COUNTING TAMELY RAMIFIED EXTENSIONS OF LOCAL FIELDS UP TO ISOMORPHISM

Transcription

1 COUNTING TAMELY RAMIFIED EXTENSIONS OF LOCAL FIELDS UP TO ISOMORPHISM Jim Brown Dpartmnt of Mathmatical Scincs, Clmson Univrsity, Clmson, SC 9634, USA Robrt Cass Dpartmnt of Mathmatics, Univrsity of Kntucky, Lxington, KY 40506, USA Kvin Jams Dpartmnt of Mathmatical Scincs, Clmson Univrsity, Clmson, SC 9634, USA Rodny Katon Dpartmnt of Mathmatics, Univrsity of Oklahoma, Norman, OK 73019, USA Salvator Parnti Dpartmnt of Mathmatics, Univrsity of Wisconsin, Madison, WI 53706, USA Danil Shankman Dpartmnt of Mathmatics, Purdu Univrsity, Wst Lafaytt, IN 47907, USA Abstract Lt p b a prim numbr and lt K b a local fild of rsidu charactristic p. In this papr w giv a formula that counts th numbr of dgr n tamly ramifid xtnsions of K in th cas p is of ordr modulo n. 1 Introduction A cntral problm in numbr thory is to classify finit fild xtnsions E/F for F a global fild. As thr ar infinitly many such xtnsions for any fixd dgr n, this is a difficult problm. It is oftn mor tractabl to instad classify local fild xtnsions and us this information to study global fild xtnsions. In particular,

2 givn a local fild K of rsidu charactristic p, it is wll known that up to isomorphism thr ar only finitly many xtnsions E/K of fixd dgr n and so such classifications ar tractabl. In this papr w provid a formula for th numbr of dgr n tamly ramifid xtnsions of K in th cas that p has ordr modulo n. On can s Thorm 11 for a prcis statmnt of th rsult. Th classification of finit xtnsions of local filds amounts to classifying unramifid, tamly ramifid, and wildly ramifid xtnsions. Unramifid xtnsions ar asy to classify as thr is only on such fild xtnsion for ach fixd dgr n. Classifying wildly ramifid xtnsions is much mor difficult and complt classifications ar only known for small dgrs (s for xampl [1,, 4]). In this papr w study th cas of tamly ramifid xtnsions, which falls btwn unramifid and wildly ramifid xtnsions in trms of difficulty. Lt n b a ramification indx and st f = n/ to b th rsidu class dgr. Th numbr of dgr n ramification indx xtnsions of K has bn calculatd by Roqutt by studying dfining polynomials for tamly ramifid xtnsions ovr th inrtia fild of K. Th radr is rfrrd to [3, Chap. 16] for a dscription of ths rsults. W tak a diffrnt approach that rlis only on lmntary counting and group action argumnts. St g = gcd(, p n/ 1). It is known that up to isomorphism th numbr of dgr n ramification indx xtnsions of K is xactly th numbr of orbits of Z/g Z undr th action of p ([3, Chap. 16]). W us lmntary mthods to calculat th siz of th orbits of Z/g Z undr th action of p, and thus th numbr of dgr n ramification indx xtnsions of K. W thn us ths orbit counts to provid a formula for th numbr of dgr n tamly ramifid xtnsions of K in th cas that p has ordr modulo n by summing ovr th numbr of orbits. In sction w prsnt two straightforward cass whr th orbit structur is asy to writ down. W thn dal with dtrmining th orbit structur of Z/gZ undr th action of p whn p has ordr l modulo g whr l is a prim. Finally, in sction 4 w us th orbit counts to giv our formulas for th numbr of tamly ramifid xtnsions of K of dgr n whn p has ordr modulo n. In this papr w adopt th following notation. W dnot th ordr of p in (Z/gZ) by ord g (p). W writ v p (n) = m if p m n. W will dnot an orbit in Z/gZ containing a undr multiplication by p by O g (a, p). W lt K(n, p) dnot th numbr of dgr n tamly ramifid xtnsions of K up to isomorphism and O(, p) th numbr of orbits of Z/g Z undr th action of p, whr w rcall from abov that g = gcd(, p n/ 1), which will b usd throughout. In particular, w hav K(n, p) = O(, p).

3 3 A coupl of straightforward cass W now giv th two simplst cass, namly whn p ±1 (mod n). Givn a nonngativ intgr k and a positiv intgr n, lt σ k (n) dnot th sum of th kth powrs of th positiv divisors of n, i.., σ k (n) = d n d k. In particular, th function σ 0 (n) is simply th numbr of divisors of n (oftn dnotd τ(n)), and th function σ 1 (n) is simply th sum of divisors of n (oftn dnotd σ(n)). Proposition 1. Lt p 1 (mod n). Thn w hav K(n, p) = σ 1 (n). Proof. Lt n. Not that sinc p 1 (mod n), w hav p 1 (mod ) so p n/ 1 0 (mod ). Thus, g =. Sinc p 1 (mod ), multiplication by p sorts Z/Z into distinct orbits. Thus, O(, p) =. This givs th rsult. Proposition. Lt p 1 (mod n). W hav if n is odd and if n is vn and w writ n = m b. K(n, p) = σ 0 (n) K(n, p) = (m + 3/)σ 0 (m) + m 1 σ 1 (m) Proof. First, suppos that n is odd and lt n. Sinc n is odd, so is and hnc so is n/. This givs p n/ 1 ( 1) n/ 1 (mod ) (mod ), i.., p n/ (mod d) for vry divisor d. Howvr, this mans if g > 1, w must hav som d > 1 with d so that d p n/ 1. This implis d p n/ 1 and d p n/ + 1, i.., d. Howvr, this is impossibl sinc n is odd. Thus, g = 1 for vry n. Thus, O(, p) = 1 for vry and so th numbr of xtnsions is xactly th numbr of divisors of n, i.., K(n, p) = σ 0 (n). Considr th cas now whn n = m p m 1 1 p m r r with m > 0. Lt n with val () < m. This implis n/ is vn and so p n/ 1 (mod n). In particular, w hav p n/ 1 (mod ). Thus, p n/ 1 and so g =. Lt a Z/Z. If 0 < a < /, thn a < and so a 0 (mod ). Thus, pa a (mod ) and hnc #O (a, p) =. If / < a < thn < a <, so a 0 (mod ). This implis

4 pa a (mod ) and thus #O (a, p) =. If / is an intgr, thn #O (/, p) = 1. Thus, in this cas th numbrs of orbits of Z/Z undr th action of p is givn by { O(, p) = + 1 vn +1 odd. Th contribution from ths cass to th total numbr of xtnsions is givn by ( ) + 1 ( ) v ()=0 0<v ()<m Th rmaining cas to dal with is whn v () = m. Hr w hav n/ is odd, so p n/ 1 (mod ). Thus, p n/ 1 (mod ) and so p n/ 1 cannot hav any odd prim divisors in common with. Howvr, if k p n/ 1, thn w hav 0 p n/ 1 (mod k ) (mod k ). This can happn only if k = 1, so g = in this cas. Sinc p 1 (mod ), this givs that p splits Z/Z into distinct orbits. Thus, w obtain ord ()=m xtnsions from this cas. Combining all of ths givs that ( ) + 1 ( ) K(n, p) = v ()=0 0<v ()<v (n) v ()=v (n) If w writ n = m b, thn w hav th following simplifications. W hav ( ) + 1 = + 1 b b v ()=0 = σ 1(b) + σ 0(b),. 4

5 5 0<v ()<v (n) ( ) + 1 = = 1 m 1 j=1 b m 1 j=1 m 1 ( j ) + 1 j + b m 1 j=1 1 b = 1 j σ 1 (b) + (m 1)σ 0 (b) j=1 ( m ) 1 = σ 1 (b) + (m 1)σ 0 (b), and v ()=v (n) = b 1 = σ 0 (b). Combining all of ths givs th rsult. Th nxt simplst cas to study is whr n is squar-fr and p is of ordr modulo n. Howvr, vn this is quit a bit mor complicatd and on dos not gt narly as clan of a formula as on gts in th prvious cas whr p ±1 (mod n). 3 Counting orbits In this sction w prsnt rsults on counting orbit sizs that will b ncssary to gnraliz th cass prsntd in th prvious sction. This sction provids th hart of th papr. Throughout this sction w writ g = m p m1 1 p m r r with m 0, m i 1, and th p i ar distinct odd prims. Lmma 3. Lt a (Z/gZ) and lt ord g (p) = k. Thn #O g (a, p) = k. Proof. W know that #O g (a, p) k as O g (a, p) {a, pa, p a,..., p k 1 a}. Suppos that #O g (a, p) < k. Thn thr xists 1 j < k so that p j a = a. Howvr, sinc a is a unit this is quivalnt to p j = 1, which contradicts ord g (p) = k. Lmma 4. Lt m 1 and st g = m. Lt p b an odd prim with ord m(p) =. W hav th following orbit structur of Z/gZ undr th action of p: 1. if m = 1, thr ar two orbits ach of siz 1;

6 . if m =, thr ar two orbits of siz 1 ({0}, {}) and on orbit of siz ({1,3}); 3. if m 3, thn w split into cass: (a) if p 1 (mod m ), thn all orbits hav siz xcpt {0} and { m 1 } ar thir own orbits; (b) if p m 1 1 (mod m ), thn all orbits hav siz xcpt {0} and { m 1 } ar thir own orbits; (c) if p m (mod m ), thn if a is vn {a} is its own orbit, and othrwis th orbit has siz. Proof. Clarly if g = thr ar xactly orbits. If g = 4, thn th only lmnt of ordr is 3, and this falls undr what w hav don abov as 3 1 (mod 4), so th orbits ar siz if a = 1, 3 and siz 1 if a = 0,. W can now assum m 3. W claim thr ar xactly 3 lmnts of ordr in (Z/ m Z) and thy ar givn by 1, m 1 ± 1. To s thr ar thr lmnts of ordr, rcall that (Z/ m Z) = C C m whr C n is a cyclic group of ordr n. Lt x b th uniqu lmnt of ordr in C and lt y b th uniqu lmnt of ordr in C m. Thn th only lmnts of ordr ar givn by (x, y), (1, y), and (x, 1). It is now simpl to s th lmnts claimd hav ordr by using th fact that m 3 so ( m 1 ± 1) = m ± m + 1 m m + 1 (mod m ) 1 (mod m ). Thus, w only nd considr ths thr lmnts whn dtrmining th orbit structur. W alrady know if p 1 (mod m ), thn th orbits hav siz xcpt for a = 0, m 1. Lt p m 1 1 (mod m ). If a = m 1, w hav pa = m m m 1 m 1 (mod m ) a (mod m ). Thus, a = 0, m 1 hav orbits of siz 1. W know all odd a hav orbits of siz, so it rmains to dal with th cas that a = j b for 1 j < m 1 and b odd. If pa a (mod m ), thn using that b is a unit modulo m w hav ( m 1 1) j j (mod m ), which is quivalnt to m (m ). Howvr, this is impossibl sinc m 3. Thus, unlss a = 0, m 1 w hav #O m(a, p) =. It now only rmains to dal with p m (mod m ). Hr w claim #O m(a, p) = 1 unlss a is odd. W hav that if a is odd thn th orbit siz is siz, so it only rmains to show that if a is vn it is its own orbit. This is asy as ( m 1 + 1)j j (mod m ). 6

7 W now rturn to th gnral cas g = m p m 1 1 p mr r. Th nxt cas to dal with is whn ord g (p) = l, l a prim, and if l m l g thn ord l m l (p) = 1. Obsrv th last rquirmnt givs that in ordr to hav an lmnt p of ordr l modulo g, it must b th cas that l (p i 1) for som i = 1,..., r. W will mak us of th following fact in th proof of Lmma 6. Lmma 5. Suppos ord g (p) = l whr l is a prim and assum if l m l g thn ord l m l (p) = 1. If ord m p i (p) = l, thn ord pi (p) = l. i Proof. Our assumption implis that l (p i 1). Suppos that it is th cas that ord pi (p) = 1. St D = (p i 1)p mi 1 i and obsrv w hav a commutativ diagram whr θ is th natural projction map taking a (mod p m i i ) to a (mod p i ), C D and (Z/p m i i Z) = θ C D (Z/p i Z) = C pi 1 ϕ 7 C pi 1 ar cyclic groups, and if w writ C D = x, thn ϕ is th map that snds x to x pm i 1 i, which is a gnrator of C pi 1. Sinc p has ordr l in (Z/p m i i Z), it ncssarily corrsponds to an lmnt of th ( form x ad/l ) for som 0 < a < l. Not that w cannot hav p i 1 ad l sinc p v m i 1 i ad l < v l (p i 1) as l p i a. Thus, w must hav that ϕ(x ad/l ) 1 in l C pi 1. Howvr, this contradicts th fact that w ar assuming θ(p) = 1. Lmma 6. Suppos ord g (p) = l whr l is a prim and assum if l m l ord l m l (p) = 1. St M = j pm j j so that ord m p j (p) = l. Lt a Z/gZ. j 1. If gcd(a, g) = 1, thn #O g (a, p) = l. g thn. If gcd(a, g) > 1, thn: (a) if M a, thn #O g (a, p) = 1; (b) if M a, thn #O g (a, p) = l. Proof. W hav alrady covrd th cas gcd(a, g) = 1. Assum now that M a. Th claim is that #O g (a, p) = 1. Lt N = g/m. W us th isomorphism Z/gZ = Z/MZ Z/NZ to writ p = (p M, p N ) and a = (a M, a N ). Not that ord M (p M ) = l and ord N (p N ) = 1 by construction of M and N. Morovr, w hav a M = 0 by assumption. Sinc ord N (p N ) = 1, w hav pa = (p M, p N ) (0, a N ) = (p M 0, p N a N ) = (0, a N ) = a. Thus, O g (a, p) = {a}, as claimd.

8 Now suppos that M a. W nd to show that p j a a (mod g) for 1 j < l. Suppos that thr is such a j, namly, w hav p j a = a (mod g). W can rwrit this as (p j M a M, p j N a N ) = (a M, a N ), i.., p j M a M = a M and p j N a N = a N. Using th first of ths quations, w hav p j M a M a M = 0, i.., a M (p j M 1) = 0. Howvr, this givs that p i (p j M 1) for som p i M for othrwis M a, i.., p has ordr lss than l modulo p i. Howvr, this contradicts Lmma 5 and th assumption that p i M. Thus, w hav #O g (a, p) = l in this cas. W can now prov th gnral rsult whn ord p (g) =. Proposition 7. Lt p b a prim with ord g (p) =. Lt M = j pm j j so that ord m p j (p) =. If ord m(p) = 1, st M = M. If ord m(p) =, thn dfin M as j follows: 1. if p 1 (mod m ) or p m 1 1 (mod m ), st M = m 1 M ;. if p m (mod m ), st M = M. If M a, thn #O g (a, p) = 1. Othrwis, #O g (a, p) =. Proof. Th proof of this proposition amounts to combining Lmma 6 and Lmma 4. W hav #O g (a, p) = unlss #O m(a, p) = 1 and #O m p i (a, p) = 1 for all i i. Howvr, ths orbits all hav siz on xactly whn M a by th prvious lmmas. Exampl 8. Lt g = 4 so m = 3, p 1 = 3, and m 1 = 1. Considr th prim p = 5. Obsrv that p has ordr modulo 4, modulo 3, and modulo 8. Morovr, p = m On asily chcks that whn acting upon Z/4Z by 5, th orbits ar givn by {0}, {1, 5}, {, 10}, {3, 15}, {4, 0}, {6}, {7, 11}, {8, 16}, {9, 1}, {1}, {13, 17}, {14, }, {18}, and {19, 3}, which agrs with th proposition sinc in this cas M = 6. Though it will not b usd in our counting argumnts, it is now asy to provid th analogous rsult to Proposition 7 for th cas ord g (p) = l for l an odd prim. W provid this rsult for compltnss. Th nxt stp is to dal with th cas whn ord g (p) = l for l an odd prim with l g but l (p j 1) for all j = 1,..., r. Not for this to b possibl w must hav l = p i for som i with m i > 1. Lmma 9. Lt p b a prim with ord g (p) = p i for som i = 1,..., r and assum p i (p j 1) for all j = 1,..., r. Lt a Z/gZ. If p i a thn #O g (a, p) = 1. Othrwis #O g (a, p) = p i. Proof. Without loss of gnrality w can assum ord g (p) = p 1. Writ h = g/p m 1 1. W can writ Z/gZ = Z/p m 1 1 Z Z/hZ. Sinc p 1 φ(h) by assumption, w hav ord h (p) = 1 and so p acts as th idntity on Z/hZ. 8

9 Suppos that p 1 a and assum thr is a j with 1 j < p 1 so that p j a a (mod g). Sinc p acts trivially on Z/hZ, this statmnt is quivalnt to p j a m p 1 = 1 a m p 1 for som j with 1 j < p 1 1. Howvr, this givs p m 1 1 (p j 1), which contradicts th fact that p ncssarily has ordr p 1 modulo p m1 1. Thus, it must b that if p 1 a, thn #O g (a, p) = l. Now assum p 1 a and writ a = p 1 c. Again w us th fact that p acts as th idntity on Z/hZ to conclud w only nd to dtrmin what happns to th p m1 1 componnt of a. Hr w mak us of th fact that if p has ordr p 1 in Z/p m 1 1 Z, thn p = bp m for som 1 b p 1 1. Th rsult is thn clar bcaus w hav pa = (bp m )(p 1 c) = p 1 c = a in th p m1 1 componnt. W now combin Proposition 7 and Lmma 9 to obtain th following rsult. Proposition 10. Lt p b a prim with ord g (p) = l for l an odd prim. Lt M = j pm j j so that ord m p j (p) = l and l p j. If l g, st M = M. If l = p j for j som 1 j m and ord m p j (p) = c, st M = cm whr c = 1, l. If M a, thn j #O g (a, p) = 1. Othrwis, #O g (a, p) = l. Proof. Not that if l g or c = 1 w ar don, so assum without loss of gnrality that l = p 1 and ord l m 1 (p) = l. First suppos that M a. St N = g/l m 1 M and considr th isomorphism Z/gZ = Z/l m 1 Z Z/M Z Z/NZ. By assumption w can writ a = (a l m 1, a M, a N ) = (a l m 1, 0, a N ). Obsrv that w hav pa = (pa l m 1, 0, pa N ) = (pa l m 1, 0, a N ) (sinc ord N (p) = 1 by assumption) = (a l m 1, 0, a N ) (by Lmma 9) = a. Thus, if M divids a w hav th orbit has siz 1 as claimd. Now suppos M a but p j a = a for som 1 j l. Howvr, this lads to th quations p j a l m 1 = a l m 1 and p j a M = a M. Sinc M a ths cannot both hold unlss j = l. 4 Main counting rsults W ar now abl to stat our main rsult. Throughout this sction w writ n = m p m 1 1 p m r r with m 0, m i 1, and th p i distinct odd prims. Considr th cas that v () = m. By assumption w hav n/ is odd and so p n/ 1 p 1 (mod ). Thus, w hav p splits Z/g Z into g orbits and so w obtain th numbr of dgr n xtnsions of K arising from this situation is givn by g. v ()=v (n) 9

10 Now suppos that v () < m. Thn w hav n/ and so g = gcd(, 0) =. It is not ncssarily th cas that ord (p) =, so w brak this into two cass. If ord (p) = 1, thn p acts on Z/Z as th idntity, hnc splits it into distinct orbits. Thus, for this cas w hav O(, p) =. If ord (p) =, w can us Proposition 7 to count th orbits in trms of M. (Not that sinc g varis in this sction w writ M g to kp track of th group Z/gZ upon which p is acting.) In this cas w hav th numbr of orbits givn by O(, p) = φ() + # {a Z/Z : gcd(a, ) > 1, a 0, M a} + # {a Z/Z : gcd(a, ) > 1, a 0, P a} + 1 whr th 1 coms from 0 always bing its own orbit. Combining all of this w obtain th following thorm. Thorm 11. Lt p b a prim with p n and ord p (n) =. For n, dfin M as in Proposition 7. Th numbr of dgr n xtnsions of K up to isomorphism is givn by K(n, p) = v ()=v (n) g + 1 ( n )) σ 1(n/) + σ 1 (gcd, p 1 + n (p 1) Proof. W immdiatly hav from th prcding discussion that K(n, p) = g v ()=v (n) v ()<v (n) p 1 (mod ) v ()<v (n) p 1 (mod ) v ()<v (n) p 1 (mod ) ( φ() M + # {a Z/Z : gcd(a, ) > 1, a 0, M a} (# {a Z/Z : gcd(a, ) > 1, a 0, M a} + 1). First, w simplify th conditions undrnath th sums. W not that ; v () < v (n) is quivalnt to n. W also not that p 1 (mod ) is quivalnt to (p 1). Finally, in th last two sums w omit th condition that a 0. In th first sum, not that P a a 0. In th scond sum w simply absorb th on to compnsat. This givs th following xprssion.. ) 10

11 11 K(n, p) = + + v ()=v (n) n (p 1) g + ( ϕ() n (p 1) + #{a Z/Z : gcd(a, ) > 1; P a} #{a Z/Z : gcd(a, ) > 1; P a} ) n (p 1) Now w not that th union of th two sts showing up in th last two sums along with th st of rsidu classs countd by ϕ() is just all of Z/Z. W us this obsrvation to rarrang th last two sums and obtain th following xprssion. K(n, p) = = + 1 v ()=v (n) n (p 1) v ()=v (n) g + n (p 1) ( + #{a Z/Z : gcd(a, ) > 1; M a}) g + 1 ( + #{a Z/Z : gcd(a, ) > 1; M a}) n 1 (#{a Z/Z : gcd(a, ) > 1; M a} ) gcd( n,(p 1)) Not that #{a Z/Z : gcd(a, ) > 1; M a} = { ϕ() if M = 1, M Sinc M 1, w hav #{a Z/Z : gcd(a, ) > 1; M a} = Thus, w may writ othrwis. M.

12 1 K(n, p) = = v ()=v (n) v ()=v (n) g + 1 g + 1 n ( + M ) 1 gcd( n,p 1) ( ) M ( n ) σ 1(n/) + σ 1 (gcd, p 1 ) + n (p 1) M, which givs th rsult. On can asily chck that this rsult rcovrs Lmma in th cas w tak p 1 (mod n). In th cas l =, whn w considr p n/ 1 modulo, this is ithr 0 if v () < m or p 1 if v () = m du to th fact that th only rmaindrs possibl upon dividing n/ by ar 0 or 1. In ithr cas it is asy to us th orbit structur to giv a count. Howvr, for gnral l w must considr rmaindrs 0, 1,..., l 1. If th rmaindr is largr than 1, it is not obvious how p will act on Z/g Z in this cas. Thus, whil w hav th rlvant orbit counting rsults for p of prim ordr l modulo n, it is not as straightforward to count th xtnsions in this cas. This will b th subjct of futur rsarch. Acknowldgmnts Th authors wr partially supportd by th grant NSF DMS funding th Clmson REU on Computational algbraic gomtry, combinatorics, and numbr thory during th summr of 013. Th authors would lik to thank th rfr for commnts that improvd th xposition of this papr. Rfrncs [1] C. Awtry. Dodcic 3-adic Filds. Int. J. Num. Th., 8: , 01. [] J. Brown, R. Cass, R. Katon, S. Parnti, and D. Shankman. Dgr 14 xtnsions of Q 7. Int. J. of Pur and Appl. Math., 100(): , 015. [3] Hlmut Hass. Numbr thory. Classics in Mathmatics. Springr-Vrlag, Brlin, grman dition, 00. Rprint of th 1980 English dition [Springr, Brlin; MR (81c:1001b)], Editd and with a prfac by Horst Güntr Zimmr.

13 [4] J. Jons and D. Robrts. A databas of local filds. J. Symbolic Comput., 41:80 97,