Bounded generation of SL 2 over rings of S-integers with infinitely many units

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1 sbang/ms/2017/1 September 19th, statmath/eprnts Bounded generaton of SL 2 over rngs of S-ntegers wth nfntely many unts Aleksander V. Morgan, Andre S. Rapnchuk and Balasubramanan Sury Indan Statstcal Insttute, Bangalore Centre 8th Mle Mysore Road, Bangalore, Inda

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3 BOUNDED GENERATION OF SL 2 OVER RINGS OF S-INTEGERS WITH INFINITELY MANY UNITS ALEKSANDER V. MORGAN, ANDREI S. RAPINCHUK, AND BALASUBRAMANIAN SURY Abstract. Let O be the rng of S-ntegers n a number feld k. We prove that f the group of unts O s nfnte then every matrx n Γ = SL 2O) s a product of at most 9 elementary matrces. Ths completes a long lne of research n ths drecton. As a consequence, we obtan that Γ s boundedly generated as an abstract group. To Alex Lubotzky on hs 60th brthday 1. Introducton Let k be a number feld. Gven a fnte subset S of the set V k of valuatons of k contanng the set V k of archmedan valuatons, we let O k,s denote the rng of S-ntegers n k,.e. O k,s = {a k va) 0 for all v V k \ S} {0}. As usual, for any commutatve rng R, we let SL 2 R) denote the group of unmodular 2 2-matrces over R and refer to the matrces ) ) 1 a 1 0 E 12 a) = and E b) = SL b 1 2 R) a, b R) as elementary over R). It was establshed n [Va] see also [L1]) that f the rng of S-ntegers O = O k,s has nfntely many unts, the group Γ = SL 2 O) s generated by elementary matrces. The goal of ths paper s to prove that n ths case Γ s actually boundedly generated by elementares. Theorem 1.1. Let O = O k,s be the rng of S-ntegers n a number feld k, and assume that the group of unts O s nfnte. Then every matrx n SL 2 O) s a product of at most 9 elementary matrces. The quest to valdate ths property has a consderable hstory. Frst, G. Cooke and P. J. Wenberger [CW] establshed t wth the same bound as n Theorem 1.1) assumng the truth of a sutable form of the Generalzed Remann Hypothess, whch stll remans unproven. Later, t was shown n [LM] see also [M]) by analytc tools that the argument can be made uncondtonal f S max5, 2[k : Q] 3). On the other hand, B. Lehl [L2] proved the result by algebrac methods for some specal felds k. The frst uncondtonal proof n full generalty was gven by D. Carter, G. Keller and E. Page n an unpublshed preprnt; ther argument was streamlned and made avalable to the publc by D. W. Morrs [MCKP]. Ths argument s based on model theory and provdes no explct bound on the number of elementares requred; besdes, t uses dffcult results from addtve number theory. 1

4 2 A. MORGAN, A. RAPINCHUK, AND B. SURY In [Vs], M. Vsemrnov proved Theorem 1.1 for O = Z[1/p] usng the results of D. R. Heath-Brown [HB] on Artn s Prmtve Root Conjecture thus, n a broad sense, ths proof develops the ntal approach of Cooke and Wenberger [CW]); hs bound on the number of elementares requred s 5. Subsequently, the thrd-named author re-worked the argument from [Vs] to avod the use of [HB] n an unpublshed note. These notes were the begnnng of the work of the frst two authors that has eventually led to a proof of Theorem 1.1 n the general case. It should be noted that our proof uses only standard results from number theory, and s relatvely short and constructve wth an explct bound whch s ndependent of the feld k and the set S. Ths, n partcular, mples that Theorem 1.1 remans vald for any nfnte S. The problem of bounded generaton partcularly by elementares) has been consdered for S- arthmetc subgroups of algebrac groups other than SL 2. A few years after [CW], Carter and Keller [CK1] showed that SL n O) for n 3 s boundedly generated by elementares for any rng O of algebrac ntegers see [T] for other Chevalley groups of rank > 1, and [ER] for sotropc, but nonsplt or quas-splt), orthogonal groups). The upper bound on the number of factors requred to wrte every matrx n SL n O) as a product of elementares gven n [CK1] s 1 2 3n2 n) , where s the number of prme dvsors of the dscrmnant of k; n partcular, ths estmate depends on the feld k. Usng our Theorem 1.1, one shows n all cases where the group of unts O s nfnte, ths estmate can be mproved to 1 2 3n2 n) + 4, hence made ndependent of k see Corollary 4.6. The stuaton not covered by ths result are when O s ether Z or the rng of ntegers n an magnary quadratc feld see below. The former case was treated n [CK2] wth an estmate 1 2 3n2 n) + 36, so only n the case of magnary quadratc felds the queston of the exstence of a bound on the number of elementares ndependent of the k remans open. From a more general perspectve, Theorem 1.1 should be vewed as a contrbuton to the sustaned effort amed at provng that all hgher rank lattces are boundedly generated as abstract groups. We recall that a group Γ s sad to have bounded generaton BG) f there exst elements γ 1,..., γ d Γ such that Γ = γ 1 γ d, where γ denotes the cyclc subgroup generated by γ. The nterest n ths property stems from the fact that whle beng purely combnatoral n nature, t s known to have a number of farreachng consequences for the structure and representatons of a group, partcularly f the latter s S-arthmetc. For example, under one addtonal necessary) techncal assumpton, BG) mples the rgdty of completely reducble complex representatons of Γ known as SS-rgdty) see [R], [PR2, Appendx A]. Furthermore, f Γ s an S-arthmetc subgroup of an absolutely smple smply connected algebrac group G over a number feld k, then assumng the truth of the Marguls- Platonov conjecture for the group Gk) of k-ratonal ponts cf. [PR2, 9.1]), BG) mples the congruence subgroup property.e. the fnteness of the correspondng congruence kernel see [Lu], [PR1]). For applcatons of BG) to the Marguls-Zmmer conjecture, see [SW]. Gven these and other mplcatons of BG), we would lke to pont out the followng consequence of Theorem 1.1. Corollary 1.2. Let O = O k,s be the rng of S-ntegers, n a number feld k. If the group of unts O s nfnte, then the group Γ = SL 2 O) has bounded generaton. We note that combnng ths fact wth the results of [Lu], [PR1], one obtans an alternatve proof of the centralty of the congruence kernel for SL 2 O) provded that O s nfnte), orgnally establshed by J.-P. Serre [S1]. We also note that BG) of SL 2 O) s needed to prove BG) for some other groups cf. [T] and [ER].

5 BOUNDED GENERATION OF SL 2 O S ) 3 Next, t should be ponted out that the assumpton that the unt group O s nfnte s necessary for the bounded generaton of SL 2 O), hence cannot be omtted. Indeed, t follows from Drchlet s Unt Theorem [CF, 2.18] that O s fnte only when S = 1 whch happens precsely when S s the set of archmedan valuatons n the followng two cases: 1) k = Q and O = Z. In ths case, the group SL 2 Z) s generated by the elementares, but has a nonabelan free subgroup of fnte ndex, whch prevents t from havng bounded generaton. 2) k = Q d) for some square-free nteger d 1, and O d s the rng of algebrac ntegers ntegers n k. Accordng to [GS], the group Γ = SL 2 O d ) has a fnte ndex subgroup that admts an epmorphsm onto a nonabelan free group, hence agan cannot possbly be boundedly generated. Moreover, P. M. Cohn [Co] shows that f d / {1, 2, 3, 7, 11} then Γ s not even generated by elementary matrces. The structure of the paper s the followng. In 2 we prove an algebrac result about abelan subextensons of radcal extensons of general feld see Proposton 2.1. Ths statement, whch may be of ndependent nterest, s used n the paper to prove Theorem 3.7. Ths theorem s one of the number-theoretc results needed n the proof of Theorem 1.1, and t s establshed n 3 along wth some other facts from algebrac number theory. One of the key notons n the paper s that of Q-splt prme: we say that a prme p of a number feld k s Q-splt f t s non-dyadc and ts local degree over the correspondng ratonal prme s 1. In 3, we establsh some relevant for us propertes of such prmes see 3.1) and prove for them n 3.2 the followng refnement of Drchlet s Theorem from [BMS]. Theorem 3.3. Let O be the rng of S-ntegers n a number feld k for some fnte S V k contanng V k. If nonzero a, b O are relatvely prme.e., ao + bo = O) then there exst nfntely many prncpal Q-splt prme deals p of O wth a generator π such that π a mod bo) and π > 0 n all real completons of k. Subsecton 3.3 s devoted to the statement and proof of the key theorem 3.7, whch s another key number-theoretc result needed n the proof of Theorem 1.1. In 4, we prove Theorem 1.1 and Corollary 1.2. Fnally, n 5 we correct the faulty example from [Vs] of a matrx n SL 2 Z[1/p]), where p s a prme 1mod29), that s not a product of four elementary matrces see Proposton 5.1, confrmng thereby that the bound of 5 n [Vs] s optmal. Notatons and conventons. For a feld k, we let k ab denote the maxmal abelan extenson of k. Furthermore, k) wll denote the group of all roots of unty n k; f k) s fnte, we let denote ts order. For n 1 prme to char k, we let ζ n denote a prmtve n-th root of unty. In ths paper, wth the excepton of 2, the feld k wll be a feld of algebrac numbers.e., a fnte extenson of Q), n whch case k) s automatcally fnte. We let O k denote the rng of algebrac ntegers n k. Furthermore, we let V k denote the set of the equvalence classes of) nontrval valuatons of k, and let V k and Vf k denote the subsets of archmedean and nonarchmedean valuatons, respectvely. For any v V k, we let k v denote the correspondng completon; f v Vf k then O v wll denote the valuaton rng n k v wth the valuaton deal ˆp v and the group of unts U v = O v. Throughout the paper, S wll denote a fxed fnte subset of V k contanng V, k and O = O k,s the correspondng rng of S-ntegers see above). Then the nonzero prme deals of O are n a natural bjectve correspondence wth the valuatons n V k \ S. So, for a nonzero prme deal p O we let v p V k \ S denote the correspondng valuaton, and conversely, for a valuaton v V k \ S we

6 4 A. MORGAN, A. RAPINCHUK, AND B. SURY let p v O denote the correspondng prme deal note that p v = O ˆp v ). Generalzng Euler s ϕ-functon, for a nonzero deal a of O, we set φa) = O/a). For smplcty of notaton, for an element a O, φa) wll always mean φao). Fnally, for a k, we let V a) = { v Vf k va) 0 }. Gven a prme number p, one can wrte any nteger n n the form n = p e m, for some non-negatve nteger e, where p / m. We then call p e the p-prmary component of n. 2. Abelan subextensons of radcal extensons. In ths secton, k s an arbtrary feld. For a prme p char k, we let k) p denote the subgroup of k), consstng of elements satsfyng x pd = 1 for some d 0. If ths subgroup s fnte, we set λk) p to be the non-negatve nteger satsfyng k) p = p λk)p ; otherwse, set λk) p =. Clearly f k) s fnte, then = p pλk)p. For a k, we wrte n a to denote an arbtrary root of the polynomal x n a. The goal of ths secton s to prove the followng. Proposton 2.1. Let n 1 be an nteger prme to char k, and let u k be such that u / k) p k p for all p n. Then the polynomal x n u s rreducble over k, and for t = n u we have kt) k ab = kt m n ) where m = gcdn, p λk)p ), wth the conventon that gcdn, p ) s smply the p-prmary component of n. We frst treat the case n = p d where p s a prme. Proposton 2.2. Let p be a prme number char k, and let u k \ k) p k ) p. Fx an nteger d 1, set t = pd u. Then kt) k ab = kt pγ ) where γ = max0, d λk)p). p n We begn wth the followng lemma. Lemma 2.3. Let p be a prme number char k, and let u k \ k) p k ) p. Set k 1 = k p u). Then ) [k 1 : k] = p; ) k 1 ) p = k) p ) None of the p u are n k 1 ) p k 1 )p. Proof. ) follows from [La, Ch. VI, 9], as u / k ) p. ): If λk) p =, then there s nothng to prove. Otherwse, we need to show that for λ = λk) p, we have ζ p λ+1 / k 1. Assume the contrary. Then, frst, λ > 0. Indeed, we have a tower of nclusons k kζ p ) k 1. Snce [k 1 : k] = p by ), and [kζ p ) : k] p 1, we conclude that [kζ p ) : k] = 1,.e. ζ p k.

7 BOUNDED GENERATION OF SL 2 O S ) 5 Now, snce ζ p λ+1 / k, we have 1) k 1 = kζ p λ+1) = k p ζ p λ). But accordng to Kummer s theory whch apples because ζ p k), the fact that k p a) = k p b) for a, b k mples that the mages of a and b n k /k ) p generate the same subgroup. So, t follows from 1) that uζ p k ) p for some, and therefore u k) p k ) p, contradctng our choce of u. ): Assume the contrary,.e. some p-th root p u can be wrtten n the form p u = ζa p for some a k 1 and ζ k 1) p. Let N = N k1 /k : k 1 k be the norm map. Then N p u) = Nζ)Na) p. Clearly, Nζ) k) p, so N p u) k) p k ) p. On the other hand, N p u) = u for p odd, and u for p = 2. In all cases, we obtan that u k) p k ) p. A contradcton. A smple nducton now yelds the followng: Corollary 2.4. Let p be a prme number char k, and let u k \k) p k ) p. For a fxed nteger d 1, set k d = k pd u). Then: ) [k d : k] = p d ; ) k d ) p = k) p, hence λk d ) p = λk) p. Of course, asserton ) s well-known and follows, for example, from [La, Ch. VI, 9]. Lemma 2.5. Let p be a prme number char k, and let u k \ k) p k ) p. Fx an nteger d 1, and set t = pd u and kd = kt). Furthermore, for an nteger j between 0 and d defne l j = kt pd j ) k pj u). Then any ntermedate subfeld k l kd s of the form l = l j for some j {0,..., d}. Proof. Gven such an l, t follows from Corollary 2.4) that [k d : l] = p j for some 0 j d. Snce any conjugate of t s of the form ζ t where ζ pd = 1, we see that the norm N kd /lt) s of the form ζ 0 t pj, where agan ζ pd 0 = 1. Then ζ 0 k d ) p, and usng Corollary 2.4), we conclude that ζ 0 k l. So, t pj l, mplyng the ncluson l d j l. Now, the fact that [k d : l d j ] = p j mples that l = l d j, yeldng our clam. Proof of Proposton 2.2. Set λ = λk) p. Then for any d λ the extenson k pd u)/k s abelan, and our asserton s trval. So, we may assume that λ < and d > λ. It follows from Lemma 2.5 that l := kt) k ab s of the form l d j = kt pj ) for some j {0,..., d}. On the other hand, l d j /k s a Galos extenson of degree p d j, so must contan the conjugate ζ p d jt pd j of t pd j, mplyng that ζ p d j l d j. Snce l d j k pd j u), we conclude from Corollary 2.4) that d j λ,.e. j d λ. Ths proves the ncluson l kt pγ ); the opposte ncluson s obvous. Proof of Proposton 2.1. Let n = p α 1 1 pαs s be the prme factorzaton of n, and for = 1,..., s set n = n/p α. Let t = n u and t = t n so, t s a p α -th root of u). Usng agan [La, Ch. VI, 9]

8 6 A. MORGAN, A. RAPINCHUK, AND B. SURY we conclude that [kt) : k] = n, whch mples that 2) [kt) : kt )] = n for all = 1,..., r. Snce for K := kt) k ab the degree [K : k] dvdes n, we can wrte K = K 1 K s where K s an abelan extenson of k of degree p β for some β α. Then the degree [K t ) : kt )] must be a power of p. Comparng wth 2), we conclude that K kt ). Applyng Proposton 2.2 wth d = α, we obtan the ncluson 3) K kt pγ ) = kt n p γ ) where γ = max0, α λk) p ). It s easy to see that the g.c.d. of the numbers n p γ for = 1,..., s s n m = gcdn, p λk)p ). concdes wth the cyclc sub- Furthermore, the subgroup of kt) generated by t n 1p γ 1 1,..., t n sp γs s group wth generator t m. Then 3) yelds the followng ncluson p n K = K 1 K s kt m ). Snce the opposte ncluson s obvous, our clam follows. Corollary 2.6. Assume that = k) <. Let P be a fnte set of ratonal prmes, and defne = p. Gven u k such that p P u / k) p k ) p for all p P, for any abelan extenson F of k the ntersecton ) E := F k u, ζ s contaned n k u, ζ ). Proof. Wthout loss of generalty we may assume that ζ F, and then we have the followng tower of feld extensons k u, ζ ) E ) u k u, ζ ). We note that the degree [ k u, ζ ) : k )] u, ζ dvdes p P p. So, f we assume that the asserton of the lemma s false, then we should be able to fnd to fnd a prme p P that dvdes the degree [ E u) : k [ u, ζ )], and therefore does not dvde the degree k u, ζ ) : E u) ]. The latter mples that p u E u). But ths contradcts Proposton 2.1 snce E u) = E k u) s an abelan extenson of k.

9 BOUNDED GENERATION OF SL 2 O S ) 7 3. Results from Algebrac Number Theory 1. Q-splt prmes. Our proof of Theorem 1.1 heavly reles on propertes of so-called Q-splt prmes n O. Defnton. Let p be a nonzero prme deal of O, and let p be the correspondng ratonal prme. We say that p s Q-splt f p > 2, and for the valuaton v = v p we have k v = Q p. For the convenence of further references, we lst some smple propertes of Q-splt prmes. Lemma 3.1. Let p be a Q-splt prme n O, and for n 1 let ρ n : O O/p n be the correspondng quotent map. Then: a) the group of nvertble elements O/p n ) s cyclc for any n; b) f c O s such that ρ 2 c) generates O/p 2 ) then ρ n c) generates O/p n ) for any n 2. Proof. Let p > 2 be the ratonal prme correspondng to p, and v = v p be the assocated valuaton of k. By defnton, k v = Q p, hence O v = Z p. So, for any n 1 we wll have canoncal rng somorphsms 4) O/p n O v /ˆp n v Z p /p n Z p Z/p n Z. Then a) follows from the well-known fact that the group Z/p n Z) s cyclc. Furthermore, the somorphsms n 4) are compatble for dfferent n s. Snce the kernel of the group homomorphsm Z/p n Z) Z/p 2 Z) s contaned n the Frattn subgroup of Z/p n Z) for n 2, the same s true for the homomorphsm O/p n ) O/p 2 ). Ths easly mples b). Let p be a Q-splt prme, let v = v p be the correspondng valuaton. We wll now defne the level l p u) of an element u O v and establsh some propertes of ths noton that we wll need later. Let p > 2 be the correspondng ratonal prme. The group of p-adc unts U p = Z p has the natural fltraton by the congruence subgroups U ) p = 1 + p Z p for N. It s well-known that U p = C U 1) p where C s the cyclc group of order p 1) consstng of all roots of unty n Q p. Furthermore, the logarthmc map yelds a contnuous somorphsm U ) p p Z p, whch mples that for any u U p \ C, the closure of the cyclc group generated by u has a decomposton of the form u = C U l) p for some subgroup C C and some nteger l = l p u) 1 whch we wll refer to as the p-level of u. We also set l p u) = for u C. Returnng now to a Q-splt prme p of k and keepng the above notatons, we defne the p-level l p u) of u O v as the the p-level of the element n U p that corresponds to u under the natural dentfcaton O v = Z p. We wll need the followng. Lemma 3.2. Let p be a Q-splt prme n O, let p be the correspondng ratonal prme, and v = v p the correspondng valuaton. Suppose we are gven an nteger d 1 not dvsble by p, a unt u O v of nfnte order havng p-level s = l p u), and nteger n s, and an element c O v such that u ns c mod p s ). Then for any t s there exsts an nteger n t n s mod d) for whch u nt c mod p t ).

10 8 A. MORGAN, A. RAPINCHUK, AND B. SURY Proof. In vew of the dentfcaton O v = Z p, t s enough to prove the correspondng statement for Z p. More precsely, we need to show the followng: Let u U p be a unt of nfnte order and p-level s = l p u). If c U p and n s Z are such that u ns c mod p s ), then for any t s there exsts n t n s mod d) such that u nt c mod p t ). Thus, we have that u ns cu s) p, and we wsh to show that u ns u d cu t) p. Snce cu t) p s open, t s enough to show that 5) u ns u d cu t) p. But snce l p u) = s and d s prme to p, we have the ncluson u d U s) p, and 5) s obvous. 2. Drchlet s Theorem for Q-splt prmes. We wll now establsh the exstence of Q-splt prmes n arthmetc progressons. Theorem 3.3. Let O be the rng of S-ntegers n a number feld k for some fnte S V k contanng V k. If nonzero a, b O are relatvely prme.e., ao + bo = O) then there exst nfntely many prncpal Q-splt prme deals p of O wth a generator π such that π a mod bo) and π > 0 n all real completons of k. The proof follows the same general strategy as the proof of Drchlet s Theorem n [BMS] - see Theorem A.10 n the Appendx on Number Theory. Frst, we wll quckly revew some basc facts from global class feld theory cf., for example, [CF], Ch. VII) and fx some notatons. Let J k denote the group of deles of k wth the natural topology; as usual, we dentfy k wth the dscrete) subgroup of prncpal deles n J k. Then for every open subgroup U J k of fnte ndex contanng k there exsts a fnte abelan Galos extenson L/k and a contnuous surjectve homomorphsm α L/k : J k GalL/k) known as the norm resdue map) such that U = Ker α L/k = N L/k J L )k ; for every nonarchmedean v V k whch s unramfed n L we let Fr L/k v) denote the Frobenus automorphsm of L/k at v.e., the Frobenus automorphsm Fr L/k w v) assocated to some equvalently, any) extenson w v) and let v) J k be an dele wth the components { 1, v v v) v = π v, v = v, where π v k v s a unformzer; then α L/k v)) = Fr L/k v). For our fxed fnte subset S V k contanng V, k we defne the followng open subgroup of J k : U S := k v U v. v S v V k \S Then the abelan extenson of k correspondng to the subgroup U S := U S k wll be called the Hlbert S-class feld of k and denoted K throughout the rest of the paper. Next, we wll ntroduce the delc S-analogs of ray groups. Let b be a nonzero deal of O = O k,s wth the prme factorzaton 6) b = p n 1 1 pnt t,

11 BOUNDED GENERATION OF SL 2 O S ) 9 let v = v p be the valuaton n V k \ S assocated wth p, and let V b) = {v 1,..., v t }. We then defne an open subgroup R S b) = v V k R v where the open subgroups R v k v are defned as follows. For v real, we let R v be the subgroup of postve elements, lettng R v = k v for all other v S, and settng R v = U v for all v / S V b). It remans to defne R v for v = v V b), n whch case we set t to be the congruence subgroup U n ) v of U v modulo ˆp n v. We then let Kb) denote the abelan extenson of k correspondng to R S b) := R S b)k ray class feld ). Obvously, Kb) contans K for any nonzero deal b of O.) Furthermore, gven c k, we let j b c) denote an dele wth the followng components: { c, v V b), j b c) v = 1, v / V b). Then θ b : k GalKb)/k) defned by c α Kb)/k j b c)) 1 s a group homomorphsm. The followng lemma summarzes some smple propertes of these defntons. Lemma 3.4. Let b O be a nonzero deal. a) If a nonzero c O s relatvely prme to b.e. co + b = O) then θ b c) restrcts to the Hlbert S-class feld K trvally. b) If nonzero c 1, c 2 O are both relatvely prme to b then c 1 c 2 mod b) s equvalent to 7) pr b j b c 1 )R S b)) = pr b j b c 2 )R S b)) where pr b : J k s the natural projecton. v V b) k v Proof. a): Snce c s relatvely prme to b, we have j b c) U S. So, usng the functoralty propertes of the norm resdue map, we obtan θ b c) K = α Kb)/k j b c)) 1 K = α K/k j b c)) 1 = d K because j b c) U S U S = Ker α K/k, as requred. b): As above, let 6) be the prme factorzaton of b, let v = v p V k \ S be the valuaton assocated wth p. Then for any c 1, c 2 O, the congruence c 1 c 2 mod b) s equvalent to 8) c 1 c 2 mod ˆp n v ) for all = 1,..., t. On the other hand, for any v Vf k and any u 1, u 2 U v, the congruence u 1 u 2 mod ˆp n v ) for n 1 s equvalent to u 1 U n) v = u 2 U n) v, where U v n) s the congruence subgroup of U v modulo ˆp n v. Thus, for nonzero) c 1, c 2 O prme to b, the condtons 7) and 8) are equvalent, and our asserton follows. We wll now establsh a result needed for the proof of Theorem 3.3 and ts refnements. Proposton 3.5. Let b be a nonzero deal of O, let a O be relatvely prme to b, and let F be a fnte Galos extenson of Q that contans Kb). Assume that a ratonal prme p s unramfed n F

12 10 A. MORGAN, A. RAPINCHUK, AND B. SURY and there exsts an extenson w of the p-adc valuaton v p to F such that Fr F/Q w v p ) Kb) = θ b a). If the restrcton v of w to k does not belong to S V b) then: a) k v = Q p ; b) the prme deal p = p v of O correspondng to v s prncpal wth a generator π satsfyng π a mod b) and π > 0 n every real completon of k. We note snce v s unramfed n F whch contans Kb), we n fact automatcally have that v / V b).) Proof. a): Snce the Frobenus Frw v p ) generates GalF w /Q p ), our clam mmedately follows from the fact that t acts trvally on k. b): Accordng to a), the local degree [k v : Q p ] s 1, hence the resdual degree fv v p ) s also 1, and therefore Frw v) = Frw v p ) fv vp) = Frw v p ). Thus, and therefore So, we can wrte α Kb)/k v)) = Frw v) Kb) = θ b a) = α Kb)/k j b a)) 1, v)j b a) Ker α Kb)/K = R S b) = R S b)k. 9) v)j b a) = rπ wth r R S b), π k. Then π = v)j b a)r 1 ). Snce a s prme to b, the dele j b a) U S, and then j b a)r 1 U S. For any v V k \ S {v}), the v -component of v) s trval, so we obtan that π U v. On the other hand, the v-component of v) s a unformzer π v of k v mplyng that π s also a unformzer. Thus, p = πo s precsely the prme deal assocated wth v. For any real v, the v -components of v) and j b a) are trval, so π equals the nverse of the v -component of r, hence postve n k v. Fnally, t follows from 9) that so π a mod b) by Lemma 3.4b), as requred. pr b j b a)) = pr b j b π)r), P roof of T heorem 3.3. Set b = bo and σ = θ b a) GalKb)/k). Let F be the Galos closure of Kb) over Q, and let τ GalF/Q) be such that τ Kb) = σ. Applyng Chebotarev s Densty Theorem see [CF, Ch. VII, 2.4] or [BMS, A.6]), we fnd nfntely many ratonal prmes p > 2 for whch the p-adc valuaton v p s unramfed n F, does not le below any valuatons n S V b), and has an extenson w to F such that Fr F/Q w v p ) = τ. Let v = w k, and let p = p v be the correspondng prme deal of O. Snce p > 2, part a) of Proposton 3.5 mples that p s Q-splt. Furthermore, part b) of t asserts that p has a generator π such that π a mod b) and π > 0 n every real completon of k, as requred. We wll now prove a statement from Galos theory that we wll need n the next subsecton. Lemma 3.6. Let F/Q be a fnte Galos extenson, and let κ be an nteger for whch F Q ab Qζ κ ). Then F ζ κ ) Q ab = Qζ κ ).

13 Proof. We need to show that BOUNDED GENERATION OF SL 2 O S ) 11 10) [F ζ κ ) : F ζ κ ) Q ab ] = [F ζ κ ) : Qζ κ )]. Let G = GalF ζ κ )/Q) and H = GalF/Q). Then the left-hand sde of 10) s equal to the order of the commutator subgroup [G, G], whle the rght-hand sde equals [F : F Qζ κ )] = [F : F Q ab ] = [H, H]. Now, the restrcton gves an njectve group homomorphsm ψ : G H GalQζ κ )/Q). Snce the restrcton G H s surjectve, we obtan that ψ mplements an somorphsm between [G, G] and [H, H] {1}. Thus, [G, G] and [H, H] have the same order, and 10) follows. 3. Key statement. In ths subsecton we wll establsh another number-theoretc statement whch plays a crucal role n the proof of Theorem 1.1. To formulate t, we need to ntroduce some addtonal notatons. As above, let = k) be the number of roots of unty n k, let K be the Hlbert S-class feld of k, and let K be the Galos closure of K over Q. Suppose we are gven two fnte sets P and Q of ratonal prmes. Let = p, pck an nteger λ 1 whch s dvsble by and for whch K Q ab Qζ λ ), and set λ = λ q. Theorem 3.7. Let u O be a unt of nfnte order such that u / k) p k ) p for every prme p P, and let q be a Q-splt prme of O whch s relatvely prme to λ. Then there exst nfntely many prncpal Q-splt prmes p = πo of O wth a generator π such that 1) for each p P, the p-prmary component of φp)/ dvdes the p-prmary component of the order of u mod p); 2) πmod q 2 ) generates O/q 2 ) ; 3) gcdφp), λ ) = λ. Proof. As n the proof of Theorem 3.3, we wll derve the requred asserton by applyng Chebotarev s Densty Theorem to a specfc automorphsm of an approprate fnte Galos extenson. Let Kq 2 ) be the abelan extenson Kb) of k ntroduced n subsecton 3.2 for the deal b = q 2. Set L 1 = Kq 2 ) )ζ λ ), L 2 = k ζ, u, L = L 1 L 2 and l = L 1 L 2. Then 11) GalL/k) = { σ = σ 1, σ 2 ) GalL 1 /k) GalL 2 /k) σ 1 l = σ 2 l }. So, to construct σ GalL/k) that we wll need n the argument t s enough to construct approprate σ GalL /k) for = 1, 2 that have the same restrcton to l. p P q Q

14 12 A. MORGAN, A. RAPINCHUK, AND B. SURY Lemma 3.8. The restrcton maps defne the followng somorphsms: 1) GalL 1 /K) GalKq 2 )/K) GalKζ λ )/K); 2) GalKζ λ )/Kζ λ )) GalQζ λ )/Qζ λ )) q Q GalQζ qλ )/Qζ λ )). Proof. 1): We need to show that Kq 2 ) Kζ λ ) = K. But the Galos extensons Kq 2 )/K and Kζ λ )/K are respectvely totally and unramfed at the extensons of v q to K snce q s prme to λ), so the requred fact s mmedate. 2): Snce Kζ λ ) = Kζ λ ) Qζ λ ), we only need to show that 12) Kζ λ ) Qζ λ ) = Qζ λ ). We have Kζ λ ) Qζ λ ) Kζ λ ) Q ab = Qζ λ ) by Lemma 3.6. Ths proves one ncluson n 12); the other ncluson s obvous. Snce q s Q-splt, the group O/q 2 ) s cyclc Lemma 3.1a)), and we pck c O so that c mod q 2 ) s a generator of ths group. We then set σ 1 = θ q 2c) GalKq 2 )/K) n the notatons of subsecton 3.2 cf. Lemma 3.4a)). Next, for q Q, we let q eq) be the q-prmary component of λ. Then usng the somorphsm from Lemma 3.82), we can fnd σ 1 GalKζ λ )/K) such that 13) σ 1ζ λ ) = ζ λ but σ 1ζ q eq)+1) ζ q eq)+1 for all q Q. We then defne σ 1 GalL 1 /K) to the automorphsm correspondng to the par σ 1, σ 1 ) n terms of the somorphsm from Lemma 3.81) n other words, the restrctons of σ 1 to Kq 2 ) and Kζ λ ) are σ 1 and σ 1, respectvely). We fx a -th root u, and for ν set ν u = u ) /ν also denoted u ν 1 ). To construct σ 2 GalL 2 /k), we need the followng. Lemma 3.9. Let σ 0 Gall/k). Then there exsts σ 2 GalL 2 /k) such that 1) σ 2 l = σ 0 ; 2) for any p P, f p dp) s the p-prmary component of then σ 2 u p dp)+1)) u p dp)+1), and consequently ether σ 2 ζ p dp)+1) ζ p dp)+1 or σ 2 acts nontrvally on all p dp)+1 -th roots of u. Proof. Snce L 1 /k s an abelan extenson, we conclude from Corollary 2.6 that 14) l k u, ζ ) k ab. On the other hand, accordng to Proposton 2.1, none of the roots p u for p P les n k ab, and the restrcton maps yeld an somorphsm Gal k u, ζ ) /k ) ) u, ζ Gal k p u, ζ ) /k u, ζ ) ). p P

15 BOUNDED GENERATION OF SL 2 O S ) 13 It follows that for each p P we can fnd τ p Gal k u, ζ ) /k u, ζ ) ) such that τ p u p dp)+1)) = ζ p u p dp)+1) and τ p u q dq)+1)) = u q dq)+1) for all q P \ {p}. Now, let σ 0 be any extenson of σ 0 to L 2. For p P, defne 1, σ 0 u p dp)+1)) = u p dp)+1) χp) = 0, σ 0 u p dp)+1)) u p dp)+1) Set σ 2 = σ 0 p P τ χp) p. In vew of 14), all τ p s act trvally on l, so σ 2 l = σ 0 l = σ 0 and 1) holds. Furthermore, the choce of the τ p s and the χp) s mples that 2) also holds. Contnung the proof of Theorem 3.7, we now use σ 1 GalL 1 /k) constructed above, set σ 0 = σ 1 l, and usng Lemma 3.9 construct σ 2 GalL 2 /k) wth the propertes descrbed theren. In partcular, part 1) of ths lemma n conjuncton wth 11) mples that the par σ 1, σ 2 ) corresponds to an automorphsm σ GalL/k). As n the proof of Theorem 3.3, we let F denote the Galos closure of L over Q, and let σ GalF/Q) be such that σ L = σ. By Chebotarev s Densty Theorem, there exst nfntely many ratonal prmes p > 2 that are relatvely prme to λ and for whch the p-adc valuaton v p s unramfed n F, does not le below any valuaton n S {v q }, and has an extenson w to F such that Fr F/Q w v p ) = σ. Let v = w k, and let p = p v be the correspondng prme deal of O. As n the proof of Theorem 3.3, we see that p s Q-splt. Furthermore, snce σ Kq 2 ) = θ q 2c), we conclude that p has a generator π such that π c mod q 2 ) cf. Proposton 3.5b)). Then by constructon π mod q 2 ) generates O/q 2 ), verfyng condton 2) of Theorem 3.7. To verfy condton 1), we fx p P and consder two cases. Frst, suppose σζ p dp)+1) ζ p dp)+1. Snce p s prme to p, ths means that the resdue feld O/p does not contan an element of order p dp)+1 although, snce s prme to p, t does contan an element of order, hence of order p dp) ). So, n ths case φp)/ s prme to p, and there s nothng to prove. Now, suppose that σζ p dp)+1) = ζ p dp)+1. Then by constructon σ acts nontrvally on every p dp)+1 -th root of u, and therefore the polynomal X pdp)+1 u has no roots n k vp. Agan, snce p s prme to p, we see from Hensel s lemma that u mod p) s not a p dp)+1 -th power n the resdue feld. It follows that the p-prmary component of the order of u mod p) s not less than the p-prmary component of φp)/p dp), and 1) follows. Fnally, by constructon σ acts trvally on ζ λ but nontrvally on ζ qλ for any q Q. Snce p s prme to λ, we see that the resdue feld O/p contans an element of order λ, but does not contan an element of order qλ for any q Q. Ths means that λ φp) but φp)/λ s relatvely prme to each q Q, whch s equvalent to condton 3) of Theorem Proof of Theorem 1.1 Frst, we wll ntroduce some addtonal notatons needed to convert the task of factorng a gven matrx A SL 2 O) as a product of elementary matrces nto the task of reducng the frst row of

16 14 A. MORGAN, A. RAPINCHUK, AND B. SURY A to 1, 0). Let RO) = {a, b) O 2 ao + bo = O} note that RO) s precsely the set of all frst rows of matrces A SL 2 O)). For λ O, one defnes two permutatons, e + λ) and e λ), of RO) gven respectvely by a, b) a, b + λa) and a, b) a + λb, b). These permutatons wll be called elementary transformatons of RO). For a, b), c, d) RO) we wll wrte a, b) n c, d) to ndcate the fact that c, d) can be obtaned from a, b) by a sequence of n equvalently, n) elementary transformatons. For the convenence of further reference, we wll record some smple propertes of ths relaton. Lemma 4.1. Let a, b) RO). 1a) If c, d) RO) and a, b) n c, d), then c, d) n a, b). 1b) If c, d), e, f) RO) are such that a, b) m c, d) and c, d) n e, f), then a, b) m+n e, f). 2a) If c O such that c amod bo), then c, b) RO), and a, b) 1 c, b). 2b) If d O such that d bmod ao), then a, d) RO), and a, b) 1 a, d). 3a) If a, b) n 1, 0) then any matrx A SL 2 O) wth the frst row a, b) s a product of n + 1 elementary matrces. 3b) If a, b) n 0, 1) then any matrx A SL 2 O) wth the second row a, b) s a product of n+1 elementary matrces. 4a) If a O then a, b) 2 0, 1). 4b) If b O then a, b) 2 1, 0). Proof. For 1a), we observe that the nverse of an elementary transformaton s agan an elementary transformaton gven by [e ± λ)] 1 = e ± λ), so the requred fact follows. Part 1b) s obvous. Note that 1) mples that the relaton between a, b) and c, d) RO) defned by a, b) n c, d) for some n N s an equvalence relaton.) In 2a), we have c = a + λb wth λ O. Then co + bo = ao + bo = O, so c, a) RO), and e + λ) takes a, b) to c, b). The argument for 2b) s smlar. 3a) Suppose A SL 2 O) has the frst row a, b). Then for λ O, the frst row of the product AE 12 λ) s a, b + λa) = e + λ)a, b), and smlarly the frst row of AE 21 λ) s e λ)a, b). So, the fact that a, b) n 1, 0) mples that there exsts a matrx U SL 2 O) whch s a product of n elementary matrces and s such that AU has the frst row 1, 0). Ths means that AU = E 21 z) for some z O, and then A = E 21 z)u 1 s a product of n + 1 elementary matrces. The argument for 3b) s smlar. ) Part 4a) follows snce e a e+ a 1 1 b) ) a, b) = 0, 1). The proof of 4b) s smlar. Remark. All assertons of Lemma 4.1 are vald over any commutatve rng O. Corollary 4.2. Let q be a prncpal Q-splt prme deal of O wth generator q, and let z O be such that zmod q 2 ) generates O/q 2 ). Gven an element of RO) of the form b, q n ) wth n 2, and an nteger t 0, there exsts an nteger t t 0 such that b, q n ) 1 z t, q n ).

17 BOUNDED GENERATION OF SL 2 O S ) 15 Proof. By Lemma 3.1b), the element zmod q n ) generates O/q n ). Snce b s prme to q, one can fnd t Z such that b z t mod q n ). Addng to t a sutable multple of φq n ) f necessary, we can assume that t t 0. Our asserton then follows from Lemma 4.12a). Lemma 4.3. Suppose we are gven a, b) RO), a fnte subset T Vf k, and an nteger n 0. Then there exsts α O k and r O such that V α) T =, and a, b) 1 αr n, b). Proof. Let h k be the class number of k. If for each v S \V k we let m v denote the maxmal deal of O k correspondng to v, then the deal m v ) h k s prncpal, and ts generator π v satsfes vπ v ) = h k and wπ v ) = 0 for all w Vf k \ {v}. Let R be the subgroup of k generated by π v for v S \ V ; k note that R O. We can pck r R so that a := ar n O k. We note that snce a and b are relatvely prme n O, we have V a ) V b) S. Now, t follows from the strong approxmaton theorem that there exsts γ O k such that vγb) 0 and vγb) 0mod nh k ) for all v S \ V k, and vγb) = 0 for all v V a ) \ S. Then, n partcular, we can fnd s R so that vγbs 1 ) = 0 for all v S \ V k. Set By constructon, γ := γs 1 O and b := γ b O k. 15) vb ) = 0 for all v V a ) S \ V k ), mplyng that V a ) V b ) =, whch means that a and b are relatvely prme n O k. Agan, by the strong approxmaton theorem we can fnd t O k such that vt) = 0 for v T V a ) and vt) > 0 for v T \ V a ). Set α = a + tb O k. Then for v T V a ) we have va ) > 0 and vtb ) = 0 n vew of 15)), whle for v T \ V a ) we have va ) = 0 and vtb ) > 0. In ether case,.e. V α) T =. On the other hand, whch means that a, b) 1 αr n, b), as requred. vα) = va + tb ) = 0 for all v T, a + r n tγ b = r n a + tb ) = r n α, Recall that we let denote the number of roots of unty n k. Lemma 4.4. Let a, b) RO) be such that a = α r for some α O k and r O where V α) s dsjont from S V ). Then there exst a O and nfntely many Q-splt prme prncpal deals q of O wth a generator q such that for any m 1mod φa O)) we have a, b) 3 a, q m ). Proof. The argument below s adapted from the proof of Lemma 3 n [CK1]. It reles on the propertes of the power resdue symbol n partcular, the power recprocty law) descrbed n the Appendx on Number Theory n [BMS]. We wll work wth all v V k and not only v V k \ S), so to each such v we assocate a symbol modulus ) m v. For v Vf k we wll dentfy m v wth the correspondng maxmal deal of O k obvously, p v = m v O for v V k \ S); the valuaton deal and

18 16 A. MORGAN, A. RAPINCHUK, AND B. SURY the group of unts n the valuaton rng O v or O mv ) n the completon k v wll be denoted ˆm v and U v respectvely. For any dvsor ν, we let ), m v be the b-multplcatve, skew-symmetrc) ) power resdue symbol of degree ν on k v cf. [BMS, p. x, y 85]). We recall that = 1 f one of the elements x, y s a ν-th power n k v n partcular, f v m v ν s ether complex, or s real and one of the elements x, y s postve n k) v ) or f v s nonarchmedean x, y / V ν) and x, y U v. It follows that for any x, y k, we have = 1 for almost all v V k. Furthermore, we have the recprocty law: ) x, y 16) = 1. v V k m v Now, let = p e 1 1 pen n be a prme factorzaton of. For each = 1,..., n, pck v V p ). Accordng to [BMS, A.17)], the values ) x, y for x, y U v m v p e cover all p e -th roots of unty. Thus, we can pck unts u, u U v for = 1,..., n so that u, u ) = ζ e p, a prmtve p e -th root of unty. Then 17) ζ := m v p e ν ν n u, u ) =1 s a prmtve -th root of unty. Furthermore, t follows from the Inverse Functon Theorem or Hensel s Lemma that we can fnd an nteger N > 0 such that 18) 1 + ˆm N v k v for all v V ). We now wrte b = βt wth β O k and t O. Snce a, b are relatvely prme n O, so are α, β, hence V α) V β) S. On the other hand, by our assumpton V α) s dsjont from S V ), so we conclude that V α) s dsjont from V β) V ). Applyng Theorem 3.3 to the rng O k we obtan that there exsts β O k havng the followng propertes: 1) 1 b := β O k s a prme deal of O k and the correspondng valuaton v b / S V ); 2) 1 β > 0 n every real completon of k; 3) 1 β βmod αo k ); 4) 1 for each = 1,..., n, we have β u mod ˆm v ), and β 1mod ˆm v ) for all v V p ) \ {v }. m v p e m v ν

19 Set b = β t. BOUNDED GENERATION OF SL 2 O S ) 17 It s a consequence of 3) 1 that b b mod ao), so by Lemma 4.12) we have a, b) 1 a, b ). Furthermore, t follows from 4) 1 and 18) that β /u k v, so u, β ) u, u ) = = ζ e m v p e m p. v p e Snce ζ defned by 17) s a prmtve -th root of unty, we can fnd an nteger d > 0 such that ) ) α, β α, β 19) 1 = ζ d n u d =, β ). b b m v By constructon, v b / V α) V ), so applyng Theorem 3.3 one more tme, we fnd α O k such that 1) 2 a := α O k s a prme deal of O k and the correspondng valuaton v a / S V ); 2) 2 α αmod b); 3) 2 α u d mod ˆmN v ) for = 1,..., n. Set a = α r. Then a O = α O s a prme deal of O and a amod b O), so a, b ) 1 a, b ). α, β ) Now, we note that = 1 f ether v V k snce β > 0 n all real completons of m v k) or v Vf k \ V α ) V β ) V )). Snce the deals a = α O k and b = β O k are prme by constructon, we have V α ) = {v a } and V β ) = {v b }. Besdes, t follows from 18) and 4) 1 that for v V p ) \ {v } we have β k v α, β ), and therefore agan = 1. Thus, the recprocty law m v 16) for α, β reduces to the relaton α, β ) α, β ) n α, β ) 20) = 1. a b m v It follows from 2) 2 and 3) 2 that α, β ) ) α, β b = b and Comparng now 19) wth 20), we fnd that β, α ) a α, β ) m v = =1 = α, β a =1 u d, β ) ) 1 b = 1. p e for all = 1,..., n. Ths mples cf. [BMS, A.16)]) that β s a -th power modulo a,.e. β γ mod a) for some γ O k. Clearly, the elements a = α r and γt are relatvely prme n O, so applyng Theorem 3.3 to ths rng, we fnd nfntely many Q-splt prncpal prme deals q of O havng a generator q γtmod a O). Then for any m 1mod φa O)) we have q m q β t b mod a O), so a, b ) 1 a, q m ). Then by Lemma 4.11b), we have a, b) 3 a, q m ), as requred.

20 18 A. MORGAN, A. RAPINCHUK, AND B. SURY The fnal ngredent that we need for the proof of Theorem 1.1 s the followng lemma whch uses the noton of the level l p u) of a unt u of nfnte order wth respect to a Q-splt deal p ntroduced n 3.1. Lemma 4.5. Let p be a prncpal Q-splt deal of O wth a generator π, and let u O be a unt of nfnte order. Set s = l p u), and let λ and m be ntegers satsfyng λ φp) and m 0modφp s )/λ). Gven an nteger δ > 0 dvdng λ and b O prme to π such that b s a δ-th power mod p) whle ν := λ/δ dvdes the order of umod p), for any nteger t s there exsts an nteger n t for whch π t, b m ) 1 π t, u nt ). Proof. Let p be the ratonal prme correspondng to p. Beng a dvsor of λ, the nteger δ s relatvely prme to p. So, the fact that b s a δ-th power mod p mples that t s also a δ-th power mod p s. On the other hand, t follows from our assumptons that λm = δνm s dvsble by φp s ), and therefore b m ) ν 1modp s ). But snce ν s prme to p, the subgroup of elements n O/p s ) of order dvdng ν s somorphc to a subgroup of O/p), hence cyclc. So, the fact that the order of umod p), and consequently the order umod p s ), s dvsble by ν mples that every element n O/p s ) whose order dvdes ν les n the subgroup generated by umod p s ). Thus, b m u ns mod p s ) for some nteger n s. Snce p s Q-splt, we can apply Lemma 3.2 to conclude that for any t s there exsts an nteger n t such that b m u nt mod p t ). Then π t, b m ) 1 π t, u nt ) by Lemma 4.12). We wll call a unt u O fundamental f t has nfnte order and the cyclc group u s a drect factor of O. Snce the group O s fntely generated Drchlet s Unt Theorem, cf. [CF, 2.18]) t always contans a fundamental unt once t s nfnte. We note that any fundamental unt has the followng property: u / k) p k ) p for any prme p. We are now n a poston to gve Proof of Theorem 1.1. We return to the notatons of 3.3: we let K denote the Hlbert S-class feld of k, let K be ts normal closure over Q, and pck an nteger λ 1 whch s dvsble by and for whch K Q ab Qζ λ ). Furthermore, snce O s nfnte by assumpton, we can fnd a fundamental unt u O. By Lemma 4.13), t suffces to show that for any a, b) RO), we have 21) a, b) 8 1, 0). Frst, applyng Lemma 4.3 wth T = S \ V k ) V ) and n =, we see that there exst α O k and r O such that V α) S V )) = and a, b) 1 αr, b). Next, applyng Lemma 4.4 to the last par, we fnd a O and a Q-splt prncpal prme deal q such that v q / S V λ) V φa O)) and αr, b) 3 a, q m ) for any m 1mod φa O). Then 22) a, b) 4 a, q m ) for any m 1mod φa O)). To proceed wth the argument we wll now specfy m. We let P and Q denote the sets of prme dvsors of λ/ and φa O), respectvely, and defne λ and as n 3.3; we note that by constructon q s relatvely prme to λ. So, we can apply Theorem 3.7 whch yelds a Q-splt prncpal prme

21 BOUNDED GENERATION OF SL 2 O S ) 19 deal p = πo so that v p / V φa O)) and condtons 1) - 3) are satsfed. Let s = l p u) be the p-level of u. Condton 3) mples that gcdφp)/λ, λ /λ) = 1 = gcdφp)/λ, φa O)) snce λ /λ s the product of all prme dvsors of φa O). It follows that the numbers φp s )/λ and φa O) are relatvely prme, and therefore one can pck a postve nteger m so that m 0mod φp s )/λ) and m 1mod φa O)). Fx ths m for the rest of the proof. Condton 2) of Theorem 3.7 enables us to apply Corollary 4.2 wth z = π and t 0 = s to fnd t s so that a, q m ) 1 π t, q m ). Snce P conssts of all prme dvsors of λ/, condton 1) of Theorem 3.7 mples that λ/ dvdes the order of umod p). Now, applyng Lemma 4.5 wth δ = and b = q, we see that π t, q m ) 1 π t, u nt ) for some nteger n t. Fnally, snce u s a unt, we have π t, u nt ) 2 1, 0). Combnng these computatons wth 22), we obtan 21), completng the proof. Corollary 4.6. Assume that the group O s nfnte. Then for n 3, any matrx A SL n O) s a product of 1 2 3n2 n) + 4 elementary matrces. Proof. Indeed, snce the rng O s Dedeknd, t s well-known and easy to show that any A SL n O) can be to a matrx n SL 2 O) by at most 1 2 3n2 n) 5 cf. [CK1, p. 683]). Now, our result mmedately follows from Theorem 1.1. Proof of Corollary 1.2. Let e + : α 1 α 0 1 ) and e : α 1 0 α 1 be the standard 1-parameter subgroups. Set U ± = e ± O). In vew of Theorem 1.1, t s enough to show that each of the subgroups U + and U s contaned n a product of fntely many cyclc subgroups of SL 2 O). Let h k be the class number of k. Then there exsts t O such that vt) = h k for all v S \ ) V k and vt) = 0 for all v / S. Then O = O k [1/t]. So, lettng U 0 ± = e ±O k ) and t 0 h = 0 t 1, we wll have the ncluson U ± h U ± 0 h. On the other hand, f w 1,..., w n where n = [k : Q]) s a Z-bass of O k then U ± 0 = e ±w 1 ) e ± w n ), hence 23) U ± h e ± w 1 ) e ± w n ) h, as requred. Remark. 1. Quanttatvely, t follows from the proof of Theorem 1.1 that SL 2 O) = U U + U nne factors), so snce the rght-hand sde of 23) nvolves n + 2 cyclc subgroups, wth h at both ends, we obtan that SL 2 O) s a product of 9[k : Q] + 10 cyclc subgroups. 2. If S = V K, then the proof of Corollary 1.2 yelds a factorzaton of SL 2 O) as a fnte product γ 1 γ d of cyclc subgroups where all generators γ are elementary matrces, hence unpotent. On the contrary, when S V K, the factorzaton we produce nvolves some dagonal semsmple) )

22 20 A. MORGAN, A. RAPINCHUK, AND B. SURY matrces. So t s worth pontng out n the latter case there s no factorzaton wth all γ unpotent. Indeed, let v S \ V K and let γ SL 2 O) be unpotent. Then there exsts N = Nγ) such that for any a = a j ) γ we have va j ) Nγ) for all, j {1, 2}. It follows that f SL 2 O) = γ 1 γ d where all γ are unpotent, then there exsts N 0 such that for any a = a j ) SL 2 O) we have va j ) N 0 for, j {1, 2}, whch s absurd. 5. Example For a rng of S-ntegers O n a number feld k such that the group of unts O s nfnte, we let νo) denote the smallest postve nteger wth the property that every matrx n SL 2 O) s a product of νo) elementary matrces. So, the result of [Vs] mples that νz[1/p]) 5 for any prme p, and our Theorem 1.1 yelds that νo) 9 for any O as above. It may be of some nterest to determne the exact value of νo) n some stuatons. In Example 2.1 on p. 289, Vsemrnov clams that the matrx ) 5 12 M = s not a product of four elementary matrces n SL 2 Z[1/p]) for any p 1mod 29), and therefore νz[1/p]) = 5 n ths case. However ths example s faulty because for any prme p, n SL 2 Z[1/p]) we have M = ) = ) ) ) 2 However, t turns out that the asserton that νz[1/p]) = 5 s vald not only for p 1mod 29) but n fact for all p > 7. More precsely, we have the followng. Proposton 5.1. Let O = Z[1/p], where p s prme > 7. Then not every matrx n SL 2 O) s a product of four elementary matrces. In the remander of ths secton, unless stated otherwse, we wll work wth congruences over the rng O rather than Z, so the notaton a bmod n) means that elements a, b O are congruent modulo the deal no. We begn the proof of the proposton wth the followng lemma. Lemma 5.2. Let O = Z[1/p], where p s any prme, and let r be a postve nteger satsfyng p 1mod r). Then any matrx A SL 2 O) of the form ) 1 p α 24) A = 1 p β, α, β Z whch s a product of four elementary matrces, satsfes the congruence ) 0 1 A ± mod r). 1 0 Proof. We note rght away that the requred congruence s obvous for the dagonal entres, so we only need to establsh t for the off-dagonal ones. Snce A s a product of four elementary matrces, t admts one of the followng presentatons: 25) A = E 12 a)e 21 b)e 12 c)e 21 d), or 26) A = E 21 a)e 12 b)e 21 c)e 12 d),

23 BOUNDED GENERATION OF SL 2 O S ) 21 wth a, b, c, d O. Frst, suppose we have 25). Then reducng t modulo bo, we obtan the followng congruence: ) 1 + da + c) a + c A E 12 a + c)e 21 d) = mod b). d 1 Lookng back at 24) and comparng the 2, 2) entres, we obtan that 1 p β 1mod b). It follows that b O,.e. b = ±p γ for some nteger γ. Smlarly, one shows that c = ±p δ for some nteger δ. Next, we observe that the sgns nvolved n b and c must be dfferent. Indeed, otherwse we would have ) A = E 12 a)e 21 ±p γ )E 12 ±p δ )E 21 d) = 1 + p γ+δ, whch s nconsstent wth 24). Thus, A looks as follows: A = E 12 a)e 21 ±p γ )E 12 p δ a1 p )E 21 d) = γ+δ ) p δ d1 p γ+δ ) ± p δ Consequently, the requred congruences for the off-dagonal entres mmedately follow from the fact that p 1mod r), provng the lemma n ths case. Now, suppose we have 26). Then A 1 = E 12 d)e 21 c)e 12 b)e 21 a), whch means that A 1 has a presentaton of the form 25). Snce the requred congruence n ths case has already been establshed, we conclude that ) A ± mod r). 1 0 But then we have as requred. 0 1 A ± 1 0 ) mod r), To prove the proposton, we wll consder two cases. Case 1. p 2 s composte. Wrte p 2 = r 1 r 2, where r 1, r 2 are postve ntegers > 1, and set r = p 1. Then 27) r ±1mod r) for = 1, 2. Indeed, we can assume that r 2 p 2. If r 2 ±1modr) then r 2 1 would be a nonzero ntegral multple of r, and therefore r r 2 + 1, hence p 2 p 2. But ths s mpossble snce p > 3. Thus, r 2 ±1mod r). Snce r 1 r 2 1mod r), condton 27) follows. Now, consder the matrx ) 1 p r1 p A = 1 p r 2 One mmedately checks that A SL 2 O). At the same tme, A s of the form 24). Then Lemma 5.2 n conjuncton wth 27) mples that A s not a product of four elementary matrces. Case 2. p and p 2 are both prmes. In ths paragraph we wll use congruences n Z. Clearly, a prme > 3 can only be congruent to ±1mod 6). Snce p > 5 and p 2 s also prme, n our ).

24 22 A. MORGAN, A. RAPINCHUK, AND B. SURY stuaton we must have p 1mod 6). Furthermore, snce p > 7, the congruence p 0 or 2mod 5) s mpossble. Thus, n the case at hand we have p 1, 13, or 19mod 30). If p 13mod 30) n Z, then p 3 7 mod 30), and therefore p 3 2 s an ntegral multple of 5. Set r = p 1 and s = p 3 2)/5, and consder the matrx 1 p 3 5p A = 3 ) s 1 p 3 Then A s a matrx n SL 2 O) havng form 24). Note that 5p 3 5mod r), whch s dfferent from ±1mod r) snce r > 6. Now, t follows from Lemma 5.2 that A s not a product of four elementary matrces. It remans to consder the case where p 1 or 19mod 30). Consder the followng matrx: A = ), and note that A SL 2 Z) and A 1 = ). It suffces to show that nether A nor A 1 can be wrtten n the form ) c + a1 + bc) 28) E 12 a)e 21 b)e 12 c)e 21 d) =, wth a, b, c, d O b + d1 + bc) 1 + bc) Set t = b + d1 + bc) and u = c + a1 + bc). Assume that ether A or A 1 s wrtten n the form 28). Then 1 + bc = 900, so b, c {±p n, ±29p n, ±31p n, ±899p n n N}. On the other hand, we have the followng congruences t bmod 30) and u cmod 30). Analyzng the above lst of possbltes for b and c, we conclude that each of t and u s ±p n mod 30) for some nteger n. Thus, f p 1mod 30) then t, u ±1mod 30), and f p 19mod 30) then t, u ±1, ±19mod 30). Snce 17 ±1, ±19mod 30), we obtan a contradcton n ether case. We observe that the argument n ths last case s nspred by Vsemrnov s argument n hs Example 2.1.) Acknowledgments. The second author was supported by the Smons Foundaton. The thrd author gratefully acknowledges frutful conversatons wth Maxm Vsemrnov.