 1 dimensional vector space over RT P there is a. B E RT such that i Xa XF B We conclude that y XA pkB Therefore X. generates Apk and the proof is complete, Theorem 1 6 The RTmodule AM is naturally isomorphic to RT M for every. Proof Since by Propositions 1 4 and 1 5 each of the P primary components of. AM is cyclic AM is itself cyclic Therefore AM is naturally isomorphic to the. quotient of RT by the annihilator ideal of AM Clearly the ideal M is contained. in that annihilator On the other hand both AM and RT M have qd elements. where d deg M Therefore M must equal the annihilator of AM and the. proof is complete, Definition 1 7 If M E RT M 0 then Af is the order of the group of.

units of RT M, License or copyright restrictions may apply to redistribution see https www ams org journal terms of use. EXPLICITCLASSFIELD THEORY 81, Corollary 1 8 The cyclic Rrmodule AM has exactly 3 A generators In fact if. X is a given generator and A G RT then Xa is a generator if and only if A and M. are relatively prime, 2 The fields k AM One knows that with one exception the prime divisors of. the rational function field k correspond one to one to the monic irreducible. polynomials P in RT The exception is the unique pole of T the infinite. prime For convenience I will use the symbol P to denote both a monic. irreducible and the prime divisor to which it corresponds No confusion should. Consider now the extension field k AM of k which arises by adjoining to k the. elements of the finite module AM Let X be a generator of AMover RT Theorem. 1 6 Since Xa is a polynomial in X with coefficients from Rt Xa G k X for every. A G RT It follows that one can obtain k AM by adjoining to k a single. generator of AM Also since AMis the set of zeros of the separable polynomial. uM over RT C k the extension k Au k is finite and Galois Further the. elements of AM are all integral over RT since by Proposition 1 1 the leading. coefficient of uM belongs to F, Let GMbe the Galois group of k AM k The action of GMcommutes with the. 7 action since the raction is given by a polynomial over tc Choose a. generator of AM Since X also generates the field extension every a G GM is. determined by its action on X We must have o X Xa for some A relatively. prime to M since o must map a generator of AMto another generator Further. this A does not depend on the choice of the generator X Therefore the map. a r A mod A is a well defined injection of GM into the group of units of. RT M One easily verifies that this injection is a group homomorphism We. have thus proved the following, Theorem 2 1 The Galois group GM is isomorphic to a subgroup of the group of.

units of RT M The Galois extension k AM k is abelian and k AM k. This last theorem does not tell the whole story since actually the map from GM. into the group of units of RT M is an isomorphism One way to prove it is to. examine the ramification at the prime divisors which correspond to the irreduci. ble factors of M just as one does in the usual cyclotomic theory. Proposition 2 2 Suppose M P where P is a monic irreducible polynomial in T. with deg P d Then every prime divisor of k except P and R is unramified in. k AM and the ramification number of P is A o o 1, Proof Let IMbe the integral closure of RT in k AM Since RT is a Dedekind. ring so is IM We must determine which finite prime divisors of k divide the. discriminant D IM of IM over RT Let be a generator of AM Then RT X is a. subring of Iu and its discriminant D X divides the divisor of Norm A where. License or copyright restrictions may apply to redistribution see https www ams org journal terms of use. 82 D R HAYES, f u is any polynomial over RT which has X as a root Take h um Then by. Proposition 1 1 h M P a constant polynomial over RT Therefore P. is the only prime divisor of RT which enters into D X Since D IM divides D X. it follows that P is the only prime divisor of RT which divides D IM Therefore. except maybe for the only ramification of the extension k AM k occurs at P. In order to calculate the ramification number at P proceed as follows Note. first that uf up x up x h for some polynomial h over RT since h. divides up by Proposition 1 1 Therefore h up up P higher terms. The roots of are obviously exactly the generators of the module AM Therefore. where A runs over a set of representatives of the group of units of RT M Now. X divides Xa in IM since u divides uA By symmetry Xa also divides X Therefore. Xa unit X Substituting this in 2 1 we find that,2 2 P unit X J. The ramification number eP of P is therefore greater than 3 A But also. e k AM k A Therefore eP k AM k A This com,pletes the proof. The main result is a corollary of this last proposition. Theorem 23 The extension k AM k has degree M and the Galois group GM. is isomorphic to the group of units of RT M, Proof By Theorem 2 1 it is enough to prove that the degree equals M For.

M P this follows from Proposition 2 2 If M has the factorization M. a n P where each P is a monic irreducible then the total ramification of. k APi at P shows that each extension k APn k for P dividing M is linearly. disjoint from the composite of the remaining ones Therefore. k AM k U lk AP y k U W M,and the proof is complete. One last result is needed for use in 4 and 7 below The analogous result in. the cyclotomic theory can be proved directly from properties of binomial. coefficients One can devise a similar direct proof which works here but we give. a proof based on Proposition 2 2, Proposition 2 4 If M P where P is a monic irreducible in RT then. f u up up is an Eisenstein polynomial over RT at P. Proof Let X be a generator of Aw From the proof of Proposition 2 2. 2 3 f u JI u XA, License or copyright restrictions may apply to redistribution see https www ams org journal terms of use. EXPLICIT CLASS FIELD THEORY 83, where A runs through a set of representatives of the group of units of RT M. Let t be the unique prime divisor of k AM lying over P From 2 2 and the total. ramification at P it follows that ordpX 1 and the same holds true of the. generator Xa Therefore 2 3 shows that the coefficients of all but the highest. order term in belong to the valuation ring at and hence are divisible by P. in RT Since the constant coefficient is P is Eisenstein and the proof is. Corollary 2 5 Suppose P is a finite prime of k which does not divide M Then the. automorphism pPof k AM which takes X in AM to Xp is that given by the Artin. Proof Let d deg P Consider a given generator X G AM Let the Artin. symbol take X to XL for suitable L G RT By definition we have XL A. mod where is a prime of k AM lying over P But also Xp X mod t by. the above proposition Now, Taking the derivative of both sides of this equation and recalling that the.

derivative of uM is just M we get,M II XB X A B, for every B in RT Since P does not divide M this means that the Xa A mod M. EXPLICIT CLASS FIELD THEORY FOR RATIONAL FUNCTION FIELDSO BY D R HAYES ABSTRACT Developing an idea of Carlitz I show how one can describe explicitly the maximal abelian extension of the rational function field over F the finite field of q elements and the action of the id le class group via the reciprocity law homomorphism

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