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Computing Class Fields Via the Artin Map

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Computing Class Fields Via the Artin Map MATHEMATICS OF COMPUTATION Volume 70, Number 235, Pages 1293{1303 S 0025-5718(00)01255-2 Article electronically published on March 24, 2000 COMPUTING CLASS FIELDS VIA THE ARTIN MAP CLAUS FIEKER Abstract. Based on an explicit representation of the Artin map for Kummer extensions, we present a method to compute arbitrary class fields. As in the proofs of the existence theorem, the problem is first reduced to the case where the field contains sufficiently many roots of unity. Using Kummer theory and an explicit version of the Artin reciprocity law we show how to compute class fields in this case. We conclude with several examples. 1. Preliminaries Let k=Q be an algebraic number field; we denote its ring of integers by ok.A congruence module m formally consists of an integral ideal m0 and a formal product (i) m m1 of real places (:) of k viewed as embeddings into R.ByI we denote the set of all ideals coprime to m0,byPm the set of principal ideals generated by elements ∗ (i) (i) α ≡ 1mod m (i.e., α ≡ 1modm0 and α > 0 for all (:) jm1) and finally by m Clm := I =Pm the ray class group modulo m. There are efficient algorithms for computing ray class groups [6, 17] provided we already know the unit and class group of k. m An ideal group H (defined mod m) is a subgroup of I containing Pm. For any ideal group H let H¯ := Im=H. Now, let K=k be a finite extension. By dK=k we denote the relative discriminant of K=k as an ideal of ok. From now on K=k will always be a finite abelian extension. In this context we denote by σp the Frobenius automorphism belonging to the unramified prime ideal p of k. We will make extensive use of the Artin map, the multiplicative extension of the function mapping unramified prime ideals to their Frobenius automorphism: Y Y dK=k vp(a) vp(a) (:; K=k):I ! Gal(K=k):a = p 7! σp ; pja pja N(p) where σp(x) ≡ x mod P for any Pjp and any x 2 oK . With this in mind we can define class fields: Definition 1.1. Let m be a congruence module, H be an ideal group and K=k be (:;K=k) an abelian extension. K is the class field belonging to H iff Gal(K=k) ' Im=H. Received by the editor April 6, 1999 and, in revised form, August 16, 1999. 2000 Mathematics Subject Classification. Primary 11Y40; Secondary 11R37. Key words and phrases. Computational algebraic number theory, class field theory, Artin reciprocity. c 2000 American Mathematical Society 1293 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1294 CLAUS FIEKER The famous existence theorem of class field theory states that for every ideal group there is a class field and every abelian extension is the class field to a certain ideal group. The main theorems of class field theory used here can be found in text books treating algebraic number theory, e.g., [11, 15, 16]. Later we will give an algorithm to compute the class field corresponding to H. The algorithm will follow closely the proof of the existence theorem as given in [15, XI x2]. First we reduce the problem to the case when k contains “sufficiently many" roots of unity. In this case we can use Kummer theory and the Artin map to compute the class fields. As a last step we have to compute a certain subfield of this large class field, again using the Artin map and elementary Galois theory. 2. The Artin map We recall the following properties of the Artin map ([15, X, x1: A2, A4, Thm. 1{3]: Theorem 2.1. Let K=k be a finite abelian extension. 1. Let K0=K=k with abelian K0=k,then 0 (:; K =k)jK =(:; K=k): 2. Let E=k be finite, then (:; KE=E)jK =(:; K=k) ◦ NE=k : 3. If m is divisible by all ramified primes, then (Im;K=k)=Gal(K=k): 4. There exists a congruence module m such that Pm ⊆ ker(:; K=k),anysuch module m is called admissible. Later on, it will be important to have an efficient method for actually computing (a;K=k) for some ideals a of k.LetK=k be a finite abelian extension, and σ1, :::, σn be the k-automorphisms of K. For arbitrary (abelian) extensions K=k it is a hard problem to compute the σi [1, 13], but for the extensions occurring in our context, we already know the whole Galois group. Since the Artin map is the multiplicative extension of the Frobenius map, we start with the investigation of Frobenius automorphisms. Let p be an prime ideal of k which is unramified in K.Wehave N(p) σp(x) ≡ x mod Pi for all prime ideals Pijp, and therefore N(p) σp(x) ≡ x mod poK for all x 2 oK . Since we assume knowledge of all automorphisms, we simply com- N(p) pute x mod poK and compare this to σ(x)modpoK for all σ 2 Gal(K=k). If we do this for sufficiently many elements we can easily identify σp. Later on we are mainly interested in Kummer extensions. In this case we can speed up the computationsp quite a bit. Let K := k(y) be a Kummer extension with generator y := n µ (for some µ 2 k). Without loss of generality we can assume that License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use COMPUTING CLASS FIELDS VIA THE ARTIN MAP 1295 i N(p) q r σi(y)=ζny holds. On the other hand we have y = µ y for N(p)=qn + r, 0 ≤ r<n. In basis representation w.r.t. the basis 1, y, y2, :::, yn−1 the congruence N(p) σi(y) ≡ y mod poK reads: i ≡ q (0,ζn; 0;::: ;0) (0;::: ;0,µ ; 0;::: ;0) mod poK (where µq is the (r +1)st component) and therefore neccessarily r = 1 holds. Since the determination of i is equivalent to the computation of σi = σp,weseethat for Kummer extensions the Frobenius automorphism can be computed in the base field. Note 2.2. For an arbitrary ideal a we have two possibilities: either use the defini- tion and factor the ideal, compute the automorphisms corresponding to the prime divisors and compose them; or establish the surjection from the ray class group into the Galois group. The main problem with the first approach is that we frequently need to factor large ideals having norms of more than 1000 digits and large prime factors. For the second approach we need the images of certain ideals generating the ray class group. Numerical experiments show that very few (compared to the group order) small ideals usually suffice to generate the ray class group, as one would expect (see [2]). So it makes sense simply to choose small random prime generators for the ray class group to avoid the factoring and use them to get the surjection. 3. Class fields In this section we reduce the problem to the case where the field contains suffi- ciently many roots of unity. Q ¯ ¯ t ¯ Let H be the product of cyclic groups of prime powerQ order: H = i=1 Hi. t Using the Artin map, it follows immediately that K = i=1 Ki with Ki belonging to H . i ∼ From now on we assume H¯ = Cpr for some prime p.LetE := k(ζpr ), then F := KE is the class field over E belonging to some ideal group HE.Since ⊆ NE=k(PmoE ) Pm we can define HE modulo moE.Using ~ ! m 7! NE=k :ClmoE I =H : aPmoE NE=k(a)H; ¯ m −1 ¯ we easily compute HE = IE =HE = Pm;ENE=k(H) as a pre-image of a homomor- phism between finite groups. Using the approximation theorem, we see that we can use any multiple of moE to define HE. Let S be a finite set of places of E such that S contains all primes dividing m and p and enough primes to generate thep (ordinary) class group of E, finally let j j pr s := S . We consider the field G := E( US),p where UpS are the S-Units of E. r r By the Dirichlet unit theorem we have G = E( p 1;::: ; p s), where 1;::: ,s−1 generate a free group and s is a generator for the torsion units of oE; hence G is r s s a Kummer extension of degree (p ) over E with Galois group isomorphic to Cpr . Analyzing the proof of the existence theorem presented in [15], we see that G contains the class field F over E that we are looking for; we are going to compute F as a subfield of G. To illustrate the relation of the various fields, we have included License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1296 CLAUS FIEKER r the following diagram (n := p , in the most simple case, i.e., Q(ζn)andk have no common subfield and (F : E)=n): Since G is abelian over E, there is an admissible module m~ such that Im~ ! s ! Gal(G=E)=Cpr is surjective and Pm~ is contained in the kernel, i.e., Clm~ Gal(G=E) is a well-defined epimorphism. Since m~ must be divisible by all ramified primes (and by m)weneedtogettheir exact powers or at least upper bounds.
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