Let p be an odd prime number and let K be an imaginary
quadratic number field whose class number is not divisible by p. For a set S of primes of K whose norm is congruent to 1 modulo p, we introduce the notion of strict circularity. We show that if S is strictly circular, then the group G(KS(p)=K) is of cohomological dimension 2 and give some explicit examples

Vogel, Denis

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Circular sets of primes of imaginary quadratic number fields

Universita¨t RegensburgMathematikCircular sets of primes of imaginaryquadratic number fieldsDenis VogelPreprint Nr. 13/2006CIRCULAR SETS OF PRIMES OF IMAGINARYQUADRATIC NUMBER FIELDSDENIS VOGELAbstract. Let p be an odd prime number and let K be an imaginaryquadratic number field whose class number is not divisible by p. Fora set S of primes of K whose norm is congruent to 1 modulo p, weintroduce the notion of strict circularity. We show that if S is strictlycircular, then the group G(KS(p)/K) is of cohomological dimension 2and give some explicit examples.1. IntroductionLet K be a number field, p a prime number and S a finite set of primesof K not containing any primes dividing p. Only little has been known onthe structure of the Galois group G(KS(p)/K) of the maximal p-extensionof K unramified outside S, in particular there has been no result on thecohomological dimension of G(KS(p)/K). Recently, Labute [La] showedthat pro-p-groups whose presentation in terms of generators and relations isof a certain type, so-called mild pro-p-groups, are of cohomological dimen-sion 2. If K = Q, Labute used results of Koch on the relation structureof G(QS(p)/Q) and ended up with a criterion on the set S for the groupG(QS(p)/Q) to be of cohomological dimension 2. Schmidt [S] extended theresult of Labute by arithmetic methods and weakened Labute’s conditionon S.The objective of this paper is to study the case where K is an imaginaryquadratic number field whose class number is not divisible by p. In the firstsection we introduce the notions of the linking number of two primes andof strict circularity of a set of primes of K, all of this in complete analogywith the case K = Q. Using Labute’s results we obtain the criterion that ifS is strictly circular then G(KS(p)/K) is a mild pro-p-group and hence ofcohomological dimension 2. In the following section we give some explicitexamples of strictly circular sets of primes, and in section 4 we study howa strictly circular set T can be enlarged to set S of primes of K, such thatG(KS(p)/K) has cohomological dimension 2 as well.2. Linking numbers and strictly circular setsLet p be an odd prime number and K an imaginary quadratic numberfield whose class number is not divisible by p, and which is different fromQ(√−3) if p = 3. Let S = {q1, . . . , qn} be a set of primes of K whosenorm is congruent to 1 mod p. For a subset T of S, we denote the maximalDate: May 12, 2006.12 DENIS VOGELp-extension of K unramified outside T by KT (p), and we put GT (p) =G(KT (p)/K).Let IK denote the ide`le group of K, and for a subset T of S let UT be thesubgroup of IK consisting of those ide`les whose components for q ∈ T are 1and for q 6∈ T are units. For q ∈ S we denote by Kq the completion of Kat q and by Uq the unit group of Kq. Furthermore, let piq be a uniformizerof Kq and let αq be a generator of the cyclic group Uq/Upq . Let Q be anextension of q to KS(p). We let σq be an element of GS(p) with the followingproperties:(1) σq is a lift of the Frobenius automorphism of Q;(2) the restriction of σq to the maximal abelian subextension K˜/K ofKS(p)/K is equal to (pˆiq, K˜/K), where pˆiq denotes the ide`le whoseq-component equals piq and all other components are 1.Let τq denote an element of GS(p) such that(1) τq is an element of the inertia group TQ of Q in KS(p)/K;(2) the restriction of τq to K˜/K equals (αˆq, K˜/K), where αq denotesthe ide`le whose q-component equals αq and all other componentsare equal to 1.For any subset T of S, class field theory provides an isomorphismIK/(UT IpKK×) ∼= GT (p)/GT (p)p[GT (p), GT (p)] = H1(GT (p),Z/pZ).Let VT denote the Kummer groupVT = {a ∈ K× | a ∈ K×mq for q ∈ T and a ∈ UqK×mq for q 6∈ T}We remark that due to [NSW], 8.7.2, we have an exact sequence0→ O×K/p→ V∅(K)→ pCl(K)→ 0.By our assumptions, this yields that V∅(K) = 0, and since VT (K) ⊂ V∅(K)we have VT (K) = 0. This implies that the dual of the Kummer groupBT (K) = VT (K)∨ is trivial. The group on the left hand side of the aboveisomorphism is therefore given byIK/(UT IpKK×) ∼= U∅/UTUp∅ =∏q∈TUq/Upq = (Z/pZ)#T(see [Ko], §11.3). In particular, the automorphism τq restricts to a generatorof the cyclic group H1(G{q}(p),Z/pZ). We use this fact for the definition ofthe linking numbers.Definition 2.1. For two primes qi, qj ∈ S, the linking number `ij ∈ Z/pZof qi and qj is defined by the formulaσqi ≡ τ `ijqj mod G{qj}(p)p[G{qj}(p), G{qj}(p)].In other words, `ij is the image of the Frobenius automorphism σqi ∈GS(p) in H1(G{qj}(p),Z/pZ) which we identify with Z/pZ by means of itsgenerator τqj . Note that `ii = 0 for all i = 1, . . . , n. The linking number`ij is independent of the choice of the uniformizer piqi of Kqi (this followsfrom the above isomorphism for the case T = {qj}), but it depends on thechoice of αqj . If αqj would be replaced by αsqj , where s is prime to p, thenCIRCULAR SETS OF PRIMES OF IMAGINARY QUADRATIC NUMBER FIELDS 3`ij would be multiplied by s. Of course, the defining equation of the linkingnumber `ij is equivalent topˆiqi ≡ αˆ`ijqj mod USIpKK×which makes it possible to calculate the linking numbers in some examples,see section 3.Let us pause here for a moment to explain the analogy to link theory.Assume we are given two disjoint knots I and J in S3. Then the linkingnumber lk(I, J) is defined as follows. The knot I is a loop in S3 − J , henceit represents an element of pi1(S3 − J). After a choice of a generator of theinfinite cyclic group H1(S3 − J), lk(I, J) is defined as the image of I underthe mappi1(S3 − J)³ piab1 (S3 − J) ∼= H1(S3 − J) ∼= Z.In the number theoretical context described above, the linking number `ijis given by the image of the Frobenius automorphism σi under the mappiet1 (X − S)³ piet1 (X − {qj})³ H1(X − {qj},Z/pZ)=H1(G{qj}(p),Z/pZ)∼= Z/pZwhere X = Spec(OK) and we have chosen a generator of the cyclic groupH1(X − {qj},Z/pZ).We denote by ΓS(p) the directed graph with vertices the primes of S anda directed edge qiqj from qi to qj if `ij 6= 0. The graph ΓS(p), together withthe `ij is called the linking diagram of S.Definition 2.2. A finite set of primes of K whose norm is congruent to 1modulo p is called strictly circular with respect to p (and ΓS(p) a non-singular circuit) if there exists an ordering S = {q1, . . . , qn} of the primesin S such that the following conditions are fulfilled:(1) The vertices q1, . . . , qn of ΓS(p) form a circuit q1q2 . . . qnq1.(2) If i, j are both odd, then qiqj is not an edge of ΓS(p).(3) `12`23 . . . `n−1,n`n1 6= `1n`21 . . . `n,n−1.We remark that condition (1) implies that n is even and ≥ 4. Note thatcondition (3) does not depend on the choice of the αqj . It is satisfied if thereexists an edge qiqj of the circuit q1q2 . . . qnq1 such that qjqi is not an edgeof ΓS(p).We will now show that G has representation of Koch type.Proposition 2.3 (Koch). The group GS(p) has a presentation of Koch type,i.e. we have a minimal presentation GS(p) = F/R where F is the free pro-p-group on generators x1, . . . , xn, and R is minimally generated as a normalsubgroup of F by relations r1, . . . , rn which are given modulo F(3) byri ≡ xN(qi)−1in∏j=1j 6=i[xi, xj ]`ij mod F(3), i = 1, . . . , n.Here F(3) denotes the third step of the descending p-central series of F .4 DENIS VOGELProof. We have already seen above that GS(p) has a minimal generatingsystem consisting of the n elements τq1 , . . . , τqn . The abelianization GS(p)abof GS(p) is a finitely generated abelian pro-p-group. If GS(p)ab were in-finite, it would have a quotient isomorphic to Zp, which corresponds to aZp-extension K∞ of K inside KS(p). By [NSW], Thm. 10.3.20(ii), a Zp-extension of K is ramified at at least one prime dividing p. This contradictsK∞ ⊂ KS(p), hence GS(p)ab is finite. In particular, GS(p) has at least asmany relations as generators. From [NSW], 8.7.11 we obtain the inequalitydimZ/pZH1(GS(p),Z/pZ) ≥ dimZ/pZH2(GS(p),Z/pZ),which implies that a minimal system of generators ofR as a normal subgroupof F consists of n elements. Such a system is given by the set of relationsri = xN(qi)−1i [x−1i , y−1i ], i = 1, . . . , n,where yi ∈ F denotes a preimage of σqi , see [Ko], §11.4. The definition ofthe +linking numbers yieldsyi ≡n∏j=1j 6=ix`ijj mod F(2).Hence we obtainri ≡ xN(qi)−1i [xi, yi] ≡ xN(qi)−1i [xi,n∏j=1j 6=ix`ijj ] ≡ xN(qi)−1in∏j=1j 6=i[xi, xj ]`ij mod F(3),which finishes the proof. ¤Since GS(p) is of Koch a type, a result of Labute, ([La], Thm. 1.6.),applies, which states that GS(p) is a mild pro-p-group if S is strictly circularwith respect to p. Then, in particular, GS(p) has cohomological dimension 2.We summarize our considerations in the followingTheorem 2.4. Let p be an odd prime number and let K be an imaginaryquadratic number field whose class number is not divisible by p, and which isdifferent from Q(√−3) if p = 3. Let S = {q1, . . . , qn} be a set of primes ofK whose norm is congruent to 1 mod p. Is S is strictly circular with respectto p, then G(KS(p)/K) is a mild pro-p-group and hence of cohomologicaldimension 2.3. Some examplesWe use the same notation as in section 1. We let S = {q1, . . . , qn}, anddenote by qi the prime of Z lying below qi.We firstly consider the case where each qi is inert in K/Q. Then piqi = qiis a uniformizer of Kqi , and an element of Uq for all primes q 6= qi of K.Hence, the ide`le pˆiqi , when considered modulo USIpKK×, is equivalent to theide`le whose q-component is equal to 1 for q 6∈ S and q = qi, and equal toq−1i for q ∈ S \ {qi}. This means that, after a choice of a generator αqj ofUqj/Upqj , `ij is given by byqi = α−`ijqj mod Upqj .CIRCULAR SETS OF PRIMES OF IMAGINARY QUADRATIC NUMBER FIELDS 5Equivalently, we can choose a primitive root ²j of κ×qj , where κqj denotes theresidue field of qj . Then `ij is the image in Z/pZ of any integer c satisfyingqi = ²−cj mod qj .In particular, `ij = 0 if and only if qi is a p-th power modulo qj . This isequivalent to qi being a p-th power modulo qj : if qi ≡ xp mod qj for somex ∈ OK , then q2i ≡ NK/Q(x)p mod qj , and the claim follows. This impliesin the case under consideration, that S = {q1, . . . , qn} is strictly circularwith respect to p if and only if SQ = {q1, . . . , qn} is strictly circular (overQ) with respect to p.Example 3.1. (cf. the example after Thm 2.1 in [S]) Let K = Q(√−359),p = 3. The class number of K equals 19. The prime numbers 7, 19, 61, 163are inert in K/Q. We setq1 = (61), q2 = (19), q3 = (163), q4 = (7)and S = {q1, q2, q3, q4}. The linking diagram has the following shape:q2²²··­­q144**q3ttq4TT JJHence, S is a circular set of primes and cdG(KS(3)/K) = 2.In the calculations above we have made use of two things: the uniformizerspiqi have been chosen inK×, and piqi has been a unit in Uq for all q ∈ S\{qi}.Another case in which this is easily achieved is the case when the idealclass group of K is trivial. Then we can take a generator of qj as theuniformizer piqj and `ij can be obtained from the same equations as abovewith qj replaced by piqj .Example 3.2. Let K = Q(i), p = 3. We putq1 = (2 + 15i), q2 = (4 + 15i), q3 = q1, q4 = q2and S = {q1, q2, q3, q4}. Then we have q1 = q3 = 229, q2 = q4 = 241, andwe setpiq1 = 2 + 15i, piq2 = 4 + 15i, piq3 = piq1 , piq4 = piq2The linking diagram has the following shape:q2··q144q3ttq4TTHence cdG(KS(3)/K) = 2. Note that, by [Ko], Ex. 11.15, G(Q{q1,q2}(3)/Q)is finite.6 DENIS VOGELThe last example raises the following question. There are no examplesknown of prime numbers q1, q2 congruent to 1 modulo p where one can showthat the cohomological dimension of G(Q{q1,q2}(p)/Q) equals 2. Is it possibleto obtain such an example by considering strictly circular sets of primes{q1, q2, q1, q2} of an imaginary quadratic number field K of class numberone, in combination with some kind of descent argument? Unfortunately,the answer to this question is negative as the following considerations show.Let q1, q2 be prime numbers congruent to 1 modulo p, and assume thereexists an imaginary quadratic number field of class number one in which q1,q2 are completely decomposed:q1OK = q1q3, q2OK = q2q4.This definition of the primes qi implies (for an appropriate choice of theprimitive roots) the following equations for the linking numbers:`12 = `34, `23 = `41, `13 = `31, `24 = `42.Since we want to avoid that the group G(Q{q1,q2}(p)/Q) is finite, we haveto make sure that the conditions of [Ko], Ex. 11.15 are not fulfilled, andtherefore we have in addition to assume that q1 is a p-th power modulo q2and that q2 is a p-th power modulo q1. It is easily seen that this puts thefollowing restraints on the linking numbers:`12 + `32 = 0, `14 + `34 = 0, `21 + `41 = 0, `23 + `43 = 0.If ρi denotes the initial form of the image of ri in the graded Lie algebraassociated to the descending p-central series of F , the above conditions yieldthe equation`23ρ1 − `12ρ2 + `23ρ3 − `12ρ4 = 0.This means that the sequence ρ1, . . . , ρ4 is not strongly free (cf. the de-finition of strong freeness in [La]), which implies, in particular, that the set{q1, q2, q3, q4} is not strictly circular, and this holds true as well if we makea different choice of the primitive roots.4. Enlarging the set of primesProposition 4.1. Let p be an odd prime number and and K an imaginaryquadratic number field whose class number is not divisible by p, and whichis different from Q(√−3) if p = 3. Let S = {q1, . . . , qn} be a set of primesof K whose norm is congruent to 1 mod p. If cdG(KS(p)/K) ≤ 2, then thescheme X = Spec(OK)− S is a K(pi, 1) for the e´tale topology, i.e. for anydiscrete p-primary G(KS(p)/K)-module M , considered as a locally constante´tale sheaf on X, the natural homomorphismH i(G(KS(p)/K),M)→ H iet(X,M)is an isomorphism for all i.Proof. We put G = G(KS(p)/K). In the same way as in the proof of [S],Prop. 3.2., the Hochschild-Serre spectral sequenceEpq2 = Hp(G,Hqet(X˜,Z/pZ))⇒ Hp+qet (X,Z/pZ),where X˜ denotes the universal p-covering of X, implies isomorphismsH i(G,Z/pZ) ∼= H iet(X,Z/pZ), i = 0, 1CIRCULAR SETS OF PRIMES OF IMAGINARY QUADRATIC NUMBER FIELDS 7and a short exact sequence0→ H2(G,Z/pZ) φ→ H2et(X,Z/pZ)→ H2et(X˜,Z/pZ)G → 0.We set X¯ = SpecOK . By the flat duality theorem of Artin-Mazur, ([Mi],III, Thm. 3.1), we haveH3et(X¯,Z/pZ) = HomX¯(Z/pZ,Gm)∨ = 0andH2et(X¯,Z/pZ)∨ = Ext1X¯(Z/pZ,Gm),the latter group sitting in an exact sequence0→ O×K/p→ Ext1X¯(Z/pZ,Gm)→ pCl(K)→ 0.Our assumptions on K impliesH2et(X¯,Z/pZ) = 0.The excision sequence for the pair (X¯,X) yields an isomorphismH2et(X,Z/pZ) =⊕q∈SH3{q}(SpecOhq ,Z/pZ),where Ohq denotes the henselization of the local ring of X¯ at q. The localduality theorem ([Mi], II, Thm. 1.8) givesH3{q}(SpecOhq ,Z/pZ) ∼= HomSpecOhq (Z/pZ,Gm)∨.As we have assumed that for all q ∈ S, the norm of q is congruent to 1modulo p, we obtain dimZ/pZH2et(X,Z/pZ) = n. Hence, by the proof ofLemma 2.3, φ is an isomorphism, and thereforeH2et(X˜,Z/pZ)G = 0.The proof is then concluded as in [S], Prop. 3.2. ¤Theorem 4.2. Let p be an odd prime number and let K be an imaginaryquadratic number field whose class number is not divisible by p, and whichis different from Q(√−3) if p = 3. Let S be a set of primes of K whosenorm is congruent 1 mod p. Assume that cdG(KS(p)/K) = 2. Let l 6∈ S bea prime whose norm is congruent to 1 modulo p, and which does not splitcompletely in the extension KS(p)/K. ThencdG(KS∪{l}/K) = 2.Proof. The proof is the same as the proof of [S], Thm. 2.3, we just have toreplace Prop. 3.2. of (loc.cit.) by Prop. 4.1. above. ¤Corollary 4.3. Assume that S contains a strictly circular subset T suchfor each q ∈ S \ T there exists an edge from q to a prime of T . Thencd(G(KS(p)/K)) = 2.Proof. We only need to remark that if we are given a prime q ∈ S suchthat the linking number of q and a certain prime l of T is nontrivial, thenq does not split completely in KT (p)/K. To see this, we fix an extension Qof q to L = K{l}(p)ab. Since the linking number of q and l is nontrivial, theFrobenius of Q in L/K generates the whole Galois group G(L/K) ∼= Z/pZ.Hence q does not split completely in L/K, which proves the claim. ¤8 DENIS VOGELExample 4.4. Let K = Q(√−359), p = 3. The prime number l = 113 isinert in K/Q, and if we put q5 = lOK , and S = {q1, q2, q3, q4, q5} whereq1, q2, q3, q4 are given as in Example 3.1, the linking diagram looks as follows:q2²²··­­q5~~}}}}}}}}q144**q3ttq4TT JJHence, by Cor. 4.3 we have cdG(KS(p)/K) = 2 (although S is not strictlycircular with respect to p).Example 4.5. Let K = Q(√−359), p = 3 and S = {q1, q2, q3, q4}, whereq1, q2, q3, q4 are given as in Example 3.1. Wet set l = (37, 14+√−359). Notethat l|37, and 37 is completely decomposed in K/Q. The unique subfield Lof degree 3 over K of the extension K(µ7)/K is a subfield of KS(p)/K, andthe prime l of K is inert in L. Therefore, we obtain by Thm. 4.2 thatcdG(KS∪{l}/K) = 2.Another result from [S] which carries over to our situation with identicalproof is given by the following theorem.Theorem 4.6. Let p be an odd prime number and let K be an imaginaryquadratic number field whose class number is not divisible by p, and which isdifferent from Q(√−3) if p = 3. Let S be a set of primes of K whose norm iscongruent to 1 mod p. Assume that G(KS(p)/K) 6= 1 and cdG(KS(p)/K) ≤2. Then scdG(KS(p)/K) = 3 and G(KS(p)/K) is a pro-p duality group.References[Ko] Koch, H.: Galoissche Theorie der p-Erweiterungen. Deutscher Verlag der Wiss.,1970 (English translation Berlin 2002)[La] Labute, J.: Mild Pro-p-Groups and Galois Groups of p-Extensions of Q (to appearin J. Reine Angew. Math.)[NSW] Neukirch, J., Schmidt, A., Wingberg, K.: Cohomology of number fields. Springer2000[Mi] Milne, J.: Arithmetic duality theorems. Academic Press 1986[S] Schmidt, A.: Circular sets of prime numbers and p-extensions of the rationals. (toappear in J. Reine Angew. Math.)Denis VogelNWF I - Mathematik, Universita¨t Regensburg93040 RegensburgDeutschlandemail: denis.vogel@mathematik.uni-regensburg.de

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Circular sets of primes of imaginary quadratic number fields

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