Finite Fields and Their Applications

Transcription

1 Fnte Felds nd Ther Applctons 18 (2012) Contents lsts vlble t ScVerse ScenceDrect Fnte Felds nd Ther Applctons Bnoml nd fctorl congruences for F q [t] Dnesh S. Thkur 1 Deprtment of Mthemtcs, Unversty of Arzon, Tucson, AZ 85721, USA rtcle nfo bstrct Artcle hstory: Receved 18 My 2011 Revsed 22 August 2011 Accepted 23 August 2011 Avlble onlne 8 September 2011 Communcted by Arne Wnterhof In memory of lte Bhmsen Josh MSC: 11T55 11R58 11B65 11A07 We present severl elementry theorems, observtons nd questons relted to the theme of congruences stsfed by bnoml coeffcents nd fctorls modulo prmes (or prme powers) n the settng of polynoml rng over fnte feld. When we look t the fctorl of n or the bnoml coeffcent n choose m nthssettng, though the vlues re n functon feld, n nd m cn be usul ntegers, polynomls or mxed. Thus there re severl nterestng nlogs of the well-known theorems of Lucs, Wlson etc. wth qute dfferent proofs nd new phenomen Elsever Inc. All rghts reserved. Keywords: Functon felds Crltz module Bernoull numbers 1. Introducton Mny strong nloges [4,6] between number felds nd functon felds over fnte felds hve been used to beneft the study of both. These nloges re even stronger n the bse cse Q, Z F q (t), F q [t]. We wll explore congruences stsfed by nlogs of fctorls nd bnoml coeffcents. When we look t the fctorl of n or the bnoml coeffcent n choose m n ths settng, though the vlues re n functon feld, n nd m cn be usul ntegers, polynomls or mxed. Thus we wll see severl nterestng nlogs of the well-known theorems of Lucs, Wlson etc. We refer to [6, Chpter 4] for hstorcl references, propertes of these nlogs nd proofs of mny thngs reclled here. E-ml ddress: thkur@mth.rzon.edu. 1 Supported n prt by NSA grnt H /$ see front mtter 2011 Elsever Inc. All rghts reserved. do: /j.ff

2 272 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) Let us fx the bsc stndrd notton. Z ntegers F q fntefeldofq elements, q powerofprmep A the polynoml rng F q [t], t vrble over F q K the functon feld F q (t) A + moncs (n t) na A d elements of A of degree d [n] =t qn t D n = n 1 =0 (tqn t q ) = [n ] q L n = n =1 (tq t) = [] e k (x) = (x ), where A runs through elements of degree < k N = q d for A d,.e., the norm of monc rreducble polynoml n A of degree d v(n) hghest k such tht N k dvdes n We use the stndrd conventon tht empty sums re zero nd the empty products re one, so tht D 0 = L 0 = Multple nlogs For n Z, n 0, we defne the frst fctorl (due to Crltz) by n!:= D n A +, for n = n q, 0 n < q. See [6, , 4.12, 4.13] for ts propertes, such s prme fctorzton, dvsbltes, functonl equtons, nterpoltons nd rthmetc of specl vlues, whch re nlogous to those of the clsscl fctorl. See lso [1,2], whch gves mny nterestng dvsblty propertes n gret generlty, n prtculr, pplyng to the frst fctorl nd to the frst nd the thrd bnomls below. (The usul fctorl wth vlues n Z wll be lwys mentoned s the clsscl fctorl.) For x A, wth x not monc, we defne the second fctorl by Π(x) := ( 1 + x ) 1 K. A+ See [6, ] for ts nlogous propertes such s the locton of poles (n -A + ), functonl equtons, nterpoltons t ll prmes nd rthmetc of specl vlues etc. Its recprocl s ntegrl! Note lso tht we bsclly excluded q = 2 wth the condtons on x. (The usul bnoml wth vlues n Z wll be lwys mentoned s the clsscl bnoml.) For n,m Z, n,m 0, we defne the frst bnoml coeffcent by ( ) n n! := m m!(n m)! A +, f n m, 0otherwse. See [6, ] for ts nlogous propertes regrdng dvsbltes. For, b A, wth, b not monc, we defne the second bnoml coeffcent by [ b ] := Π() Π(b)Π( b) K, f b not monc, 0otherwse. For A, n Z, n 0, we defne the thrd bnoml coeffcent by

3 n := D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) q n A, where x q k := x q ( 1) k / ( D L q k ), where n re the bse q dgts of n s bove. See [6, ] for nlogous propertes of ths defnton, due to Crltz, ts use Mhler type nterpolton n results of Wgner etc. We now record some results [6, Chpter 2] we wll use often. (III) (V) nd (VII) re due to Crltz. (I) [n] s the product of monc rreducble polynomls n A of degree dvdng n. (II) (D 0 D 1 D d 1 ) q 1 = D d /L d. (III) D n s the product of monc polynomls n A of degree n. (IV) L n s the (monc) lest common multple of polynomls n A of degree n. (V) x q k =e k(x)/d k, nd thus equls 0 (1 respectvely), f x A s of degree less thn (monc of degree equl to) k. (VI) Π() 1 = d =0 (1 + q ), f A d,wth not monc. (VII) If C (z) = q zq,thenc b (z) = C (C b (z)) = C b (z). (NoteC s the fmous Crltz module, but we wll not need ny more of ts relted theory.) 3. Results of Lucs type The well-known theorem of Lucs expresses the clsscl bnoml coeffcent m choose n modulo prmep s the product of m choose n, where m, n re the bse p dgts of m,n respectvely. In our cse, the modulus s prme of the functon feld, nd we get severl versons, wth dgts for the bse or N, ccordng to whch bnoml we use. Theorem 3.1. Let be prme of A of degree d. Then we hve ( ) m ( ) m (d) mod, n n (d) where m = m (d)q d nd n = n (d)q d re the bse q d -expnsons of m nd n respectvely, so tht 0 m (d), n (d)<q d. In prtculr, the bnoml s zero modulo f nd only f there s crry over of q d -dgts n the sum n + (m n),.e., n (d)>m (d) for some. Proof. Frst observe tht f there s no crry over of bse q-dgts, then ll the bnoml coeffcents bove re equl to one, becuse of the dgt expnson defnton of frst fctorl. Now suppose there s crry over t (bse q) exponents, + 1,..., j 1, but not t 1or j. Let m k q k, n k q k nd lk q k be the bse q expnsons of m, n, m n respectvely. Then n k + l k s m + q, m k + q 1or m j 1 ccordng s whether k s, + 1 k j 1ork = j. Thus, usng the dgt expnson nd the defnton of the fctorls, we see tht the contrbuton of ths block of dgts to the bnoml coeffcent expresson s D j D q 1 j 1 Dq 1 +1 Dq =[j] []. On the other hnd, the congruence clss of [k] modulo depends on the congruence clss of k modulo d, nd both re zero f d dvdes k. Theorem 3.2. Let be prme of A of degree d. Then we hve mod, n n (d)

4 274 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) where = (d) (n = n (d)q d respectvely) s the bse - (q d = N respectvely)-expnsons of m nd n respectvely, so tht 0 m (d), n (d)<q d. In prtculr, the bnoml s zero modulo f nd only f the q-degree of n (d) s greter thn the t-degree of,forsome. Proof. All congruences below re modulo nd 0 j < d. By (V) nd (III), q 0for d. Hence by comprng the coeffcents n (VII), for b =, wehve() On the other hnd, we hve (b) q k+d q kd+ j = +l=k+d q q l q q k. = k k q kd+ j q kd+ j k k q j, where the frst equlty s by the F q -lnerty of the bnoml coeffcent n the top vrble nd (V), the lst by () nd the mddle by usng the defnton of the bnoml nd notcng tht /L kd+ j 0, s the numertor hs vluton t, nd > k = the vluton of the denomntor by (I) nd defnton of L.Ifwewrtethebseq-expnson of n (d) = n j q j, combnng we hve n = j q d+ j nj nj = q d+ j n (d)q d n (d), s clmed. Theorem 3.3. Let be prme of A of degree d. Then we hve [ ] [ ] mod, b b where = nd b = b re the bse -expnsons of nd b respectvely (so tht N, N b < N ) nd when ll the bnomls re defned. In prtculr, under these condtons the bnoml on the left s zero modulo f nd only f b s monc for some. Proof. We hve [ b ] = (1 + b )(1 + b) q q 1 + q. In the proof of the lst theorem we sw tht modulo, wehve q d+ j q j, so tht the product bove decomposes over dgts bse q d nd we see tht the left sde s the product over of [ b ] s clmed. Remrks 3.4. Note tht under the condtons of the theorem, the bnoml on the left s -ntegrl. We cn stll get some nformton by relxng these condtons (of no poles ) by nterpretng zero n the numertor or the denomntor of the bnoml expressons on the rght s dvsblty by of the numertor or denomntor.

5 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) Results of Wlson type The clsscl Wlson/Lebntz/Lgrnge theorem [3] sys (p 1)! 1 mod p, for the clsscl fctorl nd p prme. Let be monc prme of degree d. Note tht N 1 = q d 1 = d 1 =0 (q 1)q. Theorem 4.1. We hve for the frst fctorl, (N 1)!( 1) d 1 mod. Proof. We gve two proofs. Frst we observe tht the product of non-zero elements of degree less thn d s 1 modulo by prng nverses modulo wth elements, s n the clsscl Wlson proof, snce 1 nd re the only elements of order dvdng two n the cyclc group (A/ A), whch get pred to themselves. Next, note tht by (III) of Secton 2, we see tht (N 1)!=(D 0 D d 1 ) q 1 dffers from ths product by multplcton of sgns whch contrbute ( 1) 1+q+ +qd 1 = ( 1) d,snce θ F q θ = 1. Next we gve nother proof wth formuls. By (II) of Secton 2, we hve (N 1)! ( 1) d 1 = D d + ( 1) d =[d 1] q 1 [1] q d ( 1) d. L d The proof follows from the observton tht for 0 < < d, wehve[] qd 1 (t qd t qd )/[] 1 modulo, snce by Fermt s lttle theorem t qd t. Theorem 4.2. If p s odd, nd θ F q, θ 1,wehve Π( 1) 1/2 mod, Π( + θ) ( 2(1 + θ) ) 1 mod. Proof. We use ( (VI). The d-th term of the product gves 2 1. The constnt θ or 1 mttersonlyn ) 0-th term, nd q 0mod, for < d. Ths mples the clm esly. 5. Prmlty crter Now we wll study nlogs ( of ) the clsscl prmlty crter tht (n 1)! 1modn f nd only n+m f n s prme, f nd only f n 1modn, for0 m < n, where the fctorl nd bnoml re clsscl. Theorem 5.1. We hve: () For A, nd the frst fctorl, (N 1)!( 1) deg() 1 mod, f nd only f s prme. () For the second bnoml, we hve (ssumng sgn condtons gvng exstence) [ ] +b 1 mod, for ll b, N b < N, f nd only f s prme. Proof. The f prt of () s just the Wlson theorem nlog bove. Conversely, f s not prme, t hs fctor of degree less thn deg, whch thus dvdes (N 1)! by (III) of Secton 2, nd so cnnot dvde the fctorl ±1. Ths proves ().

6 276 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) Let deg = d nd deg b = l. Then by (VI) of Secton 2, [ +b ] d =0 = (1 + q ) l =0 (1 +b q ) d =0 (1 + q +b q ). By (V), one hs cncellton nd ll products need only run to l. Now, f s prme, then by (III), (IV) (or by (III) nd (V)) nd the defnton of the thrd bnoml, s dvsble by, f < d. Somodulo, the frst product n the numertor contrbutes 1 nd the q second nd the denomntor re the sme, provng f prt of (). Conversely, f s not prme nd f k s the degree of the smllest degree prme, sy c dvdng, thenweuseb of degree l = k. All the lower degree terms cncel s before, but q k, whch s e k()/d k by (V) s not zero modulo, becuse the only term n the product for the numertor whch hs fctor common to s tself, nd c dvdes the denomntor by (II). Remrks 5.2. () On the ( other ) hnd, we do not get such bnoml prmlty crteron for the frst n+m bnoml. In fct, for t, n = 1, for ll 0 m < n, fndonlyfn = q k, s there s no crry over modulo q. Smlrly, for the thrd bnoml, we hve wth A +, +b N =1, for 0 N b < N, by(v). () The converse of lst theorem of prevous secton does not hold, s the concluson of the theorem holds lso e.g., f we replce by composte =[1] 2, when q = 3. As explned n the proof, the concluson then s equvlent by (VI) to deg() 1 =1 (1 + ) 1modulo. It s esy to see tht when q q = 2 = (t 2 + t + 1) 2 stsfes ths, s modulo, the terms correspondng to = 1, 2, 3 respectvely re congruent to 1, 1 + /L 2 nd 1 + /L 3 respectvely. But we need p 2. I thnk Alejndro Lr Rodrguez for mkng serch through s of smll degree, usng SAGE when q = 3 nd fndng the exmple bove, unque such for degree t most 7. (No exmple exsts of degree t most 6 for q = 5.) 6. Refned Wlson theorems Interestngly, whle the product of the reduced system of representtves of smllest postve (monc respectvely) modulo p ( respectvely) s (p 1)! (((N 1)/(q 1))! respectvely), f we use smllest sze representtves, t s (±(p 1)/2)! (±(N 1)! respectvely)! Also, whle smple countng gves ((p 1)/2)! 2 ( 1) (p 1)/2 (p 1)!, nourcse((n 1)/w)! w = (N 1)! for ll w dvdng q 1, just from the defntons. Another nterestng dfference s tht we hve the sme N 1forseverl s (nmely of sme degree). Whle the postve elements smller thn p gve reduced system of representtves modulo p, the monc elements of smller sze (degree) thn do not gve the full reduced system nd nsted, ther product s D 0 D d 1, when the degree of the prme s d. One cn sk ts congruence clss. In fct, f we consder the smllest bsolute vlue sze representtves clssclly the relevnt product s ((p 1)/2)!. Forp 3 mod 4, t s (see e.g. Hrdy Wrght) congruent modulo p to ( 1) n, where n s the number of qudrtc non-resdues less thn p/2. So we now nvestgte ( ) N 1 S := S d := S q,d, :=!=D 0 D d 1 mod. q 1 We hve proved tht S s q 1-th root n (A/ A) of ( 1) d 1. But whch one? Let us strt wth some trvl observtons. () If d = 1orq = 2, then S = 1. () If p = 2ord s odd, then S F q. On the other hnd, q = 3, d = 2, = t2 + 1, then S = t. () If d s odd, ((N 1)/2)!=S (q 1)/2 (whch s S, fq = 3) s ±1, closer to the clsscl cse.

7 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) (v) If q = 3 nd d odd, then 1 s qudrtc non-resdue modulo, nd so prllel to the clsscl cse, by the sme rgument, S = ( 1) n, where n s the number qudrtc non-resdues modulo mongmoncsofdegreelessthnd. (v) If F p r [t] F q [t], nd p = 2ord s odd, then S F p r [t] F q = F p r.inprtculr,s = 1, f p = 2, r = 1. For exmple, f = t 2 + t + 1 nd q = 2 s,wths odd, then S = 1. Theorem 6.1. Let θ,r F q. The monc prmes (t) for whch S = r nd those for whch S = r/θ d(d 1)/2 re n bjecton v (t) (θt)/θ d. If gcd(q 1, d(d 1)/2) = 1, then the prmes of degree d re equdstrbuted n ech congruence clss S = r, s r runs through q 1-th roots of ( 1) d 1. Proof. Note tht f we replce t by θt, [n] gets replced by θ[n], nd so notng tht q 1modq 1, we see tht D 0 D d 1 =[d 1][d 2] 1+q [1] 1+q+ +qd 2 gets multpled by θ (d 1) = θ d(d 1)/2 provng the frst clm. Gven the gcd condton, s θ runs through ll elements of F q,so does θ d(d 1)/2, provng the second clm. Remrks 6.2. We cn derve mny specl conclusons. For exmple, f d 3 mod 4, by choosng θ = 1, we see tht prmes re equdstrbuted n S = r nd S = r. Theorem 6.3. (1) Let θ F q.ifs q,d, (t) = r(t), thens q,d, (t+θ) = r(t + θ). In prtculr, f p = 2 or d odd, then S q,d, (t) = S q,d, (t+θ). (2) Let d be odd or p = 2, so tht S cn be consdered n F q.letσ be n utomorphsm of F q.then S σ q,d, = S q,d, σ. In prtculr, prmes re equdstrbuted n ll congruence clsses n Gl(F q/f p ) orbt. Proof. Snce [n] =t qn t = (t + θ) qn (t + θ), nd D s re the products of these, the frst sttement follows. When p = 2, or d s odd, then s we hve seen r F q, so tht r(t) = r(t + θ) nd the concluson (1) follows. To see (2), we hve only to note tht D 0 D d 1 F p [t]. Remrks 6.4. Sometmes, (t + θ)= (t). Forexmple,fq = p nd = defned below. Theorem 6.5. If d s odd nd not dvsble by p, then the number of prmes of degree d for whch S s n prtculr congruence clss r s multple of q. Proof. The product P r of prmes of degree d for whch S r modulo s the gretest common dvsor of [d] nd D 1 D d 1 r, nd thus polynoml (sy of degree k) n[1] wth F q -coeffcents. Hence, the number N r := deg(p r )/d = q(k/d) of such prmes s multple of q by the hypothess. Remrks 6.6. Ths q-dvsblty mkes one wonder whether there s ny F q -vector/ffne spce structure lurkng behnd. 7. Congruences modulo prme powers The usul proof of the Wlson theorem (tht the product of elements n (Z/pZ) s 1) generlzes mmedtely to the proof of well-known fct tht the product of elements n fnte beln group s the product of ts elements of order 2, nd s thus 1 f there s more thn one element of order 2 nd s the element of order 2 otherwse. Hence, the product for (Z/p n Z) s 1 for odd p or p n = 4 nd 1 otherwse, whle for (A/ n A) t s 1 unlessq = 2, deg = 1 nd n = 2, 3. See [6, p. 7] for more detls.

8 278 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) We sk now when does the Wlson type congruence works modulo hgher power. Note (5 1)! 1 mod5 2. Theorem 7.1. Let q = p, F q,sotht := t p t sprmena.wehve [1][2] [p 1] 1 mod q 1, (N 1)!1 mod q 1. Proof. Modulo such :=, we hve [1], so tht [n] [1] pn [1] n mplyng [1][2] [p 1] p (p 1) 1, by the usul Fermt nd Wlson theorems! In fct, [1]= +, [2]= p etc. mples [n] + n mod p.somodulo p,wehve [1][2] [p 1] ( + n) θ Fq ( + θ) p 1 1, provng the frst clm, s well s the fct tht the power of cnnot be mproved. From the frst prgrph, we lso see tht [p 1] p [1] p p 1 1mod p, so dvdng by the frst clm, we get the second clm tht (N 1)!=[p 1] p 1 [p 2] p2 1 [1] p p 1 1 1mod p 1. Remrks 7.2. If q = 2 s nd d = 2, then (N 1)!, s s seen from [2]=[1] q +[1]= ([1]+θ), where θ runs through elements of F q. Ths lso shows drectly tht S s equdstrbuted n F q n ths cse. Theorem 7.3. Wehve,forthefrstbnoml, ( N k ) ( ) n n N k m m mod v(n)+q 1. Proof. By (I), dvdes [r] f nd only f d dvdes r. Agn by (I), qr dvdes [d] qr =[r + d] [r]. If further d dvdes r, then[r + d]/ [r]/ mod qr 1. As explned n the proof of Lucs theorem for the frst fctorl, f there s no crry over n m + (n m), then both sdes re one. Otherwse crry over produces products of consecutve brckets t the crry over plces. More precsely, let k (r respectvely) be smllest such tht n k (m r respectvely) s non-zero. If r < k, thsproduces[k] [r + 1]. By the two observtons bove, t s congruent to [k + d] [r d] modulo s := k/d r/d +q r+1 1-th power of. Now v(n) = k/d nd s v(n) + q 1. Any other [] occurrng v crry over (e.g. f r k) hs k nd thus [ + d] [] mod qk nd q k v(n) + q 1. Remrks 7.4. The cse n = q, m = 1, = t lredy shows tht the power n congruence s the best possble n generl. But t cn be mproved wth more nformton s below. Theorem 7.5. Wehve,forthefrstbnoml, ( ) ( ) N n n N m m mod w, where w = mx(0, k/d r/d ) + q mx(r,u)+1 1,whereq k n, q r m, q u m n. (Note k/d =v(n) nd r/d =v(m).) Proof. The proof follows the sme des s n the prevous proof, so we just sketch the chnges nd the cses. () If r < k, thenu = r nd the crry over produces [k] [r + 1], whch s dvsble by the dfference of the floors power of s before nd when dvded by them reduces the power by one, s n the prevous proof, ledng to power dfference of floors +q r+1 1 s before nd s clmed. () If r > k, thenu = k nd only possble crry overs led [] s wth r ledng congruences to

9 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) q r+1 1 power s clmed. () r = k nd u = r = k. Ths leds to sme stuton s n (). Fnlly, n cse (v) r = k, u > ( ) ( n r, we use symmetry m = n n m) nd reduce to the prevous cse ledng to q u+1 1-th power s clmed. Remrks 7.6. Good exmple, where the reder cn proftbly mke drect check very esly s = t, n = q k nd m = q j. Theorem 7.7. We hve, for the thrd bnoml, denotng vluton t by v,ndv:= v (), wehve q k N q k mod mx(qk+1,v k/d ). Proof. Comprng the coeffcents n (VII), we get q k+d = j+l=k+d q j q l q j = j+l=k+d q l q l q j. Now the defnton of the thrd bnoml nd (II), (III) mmedtely gves tht v ( ) = q j 1, f j < d, so tht the frst sum expresson gves q k q d q k mod qk+1, rrespectve of v. (Usng dgt expnsons, we cn replce q k n the two bnomls by ny of ts multple n the congruence bove.) But f ths vluton s bg, we cn do better by usng the second sum expresson nsted. Frst note tht by Fermt s lttle theorem nd (IV), the term for l = d s congruent to k. Now q f we look t terms n the defnton of the thrd bnoml q k, they hve vlutons q j v v (D j ) q j (k + j)/d, whch hve unque mnmum t j = 0, f ( the mnmum ) v k/d s q k+1, whch s the only cse of nterest for us. Ths gves the vluton of k nd the second sum expresson q thus gves the requred congruence to w-th power, where w = 1 + mn(q d j (v (k + j)/d )), nd 1 j d. Agn under our ssumpton tht v k/d q k+1, the mnmum s unque t j = d nd s v k/d s clmed. Remrks 7.8. () Note tht both the sdes of the congruence re zero, for k > deg(). () The sum formul n the defnton of the thrd bnoml, together wth q d =1ledto /L d ( 1) d modulo q, f d > 1, nd modulo q 1, f d = 1. By (I) nd (IV), ths cn be reformulted s nterestng congruence syng tht the monc lest common multple of ll, except, elements of degree d, whch s lso equl to the product of P d/ deg(p) over monc prmes P (of degree t most only mtter) s congruent to ( 1) d modulo q (respectvely q 1 )fd > 1(respectvely d = 1). Let us provde n nlog of the theorem [3] on the clsscl fctorl tht (np)!n!(p!) n mod p n+3. Theorem 7.9. We hve for the frst fctorl, (nn )!n! ( (N )! ) n mod n+q.

10 280 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) Proof. Let n = n q be the bse q expnson. Hence (nn )!= D n +d = ( ) [d + ][d + 1]q [d + 1] q 1 n n q D. d Now, by (III), n dvdes D n d nd when you dvde out ths power, we use [d + j][j] mod q j. Hence the clm follows, when we notce tht = 0 gves dentty, so, n generl, the mnmum power n the dvded out congruence holds for q-th power of correspondng to = 1. The specl cse n = q shows tht the power n the congruence s best possble n generl. 8. Hgher powers nd Bernoull numbers See [3] nd [5] (nd ts revew on MthScNet by Evns) for references for mny more results (some mentoned below) by Lgrnge, Wolstenholme, Ferrers, Glsher, Crltz etc. generlzng these results nd on congruences modulo hgher powers of prmes for the clsscl bnoml. In prtculr, congruences nlogous to those n Secton 7 work, for gven m, n, tohgherpower of prme p f nd only f p dvdes ( Bernoull ( number B p 3 = B p 1 2 Q. In fct, for the usul np bnoml coeffcents nd fctorls, () mp) / n m) 1+mn(m n)b p 3 p 3 /3modp t+4, where p 5 nd p t mn(m ( n). ) Crltz proved ths wth p t+4 replced by p 4. Its often quoted mmedte consequences np 1 re (b) p 1 1 n(n 1)p 3 B p 3 /3modp 4 nd (c) (np)!/(n!(p!) n ) 1 (n 3 n)p 3 B p 3 /9modp 4 (. Crltz lso proved (d) (p 1)/2) ( 1) (p 1)/2 4 p 1 (1 + p 3 B p 3 /12) mod p 4. Let us look t the F q [t] stuton. For n = q + 1, m = 2 the congruence n the cse (Theorem 7.3) of the frst bnoml reduces to [d + 1][1] nd q s the exct power of dvdng [d + 1] [1]=[d] q for ny of degree d, by (I). So there s no extr dvsblty, t lest n the full generlty of the clsscl cse. In the cse (Theorem 7.7) of the thrd bnoml lso, f, for exmple, we tke = t, then the power of n the dsplyed congruence n the proof s the best by rgument there, for ny t. Hence there ( s no extr dvsblty s n (). As for (b), no crry over mples tht for the frst nn 1) bnoml, we hve N 1 = 1, n fct. (Note tht mmedte rgument for deducng (b) from () n ( N the clsscl cse fls n ths cse.) Anlog of (d): For the frst bnoml, (N 1)/(q 1)) s dentclly one, gn for no crry over reson. Smlrly, we do not get extr dvsblty for (c). On the other sde wth Bernoull, see [6, Sec nd 5.3.9] or [4] for more on these Bernoull Crltz numbers B n F q (t) nd mny nloges they stsfy, such s ther genertng functon, occurrence n specl Crltz zet even vlue, nlog of the von-studt theorem etc. Anlog of B n /n s then B n (n 1)!/n! [6, Sec. 4.16], where the fctorl s the frst fctorl. In ths cse, no nce functonl equton s known for the Crltz Goss zet functon [4,6], so tht there re lso Bernoull Goss numbers β(n) (these, rther thn Bernoull Crltz numbers, stsfy nlogs of Kummer congruences ledng to -dc nterpolton) comng from the specl vlues t negtve ntegers. Now, let us recll the Crltz evluton [6, Thm ] of the specl Bernoull Crltz numbers B qh q = ( 1)h (q h q )!. From the evluton nd (I), we see tht never dvdes the Bernoull Crltz number B N q = B (N 1) (q 1) K. We hve not yet fully nvestgted wht hppens for them n generl, but we hve checked tht for q = 3, does not dvde ζ(1 (q d q)), fd 5. On the other hnd, out of 8 prmes of degree 3nF q [t], two dvde the Bernoull Goss number β(3 3 3), wth close connecton to the clss group component relted to B 3 3 3, v nlogs of the Herbrnd Rbet [6, Theorems 5.2.4, 5.3.8], [4]. L q h

11 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) Next, let us look t the Lgrnge, Wolstenholme, Ferrers, Glsher results generlzng Wlson s theorem to other elementry symmetrc functons: If we wrte (x 1)(x 2) (x p + 1) = x p 1 A 1 x p A p 1,thenmodulop (for ndces strctly between 0 nd p 1), () A r 0, () A 2r /p B 2r /(2r), () A 2r+1 /p 2 (2r + 1)B 2r /(4r). The Crltzn nlog would be (x ) = d =0 xqd 1 F qd qd, where the product s over non-zero polynomls of norm less thn N. Thus by the defnton of the thrd bnoml nd (V), we hve F qd q d = D d( 1). D d L qd So by (I) (IV), these coeffcents re dvsble by, but not ts hgher powers, n nlogy s well s contrst wth the clsscl cse mentoned bove. Comprng the Bernoull Crltz evluton bove, usng n dentty smlr to (II), n esy clculton shows tht we hve, for 0 < < d, F q d q d / = (L d/ )B q d q d /L d ( 1) d B q d q d /L d, where the lst congruence s modulo q (or q 1,fd = 1) nd follows by Remrk 7.8(). 9. Further questons, observtons nd prtl results (A) Snce the second congruence n Theorem 7.1 works wth the frst power of for ll, trses smlr queston for the frst congruence. By n rgument smlr to Eucld s rgument for nfntude of prmes, by (I), [1] [s 1]+1 cn only be dvsble by prmes of degree s nd we re skng when t s dvsble by the lowest possble degree s. Fors = p = q, tseems,butnotyetprovedtht the only prmes whch enter re the s dentfed n Theorem 7.1. Smll mount of dt tht we clculted shows tht t works for some other prmes (wht s ther chrcterzton?), but only n degrees dvsble by the chrcterstc p. More precsely, t suggests the guess tht the gretest common dvsor G of A := [1] [s 1]+1 nd B := [s] s non-trvl, only f p s. Here s the proof for s 4 nd ll q: For s = 1 t s vcuously true. For s = 2, t follows snce D 1 =[1] s congruent to q 1-th root of 1, nd s thus congruent to one only f p = 2. For s = 3, 4, t follows lso by the followng clculton. (We speculte, but cnnot prove yet, tht the method of ths proof generlzes to ll s.) Let x := [1], then[s]=x qs 1 + +x q + x, so tht gcd G of A, B s polynoml n x. IfG s polynoml n x of degree w > 0prmetos, theng s polynoml of degree wq n t, on the other hnd ts prme fctors re ll of degree s. Sowq s multple of s, hencep dvdes s. For s = 2, A q B = x 1. For s = 3, x q B A q A = x 2 2 dvsble by G, so tht w = 1or2. For s = 4, let C = A q x q (x q2 + x q )B, D = (A C)/x 2, E = B D q = x q2 + x, F = x q+1 E + C, nd G = F (x q2 + x q + x)f x q A = x q2 + x. ThenE + G = 2x, sow = 1, unless p = 2. (B) The next unresolved queston s wht the dstrbuton of prmes correspondng to the dfferent possbltes for S n Secton 6 s, when the hypothess of Theorem 6.1 does not pply. Here re some observtons from the smll numercl dt gthered by clcultng P r nd N r, the number of prmes correspondng congruence clss r (see proof of Theorem 6.5 for the notton), usng mxm. () For d = 3, we focus on the prmes q 61 nd q 1 modulo 3 (not fully hndled by Theorem 6.1) nd gve the vector of entres N r /q (see Theorem 6.5) correspondng to 1 r (q 1)/2 (ths rnge s enough by Remrk 6.2) s gven by q = 7, [3, 4, 1]; q = 13, [3, 4, 4, 7, 3, 7]; q = 19, [9, 4, 4, 7, 4, 7, 9, 9, 7]; q = 31, [12, 12, 13, 12, 7, 13, 13, 12, 7, 7, 7, 13, 7, 13, 12];

12 282 D.S. Thkur / Fnte Felds nd Ther Applctons 18 (2012) q = 37, [9, 16, 13, 13, 13, 9, 13, 9, 16, 9, 9, 16, 13, 9, 16, 16, 16, 13]; q = 43, [12, 12, 19, 12, 19, 19, 13, 12, 13, 19, 12, 19, 13, 13, 13, 12, 13, 13, 19, 19, 12]; q = 61, [21, 16, 21, 25, 16, 16, 16, 21, 21, 25, 21, 25, 16, 25, 16, 16, 25, 16, 25, 21, 16, 16, 21, 21, 25, 25, 21, 21, 25, 25]. Here, ll the repet entres cn be explned by the bjecton n Theorem 6.1. () Let d = 3 nd r = 1. For q = 3 n, N 1 s q(q + 1)/3 (nd thus not multple of q), t lest for n 4. For q = 4, 16, 64, 256 N 1 /q s 3, 3, 27, 75 respectvely nd for q = 25, 49 t s 12, 21 respectvely. () For q = 4, d = 4, S = 1 clss s empty nd the prmes re equdstrbuted n clsses for S = ζ 3 nd S = ζ 2 3 (equdstrbuton s n ccordnce wth Theorem 6.3(2)). An ndependent chrcterzton of the dstrbuton of these numbers nd of the prmes themselves n congruence clsses would be nterestng. References [1] Mnjul Bhrgv, P -orderngs nd polynoml functons on rbtrry subsets of Dedeknd rngs, J. Rene Angew. Mth. 490 (1997) [2] Mnjul Bhrgv, The fctorl functon nd generlztons, Amer. Mth. Monthly 107 (9) (2000) [3] Leonrd Dckson, Hstory of the Theory of Numbers (1919), vol. I, Dover edton, 2005, Chpter 9. [4] Dvd Goss, Bsc Structures of Functon Feld Arthmetc, Ergeb. Mth. Grenzgeb. (3) (Results n Mthemtcs nd Relted Ares (3)), vol. 35, Sprnger-Verlg, Berln, [5] Andrew Grnvlle, Arthmetc propertes of bnoml coeffcents I bnoml coeffcents modulo prme powers, n: Orgnc Mthemtcs, Burnby, BC, 1995, n: CMS Conf. Porc., vol. 20, Amer. Mth. Soc., Provdence, RI, 1997, pp [6] Dnesh S. Thkur, Functon Feld Arthmetc, World Scentfc Publshng Co. Inc., Rver Edge, NJ, 2004.