Gauss and Jacobi Sums and the Congruence Zeta Function

Transcription

1 U.U.D.M. Project Report 2018:19 Gau and Jacob Sum and the Congruence Zeta Functon Enar Waara Examenarbete matematk, 15 hp Handledare: Andrea Strömbergon Examnator: Martn Herchend Jun 2018 Department of Mathematc Uppala Unverty

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3 GAUSS AND JACOBI SUMS AND THE CONGRUENCE ZETA FUNCTION BACHELOR S THESIS DEPARTMENT OF MATHEMATICS UPPSALA UNIVERSITY EINAR WAARA 1. Introducton 2 2. Prerequte Theory Multplcatve Character Bac Noton from Algebrac Geometry Trace and Norm n Fnte Feld 6 3. Gau and Jacob Sum Gau Sum Jacob Sum and Applcaton 8 4. The Zeta Functon The Zeta Analogy The Ratonalty of the Zeta Functon 19 Content Date: January

4 1. Introducton The German mathematcan Carl Fredrch Gau wa the frt to ntroduce the noton of what now called quadratc Gau um, whch are um of the form g a = t F p t/pe 2πat/p, where F p the fnte feld of order p and /p the quadratc recprocty ymbol. Among many other thng, Gau proved that g 1 take the value p or p when p 1 4 or p 3 4 repectvely. Thee um can be, and have been, greatly generalzed and appear very often n area related to modern number theory, n the tudy of theta functon and the dcrete Fourer tranform for ntance. In th expoton we preent an ntroducton to the theory of Gau and Jacob um and ther nterrelaton and apply thee noton to tudy the congruence zeta functon. We gve a proof of the Hae-Davenport relaton, whch a lftng relaton relatng the Gau um of two dfferent fnte feld. We outlne a proof of a pecal cae of the frt part of the Wel conjecture named after André Wel proved n 1959 by Bernard Dwork , whch tate that any algebrac et V ha a ratonal zeta functon Z V u. We tate and prove a characterzaton of ratonal zeta functon Z f u and ue th characterzaton to prove that Z f u ratonal when fx 0..., x n = a 0 x m 0 + a 1x m a nx m n where a 0,..., a n F and F a fnte feld of order q 1 m. We alo further dplay the potency of Gau and Jacob um by provdng a hort proof of the famou law of quadratc recprocty, a theorem frt proved by Gau who refereed to t a the golden theorem and who ubequently provded fve addtonal proof, the fnal of whch wa publhed n Th expoton greatly npred by the excellent clac IR. K. Ireland and M. Roen, A Clacal Introducton to Modern Number Theory, Sprnger-Verlag, Th the man reference for the preent the. 2

5 2. Prerequte Theory Here we preent ome bac defnton and reult from elementary number theory for future reference. We tart wth the concept of a quadratc redue. Defnton 1. Let a Z and uppoe a, m = 1. Then a called a quadratc redue modulo m f x 2 a m ha a oluton,.e. a a perfect quare modulo m. If a not a quadratc redue, we ay that a a quadratc nonredue. For the remander of th ecton, let p be an odd prme number. Next we defne a convenent ymbol for dealng wth quadratc redue. Defnton 2. The Legendre ymbol /p defned by 1 f a a quadratc redue modulo p a/p = 1 f a a quadratc nonredue modulo p 0 f p a Propoton 2.1. IR, I a p 1/2 a/p p II ab/p = a/pb/p III a b p = a/p = b/p Proof. Aume that p a and p b f not, the proof trval. Snce a p 1 1 p we have a p 1/2 + 1a p 1/2 1 = a p p and thu a p 1/2 ±1 p. By a well-known fact we have a p 1/2 1 p f and only f a a quadratc redue modulo p. The econd part ealy proved ung the frt. The thrd part trval. Corollary 2.2. IR, corollary Modulo p there an equal number of quadratc redue a there are quadratc nonredue. Proof. The equaton a p 1/2 1 p ha p 1/2 oluton, hence there are p 1/2 quadratc redue and p 1 p 1/2 = p 1/2 quadratc nonredue. Corollary 2.3. IR, 6.3 lemma 2 p 1 t=0 t/p = 0. Proof. p 0 o 0/p = 0 by defnton. Thu there are p 1 p odd o that p 1 even term of ±1 left, and we know thee term cancel by the prevou corollary. Let F be a fnte feld of order q. Propoton 2.4. IR, The equaton x n = α F olvable f and only f α q 1/d = 1, where d = n, q 1. If the equaton olvable, then there are d oluton. Proof. The multplcatve group F of F cyclc. Let g be a generator of F. Set α = g a and x = g b for ome a, b {1, 2,..., q 1}. Subttutng th nto the equaton yeld g nb = g a f and only f g nb a = 1, thu q 1 nb a.e. nb a q 1, whch prove the reult by bac fact about congruence Multplcatve Character. The defnton of Gau and Jacob um rely on the noton of a multplcatve character, we therefore provde a bref ntroducton of th topc here. We follow [IR, Ch. 8.1]. Defnton 3. A multplcatve character on F p a group homomorphm from the multplcatve group of F p to the multplcatve group of the complex number: χ : F p C \ {0},. 3

6 The Legendre ymbol can therefore be regarded a pecal cae of a character. The et of character on F p conttute a group under the followng defnton. Let λ and χ be character. We defne λχ a the map a λaχa, and χ 1 a a χa 1. The dentty wll be denoted by ɛ and characterzed by ɛa = 1 for all a F p. It wll be ueful to extend the doman and let χ0 = 0 for all χ ɛ whle ɛ0 = 1. Notce that th n accordance wth the Legendre ymbol. We hall denote the group of character on F p by Ω p. Snce we are now dealng wth a group, we naturally defne the order of a character a the mallet potve nteger n uch that χ n = ɛ. The Legendre ymbol thu a character of order two. It turn out that Ω p a cyclc group of order p 1. In order to prove th, we need a few bac reult ealy proved from the defnton. Propoton 2.5. Let χ Ω p and a F p, then I χ1 = 1 II χa a p 1t root of unty III χa 1 = χa 1 = χa Proof. χ1 = χ1 2 = χ1 2 and nce χ1 0 we have χ1 = 1. Alo, nce Fp = p 1 we have a p 1 = 1 and thu χ1 = χa p 1 = χa p 1 = 1. Latly χa 1 χa = χa 1 a = χ1 = 1 hence χa 1 = χa 1. The complex number χa le on the unt crcle n C nce χ1 = χa p 1 = χa p 1 = χa p 1 = 1 mple χa = 1. Due to the defnton of complex multplcaton we have χa 1 = χa. Propoton 2.6. Let χ Ω p, then t F p χt = { 0 f χ ɛ p otherwe Proof. The equalty trval when χ = ɛ. If χ ɛ, then there ome a Fp uch that χa 1. Then χa t F p χt = t F p χat = t F p χt, hence t F p χt = 0. Propoton 2.7. Ω p a cyclc group of order p 1. Proof. We how that there ext a generator to Ω p. The multplcatve group Fp cyclc. Let g Fp be a generator. If a Fp then a = g l for ome l {0, 1,..., p 1}. Thu any character χ Ω p completely determned by t acton on g nce χa = χg l = χg l. In general the number of n th root of unty n. Hence nce χg C \ {0} a p 1t root of unty, t follow that Ω p p 1. We clam that a generator to Ω p gven by λg k = e 2πk/p 1. Frtly λ well defned nce λg k+p 1 = λg k e 2π = λg k. Snce λg k 1 g k 2 = λg k 1 λg k 2 we alo know that λ character, hence λ Ω p. We now how that λ a character of order p 1. Aume that λ n = ɛ. Then λ n g = λg n = e 2πn/p 1 = ɛg = 1 thu p 1 n. Now, λ p 1 a = λa p 1 = λa p 1 = λ1 = 1 thu λ p 1 = ɛ. There are no maller potve nteger whch dvde p 1 than p 1 telf. Thu p 1 the mallet potve nteger n uch that λ n = ɛ and therefore the order of λ p 1. Hence we know that ɛ, λ, λ 2,..., λ p 2 are all dtnct member of Ω p and nce there are p 1 of them and Ω p p 1 we know that Ω p = p 1. Thu λ ndeed a generator, and hence Ω p cyclc. Remark. Let a F p \ {1} then a = g l for ome l {1, 2,..., p 2}, o p 1 l. Thu λa = λg l = e 2πl/p 1 1. We conclude that gven a F p \ {1} there alway ome character χ uch that χa 1. Th crucal for the followng corollary. Corollary 2.8. If a F p \ {1} then χ Ω p χa = 0. 4

7 Proof. Let z be the complex number χ Ω p χa. Snce a 1 there by the remark above ome character χ uch that χa 1, thu λaz = λaχa = λχa = z. χ Ω p χ Ω p The lat equalty hold nce {λχ} χ Ωp = Ω p. Th follow from the mplcaton λχ 1 = λχ 2 = χ 1 = χ 2. Thu, nce λa 1, we ee that z = 0. We hall ee how character can be ued n the tudy of equaton n fnte feld. The followng propoton an early tool n th drecton. Propoton 2.9. If a F p, n p 1 and the equaton x n = a unolvable, then there a character χ uch that χ n = ɛ and χa 1. Proof. Let g and λ be a n the proof of propoton 2.7. We know that a = g l for ome l and by aumpton x n = a not olvable, thu n l otherwe g l/n would be a oluton. Let χ = λ p 1/n. Then χg = λg p 1/n = e 2π/n and thu χa = χg l = e 2πl/n 1. Latly χ n = λ p 1 = ɛ nce λ wa determned to be a generator to Ω p. Let Nx n = a denote the number of oluton to the equaton x n = a n F p. The followng propoton wll be a ueful tool for countng the number of oluton to more complcated equaton n F p and wll be ued often. Propoton If n p 1, then Nx n = a = χ n =ɛ χa where the um taken over all character χ uch that χ n = ɛ. Proof. If χ a character uch that χ n = ɛ, then χg C a n th root of unty nce χg n = χ n g = ɛg = 1. Thu there are at mot n uch character n Ω p. But n the prevou propoton we defned χ = λ p 1/n whch meant that χg = e 2π/n, thu ɛ, χ, χ 2,..., χ n 1 are n dtnct character, all of order n. Thu, thee are the character n the um of conderaton. To prove the equalty we conder three cae. Frt, f a = 0, clearly there only one oluton to x n = a, namely x = 0. Snce χ0 = 0 except for when χ = ɛ whch map 0 to 1 the equalty hold n th cae. Next, aume a 0 and that the equaton x n = a olvable. Let b be a oluton, then χa = χb n = χb n = ɛb = 1 and thu χ n =ɛ χa = n = Nxn = a. Latly, aume that a 0 and that the equaton x n = a unolvable. By propoton 2.9, there a ρ Ω p uch that ρ n = ɛ and ρa 1. We have {χ : χ n = ɛ} < Ω p, thu ρa χ n =ɛ χa = χ n =ɛ χρa = χ n =ɛ χa, hence ρa 1 χ n =ɛ χa = 0 whch prove the reult Bac Noton from Algebrac Geometry. In order to dcu the congruence zeta functon, we need ome fundamental defnton from algebrac geometry. We follow [IR, Ch. 10.1]. For the remander of th ecton, let F be a feld of order q. Defnton 4. A n F = {a 1, a 2,..., a n : a F for = 1, 2,..., n} called the affne n-pace over F. A n F can be condered to be a vector pace over F wth the uual defnton of calar multplcaton and vector addton. Let u conder A n+1 F \ {0, 0,..., 0} and defne an equvalence relaton on th et by a 0, a 1,..., a n b 0, b 1,..., b n γ F uch that a = γb for = 0, 1,..., n. Defnton 5. P n F = An+1 F \ {0, 0,..., 0} called the projectve n-pace over F. The equvalence cla of a A n+1 F \ {0, 0,..., 0} denoted by [a]. 5

8 The ze of the affne n-pace q n, whle the ze of the projectve n-pace qn+1 1 q 1 = q n + q n q + 1. A homogeneou polynomal fx 1, x 2,... x n = 1, 2,..., n a n x 1 1 x x n n n F [x 1, x 2,..., x n ] one where each nonzero term have the ame degree. For the ret of th ecton, let F K be a feld extenon. Gven ome nonzero not necearly homogeneou polynomal f F [x 1, x 2,... x n ] we can vew t a a functon f : A n K K by evaluatng f at pont n A n K n the uual way. Defnton 6. H f K = {a A n K : fa = 0} called the affne hyperurface of f n A n K. Let g F [x 0, x 1,..., x n ] be ome homogeneou polynomal of degree d. Defnton 7. H g K = {[a] P n K : ga = 0} called the projectve hyperurface of g n P n K. The et H g K well defned nce f γ F we have gγx = γ d gx. Th becaue g homogeneou of degree d. Thu f ga = 0 for ome repreentatve a n [a], then g map all element n that equvalence cla to 0. We defne a mlar et a above, but for everal polynomal ntead of only one. In word, t the et of common zero to a fnte et of polynomal. Let f 1, f 2,..., f m F [x 1, x 2,..., x n ]. Defnton 8. V = {a A n K : f j a = 0 for j = 1, 2,..., m} called an algebrac et defned over K Trace and Norm n Fnte Feld. In th hort ecton we provde ome bac defnton and reult whch we wll need later. We follow [IR, Ch. 11.2]. Let F be a feld of order q = p n and let E F be a feld of order q. Defnton 9. The trace of α E from E to F defned a tr E/F α = 1 =0 αq and the norm of α from E to F defned a N E/F α = 1 =0 αq. Propoton Let α, β E and a F, then I tr E/F α F II tr E/F α + β = tr E/F α + tr E/F β III tr E/F aα = a tr E:F α IV tr E/F : E F a urjectve mappng. V N E/F α F VI N E/F αβ = N E/F αn E/F β VII N E/F aα = a N E/F α VIII N E/F : E F a urjectve mappng. Proof. We prove the lat part. Note that N E/F a group homomorphm, o that α kern E/F f and only f N E/F α = 1 =0 αq = α q 1/q 1 = 1. By propoton 2.4 we know that the equaton x q 1/q 1 = 1 ha q 1/q 1 oluton n E. The frt omorphm theorem gve mn E/F = E / kern E/F = q 1 = F, hence N E/F ndeed urjectve. 6

9 3. Gau and Jacob Sum We tart by ntroducng the noton of Gau and Jacob um over F p, after whch we wden our vew and conder Gau and Jacob um over arbtrary fnte feld of order q = p r. We then gve a hort proof of the law of quadratc recprocty ung Gau and Jacob um. We follow [IR, Ch ] Gau Sum. Let χ Ω p, a F p and ζ = e 2π/p. Defnton 10. A um of the form t F p χtζ at called a Gau um belongng to the character χ and wll be denoted g a χ, and g 1 χ n partcular wll be denoted gχ. Lemma 3.1. { p 1 p f a 0 p t=0 ζat = 0 otherwe Proof. If a 0 p then a/p an nteger o ζ a = 1, hence p 1 t=0 ζat = p. Otherwe, f a 0 p, then ζ a 1 and p 1 t=0 ζat = ζap 1 ζ a 1 = 0. Notce that, a uggeted by the notaton, the Gau um depend on two object: a and χ. The followng propoton gve ome nght n how the Gau um behave wth repect to thee n extreme cae. Propoton 3.2. I a 0 and χ ɛ = g a χ = χa 1 g 1 χ II a 0 and χ = ɛ = g a ɛ = 0 III a = 0 and χ ɛ = g 0 χ = 0 IV a = 0 and χ = ɛ = g 0 ɛ = p Proof. We prove each tatement n turn. Notce that χag a χ = χa t F p χtζ at = t F p χatζ at = g 1 χ. Multplyng both de by χa 1 = χa 1 prove the frt tatement. Next we ee that g a ɛ = t F p ɛtζ at = t F p ζ at = 0 by lemma 3.1, th prove the econd tatement. Latly, g 0 χ = t F p χt, hence the thrd and fourth tatement both follow from propoton X. Propoton 3.3. If χ ɛ, then gχ = p. Proof. Our trategy wll be to evaluate the um S = a F p g a χg a χ n two dfferent way. Frt, uppoe that a 0, then we have g a χg a χ = χa 1 gχχa 1 gχ = χa 1 gχχagχ = gχ 2. By propoton 3.2 we have g 0 χ = 0, thu S = p 1 gχ 2. Next we ue the defnton 10 drectly to get g a χg a χ = χxχyζ ax ay x y 7

10 thu, by lemma 3.1 we have χxχyζ ax ay = a x y x χxχyδx, yp = p 1p y where δx, y = 1 f x y p and δx, y = 0 f x y p. Equatng thee two dfferent way of expreng S prove the reult Jacob Sum and Applcaton. We defne the Jacob um and how how t can be ued to count the oluton to certan equaton over F p. Let χ 1,..., χ k Ω p. Defnton 11. A um of the form Jχ 1,..., χ k = t t k =1 χ 1t 1 χ k t k called a Jacob um. We tart by conderng the famlar equaton x 2 + y 2 = 1 where x, y F p. Let Nx 2 + y 2 = 1 denote the number of oluton to th equaton. Frt, notce that Nx 2 + y 2 = 1 = a+b=1 Nx2 = any 2 = b = a+b=1 1 + a/p1 + b/p = p + a a/p + b b/p + a+b=1 a/pb/p = p + a+b=1 a/pb/p where the lat equalty follow from corollary 2.3. We are left wth evaluatng the um 1 a/pb/p. a+b=1 Let u paue th problem for a moment and conder Nx 3 + y 3 = 1 ntead. Agan, notce that Nx 3 + y 3 = 1 = a+b=1 Nx3 = any 3 = b. Now, f p 2 3 th um ealy evaluated to be p by propoton 2.4. However f p 1 3 then t not a mple. Let χ Ω p \ {ɛ} be a character of order 3. Snce 2, 3 = 1, χ 2 alo a character of order 3. Snce Ω p cyclc, the number of element of order d p 1 φd, where φ denote Euler totent functon. Thu there are φ3 = 2 character of order 3, and nce 3 prme, a character mut have order 3 or 1 for t order to dvde 3. Thu by propoton 2.10 we have 2 Nx 3 + y 3 = 1 = 2 2 χ a χ j b a+b=1 =0 = j a+b=1 j=0 χ aχ j b In ummary, wth both thee example we are eventually forced to deal wth a Jacob um. Th motvate the noton of uch um and we hall ee how the properte of thee um wll allow u to fnally evaluate at leat partally 1 and 2. The followng theorem whch we oon hall generalze, provde u wth the neceary tool for th purpoe. Part four how how the Jacob and Gau um are related n an urprngly mple way. Theorem 3.4. Let χ, λ Ω p \ {ɛ}. Then I Jɛ, ɛ = p II Jɛ, χ = 0 III Jχ, χ 1 = χ 1 IV χλ ɛ = Jχ, λ = gχgλ gχλ 8

11 Proof. The frt part trval nce the number of par a, b uch that a + b = 1 p and the econd part follow from propoton 2.6. For part three, notce that Jχ, χ 1 = χaχ 1 b = χ1χ χab 1 a+b=1 a+b=1 b 0 = a 1 χ a1 a 1. Let c = a1 a 1, then f c 1 we have a = c1 + c 1. Hence, when a vare over F p \ {1}, c vare over F p \ { 1}. Thu, Jχ, χ 1 = χc = χc χ 1 = χ 1 c F p c 1 by propoton 2.6. To prove the lat part, notce that gχgλ = χxζ x x y If t = 0 then λyζ y = χxλyζ x+y x,y = χxλy ζ t. t x+y=t χλx = 0. χxλ x = λ 1 x+y=0 x If t 0, let x = tx and y = ty. Then we have χxλy = χtxλty = χλtχxλy = χλtjχ, λ. Thu, x+y=t X+Y =1 gχgλ = Jχ, λ t Corollary 3.5. χ, λ, χλ ɛ = Jχ, λ = p. X+Y =1 χλtζ t = Jχ, λgχλ. gχgλ p p Proof. Jχ, λ = gχλ = = p by propoton 3.3. p Ung thee dentte, we can defntvely evaluate 1 and make further progre wth 2. For 1 mply note that an applcaton of part three yeld a+b=1 a/pb/p = 1/p = 1 p 1/2. The lat equalty follow from the frt part of propoton 2.1 nce 1 p 1/2 1/p p p 1 p 1/2 1/p 1 p 1/2 1/p = 0. Hence, the number of oluton to x 2 + y 2 = 1 n F p turn out to be Nx 2 + y 2 = 1 = p 1 p 1/2. We can ue the frt three part of propoton 3.4 to mplfy 2 a 3 χ aχ j b = p χ 1 χ Jχ, χ + Jχ 2, χ 2. j a+b=1 9

12 Snce χ a character of order 3 we have χ 1 = χ 1 3 = χ 3 1 = ɛ 1 = 1 and χ 2 = χ 1 = χ. Hence Jχ, χ + Jχ 2, χ 2 = Jχ, χ + Jχ, χ = χab + χab = 2ReJχ, χ. Thu 3 mplfe to 4 a+b=1 a+b=1 Nx 3 + y 3 = 1 = p 2 + 2ReJχ, χ. Clearly, th not a prece a the value of Nx 2 + y 2 = 1. However, we can now gve a etmate to the number of oluton to x 3 + y 3 = 1 n F p. By corollary 3.5 we have 5 Nx 3 + y 3 = 1 p p. whch tell u that, the number of oluton to x 3 +y 3 = 1 approxmately p 2 wth an error term 2 p, hence for large prme p there wll alway ext many oluton to th equaton. To begn generalzng ome of the reult above, the followng um wll be ueful k J 0 χ 1,..., χ k := χ t. t t k =0 =1 Th the ame a the Jacob um except that t = 0 ntead of 1. Alo, let ψ : F p C be the map t ζ t, then ψ t = ψt. Propoton 3.6. I J 0 ɛ,..., ɛ = Jɛ,..., ɛ = p k 1 II If ome but not all χ = ɛ, then J 0 χ 1,..., χ k = Jχ 1..., χ k = 0 III Suppoe χ k ɛ, then { 0 f k =1 J 0 χ 1,..., χ k = χ ɛ p 1χ k 1Jχ 1,..., χ k 1 otherwe IV χ 1,..., χ k, χ 1 χ k ɛ = k k gχ = Jχ 1,..., χ k g χ. =1 =1 Proof. If k 1 of the term n t choen, the lat term unquely determned by the requrement that the whole um equal 0 or 1. Thee k 1 term can be choen n p k 1 way, th prove part one. For the econd part we can aume we have ordered the character n uch a way that χ 1,..., χ ɛ and χ +1,..., χ k = ɛ. Then, for J 0 χ 1,..., χ k we have k χ t = χ t = p k 1 χ t = 0 t t k =0 =1 =1 =1 t F p t 1,...,t k 1 where the lat equalty follow from propoton 2.6. The proof for Jχ 1,..., χ k = 0 mlar. To prove part three we rewrte J 0 χ 1,..., χ k a J 0 χ 1,..., χ k = k 1 6 χ t χ k t ,t k 1 = =1 where 0 nce th term get deleted by χ k 0 = 0. Thu we may defne T by t = T. Th gve 7 k 1 t t k 1 = =1 χ t = k 1 j=1 χ j k 1 T T k 1 =1 =1 χ T = k 1 j=1 χ j Jχ 1,..., χ k 1, 10

13 ubttutng the lat expreon n 7 nto 6 yeld J 0 χ 1,..., χ k = χ 1 χ k 1 1Jχ 1,..., χ k 1 0 χ 1 χ k. If k =1 χ ɛ then 0 χ 1 χ k = 0, otherwe the um equal p 1 by propoton 2.6. Alo, χ 1 χ k 1 1 = ±1 and χ k 1 = ±1. In the cae when k =1 χ ɛ then χ 1 χ k 1 1χ k 1 = ɛ 1 = 1 hence we can replace χ 1 χ k 1 wth χ k. To prove part four, notce that gχ = 8 χ t ψt = t t t k = χ t ψ. χ t = 0 Snce by aumpton χ 1,..., χ k ɛ, part three above mple that t t k = f = 0, o we can uppoe 0 and defne t = T a before. Then 9 χ t = χ T = χ 1 χ k Jχ 1,..., χ k, t t k = T T k =1 χ 1 χ k ubttutng the lat expreon n 9 nto 8 gve gχ = Jχ 1,..., χ k χ 1 χ k ψ = Jχ 1,..., χ k g 0 Part four ha a couple of corollare of nteret. Corollary 3.7. Suppoe that χ 1,..., χ k ɛ and χ 1, χ k = ɛ then k gχ = pχ k 1Jχ 1,..., χ k 1. =1 Proof. Ue part four of propoton 3.6 and multply both de by gχ k. Note that f k = 2 we have Jχ 1 = 1 n the rght hand de, by defnton. Corollary 3.8. Under the ame aumpton a n corollary 3.7 we have Jχ 1,..., χ k = χ k 1Jχ 1,..., χ k 1. Proof. The cae when k = 2 the content of theorem 3.4 part three, hence uppoe k > 2. Ue equaton 8 and 9 above combned wth the aumpton to get gχ = J 0 χ 1,..., χ k + χ t ψ 0 t t k = = J 0 χ 1,..., χ k + 0 ɛjχ 1,..., χ k ψ χ. = J 0 χ 1,..., χ k + Jχ 1,..., χ k 0 ψ = J 0 χ 1,..., χ k Jχ 1,..., χ k where the lat equalty follow from part two of propoton 3.2. By corollary 3.7 and part three of propoton 3.6 we have pχ k 1Jχ 1,..., χ k 1 = J 0 χ 1,..., χ k Jχ 1,..., χ k whch conclude the proof. Theorem 3.9. If χ 1,..., χ k ɛ, then = p 1χ k 1Jχ 1,..., χ k 1 Jχ 1,..., χ k. 11

14 I χ 1 χ k ɛ = Jχ 1,..., χ k = p k 1/2. II χ 1 χ k = ɛ = J 0 χ 1,..., χ k = p 1p k/2 1 and Jχ 1,..., χ k = p k/2 1. Proof. Snce each χ ɛ we have by propoton 3.3 gχ = p for each {1, 2,..., k}. Thu Jχ 1,..., χ k = gχ p k gχ 1,..., χ k = = p k 1/2. p For the econd part note that J 0 χ 1,..., χ k = p 1χ k 1Jχ 1,..., χ k 1. by propoton 3.6 part three. Snce χ 1 χ k 1 ɛ we have by corollary 3.7 and 3.8 and J 0 χ 1,..., χ k = p 1 gχ p Jχ 1,..., χ k = χ k 1Jχ 1,..., χ k 1 = = p 1 p k p = p 1p k/2 1 gχ p k = p p = p k/2 1. We now conder the mot general cae needed for our purpoe, namely the number of oluton N to the equaton a 1 x n a 2x n a k x n k k = b where a 1,..., a k Fp and b F p. Here the man theorem of th ecton, and we wll ee later how t wll be ueful for our tudy of the congruence zeta functon. Theorem I b = 0 = N = p k 1 + k =1 χ a 1 J 0 χ 1,..., χ k where the um over all k-tuple χ 1,..., χ k where χ ɛ, χ n = ɛ for = 1, 2,..., k and χ 1 χ k = ɛ. If M the number of uch k-tuple, then N p r 1 Mp 1p k/2 1. II b 0 = N = p r 1 + χ 1 χ k b k =1 χ a 1 Jχ 1,..., χ k where the um over all k-tuple χ 1,..., χ k where χ ɛ and χ n = ɛ for = 1, 2,..., k. If M 1 the number of uch k-tuple wth χ 1 χ k = ɛ and M 2 the number of uch k-tuple wth χ 1 χ k ɛ, then N p r 1 M 1 p k/2 1 + M 2 p k 1/2. Proof. Let Lu = Lu 1,..., u k = k =1 a u and N u = Nx n = u. Smlarly a before, notce that 10 N = N 1 u 1 N 2 u 2 N k u k Lu=b where the um over all k-tuple u = u 1,..., u k uch that Lu = b. By propoton 2.10 we know that N u = χ χ u where χ range over all character of order dvdng n. Inertng th nto equaton 10 gve N = χ 1 u 1 χ k u k = 11 χ 1 u 1 χ k u k. χ 1 χ k Lu=b χ 1,...,χ k Lu=b 12

15 We conder the two cae b = 0 and b 0 eparately. If b = 0 we defne t = a u, then χ 1 u 1 χ k u k = χ 1 t 1 χ 1 a 1 1 χ kt k χ k a 1 k 12 Lu=b If b 0 we defne t = b 1 a u, then χ 1 u 1 χ k u k = 13 Lu=b t t k =0 = χ 1 a 1 1 χ ka 1 k t t k =0 = χ 1 a 1 1 χ ka 1 k J 0χ 1,..., χ k. t t k =1 χ 1 t 1 χ k t k χ 1 a 1 1 χ 1bχt 1 χ k a 1 k χ kbχt k = χ 1 χ k bχ 1 a 1 1 χ ka 1 k Jχ 1,..., χ k. Now, n the frt cae 11 become 14 χ 1 a 1 1 χ ka 1 k J 0χ 1,..., χ k χ 1,...,χ k If χ = ɛ for all, then J 0 χ 1,..., χ k = p k 1. If ome but not all χ = ɛ then J 0 χ 1,..., χ k = 0 and f k =1 χ ɛ then J 0 χ 1,..., χ k = 0, all th follow drectly from propoton 3.6. Gven thee fact we are left wth the expreon for N a tated n the theorem. If b 0 the proof mlar. Both etmate follow drectly from the fact that for any χ Ω p and a Fp χa = 1 nce χa a p 1t root of unty by propoton 2.5. Ung the tool outlned n the earler ecton, we can generalze the noton of Gau and Jacob um n the followng way. Let F be an arbtrary fnte feld of order q = p r. Let ψ : F C be the map α ζ tr F/Fp α p, where F p = Z/pZ and ζ p = e 2π/p. The multplcatve group of any fnte feld cyclc, thu the propoton of multplcatve character whch hnge on th fact naturally extend to feld of arbtrary order, mply replace p wth q. Gven the followng defnton, the ame extenon to arbtrary fnte feld alo vald for Gau um, Jacob um and the nterrelaton we aw between them earler n th ecton. Let χ be a character of F and α F. Defnton 12. A um of the form t F χtψαt called a Gau um on F belongng to the character χ and wll be denoted g α χ. Note that th defnton reduce to the prevou one f F = F p. To generalze the defnton of Jacob um, mply let the character be character on F ntead of F p. Next we prove a theorem whch wll be of great ue later on. Conder the equaton fy 0,..., y n = n =0 a y m = 0 where a F. Th a homogeneou equaton, o t defne a hyperurface H f F P n F. Theorem Suppoe F a fnte feld of order q 1 m. Then n 1 H f F = q + 1 q 1 =0 χ 0,...,χ n J 0 χ 0,..., χ n n j=0 χ j a 1 j where the um over all n+1-tuple χ 0,..., χ n uch that χ ɛ, χ m = ɛ and χ 0 χ n = ɛ. Moreover, under the ame condton we alo have 1 q 1 J 0χ 0,..., χ n = 1 q 13 n gχ. =0

16 Proof. The frt part a corollary of the frt part of theorem 3.10, we have H f F = q n + 15 χ a 1 J 0 χ 0,..., χ n. χ 0,...,χ n where the um over all n + 1-tuple χ 0,..., χ n a gven by the theorem. The number H f F obtaned from dvdng 15 by q 1. For the econd part, recall that 16 J 0 χ 0,..., χ n = χ n 1q 1Jχ 0,..., χ n 1 where we have extended propoton 3.6 to feld of arbtrary order q a dcued. By the ame propoton we have 17 Jχ 0,..., χ n 1 = gχ 0 gχ n 1 gχ 0 χ n 1. Multply the numerator and denomnator of the rght hand de of 17 by gχ n and apply corollary 3.7 to get 18 J 0 χ 0,..., χ n 1 = gχ 0 gχ n gχ 0 χ n 1 gχ n = gχ 0 gχ n, χ n 1q ubttutng 18 nto 16 conclude the proof A Proof of the Law of Quadratc Recprocty. Here we provde an elegant proof of the famou law of quadratc recprocty, ung what we know about Jacob and Gau um. Theorem 3.12 Law of Quadratc Recprocty. Let p and q be two dtnct odd prme number, then p/qq/p = 1 p 1 q Proof. Let χ denote the character on F p of order two. Then χ q+1 = ɛ nce q + 1 even. By corollary 3.7 we have gχ q+1 = q+1 gχ 2 2 = p q+1 p where there are q component n Jχ,..., χ. Thu, Jχ,..., χ = t t q=1 q+1 2 = p 1 p 1 2 Jχ,..., χ χt 1 χt q = p q 1 p If t 1 =... = t q, then t = 1/q and χ1/q χ1/q = χ1/q q = χq q = χq. If t t j for at leat two ndexe, then there are q term n Jχ,..., χ, all of equal value, found by permutatng t 1,..., t q cyclcly. Thu n modulo q all of thee term vanh, hence Thu, p q 1 p q 1 2 χq q. q p 1 q p/q q/p q by propoton 2.1. Snce q dvde the dfference n 19, we mut n fact have equalty. Multplyng both de by p/q conclude the proof. 14

17 4. The Zeta Functon E. Artn ntroduced the concept of the congruence zeta functon n In 1949 A. Wel formulated a et of conjecture known a the Wel conjecture related to the zeta functon. The frt part of the Wel conjecture tate that any algebrac et ha a ratonal zeta functon, th wa proved n 1959 by B. Dwork. We defne the congruence zeta functon, how how t and the Remann zeta functon atfy analogou relaton, and prove that under certan crtera Z f u ratonal. We follow [IR, Ch. 11.1] and [IR, Ch ]. It can be hown that f F a fnte feld of order q then there ext a feld F F of order q where 1 an nteger. Gven a homogeneou polynomal fx F [x 0, x 1,..., x n ] we let N denote the number H f F. We wh to tudy the number N by tudyng the power ere =1 Nu of a complex varable u. We can and wll vew th a a formal power ere, whch enable the excluon of all queton of convergence. However, th not neceary. Notce that If u < q n and 1, then N P n F = qn+1 1 q 1 = n q k < n + 1q n. k=0 20 N u < n + 1 =1 =1 x where 0 x = q n u < 1. Thu 20 converge,.e. u < q n. =1 N u converge abolutely for all Defnton 13. The zeta functon of the hyperurface defned by f the ere gven by Z f u = exp =1 N u If the zeta functon expanded about the orgn we ee that the contant term 1. We can therefore aume that Z f u = pu qu where p0 = q0 = 1. To ee th, aume that not the cae. Then Z f 0 = p0 q0 = 1 f and only f p0 = q0 = c. Now f pu = n =0 a u and qu = m =0 b u then pu qu = c n =0 c 1 a u n c m =0 c 1 b u = =0 c 1 a u m =0 c 1 b u = p u q u where p u and q u have the dered property. Hence the zeta functon can be expreed n the form Z f u = 1 α u 21 j 1 β ju where α, β j C. We now characterze ratonal zeta functon, th wll be ueful later when we conder Z f u for partcular homogeneou polynomal f. Propoton 4.1. Zu { pu qu : pu, qu C[u]} f and only f there ext α, β j C uch that N = j β j α 15.

18 Proof. We tart by aumng that there ext α, β j C uch that N = j β j α. By nertng th expreon for N we get N u j = β j α u =1 =1 j = β ju α u =1 = β j u α u. j =1 =1 Further mplfy the lat expreon above by ung the dentty 22 ln1 α u ln1 β j u j =1 z = ln1 z to get Fnally, to fnd the zeta functon, we exponentate 22 to get Z f u = exp ln1 α u ln1 β j u = exp ln1 α u exp j j ln1 β ju exp ln1 α u = j exp ln1 β j u = 1 α u j 1 β ju whch a ratonal functon of u. We now how the other drecton. Suppoe therefore that the zeta functon ratonal o that 23 Z f u = 1 α u j 1 β ju where α, β j C. Take the logarthmc dervatve of both de of 4.1 to get Z f u Z f u = d du ln 1 α u j 1 β ju = d ln1 α u ln1 β j u du j = α 1 α u β j 1 β j j u. Multply the frt and lat expreon above by u and expand the denomnator n the lat expreon n a geometrc ere to get u Z f u Z f u = α u 1 α u β j u 1 β j j u = α u α u =0 j = j u j =1β α u =1 = βj α u. =1 j 16 β j u β j u =0

19 We now compare th lat power ere wth an alternate way of computng u Z f u defnton Z f u = exp by u we get hence we have and thu =1 Nu Z f u.. By takng the logarthmc dervatve and then multplyng u Z f u Z f u = u d du N u = =1 βj j =1 N = j α β j u = α. N u, =1 N u Next we how that N ndependent of the choce of F, o that N only depend on. Th a conequence of the followng propoton. Propoton 4.2. Let E and E be feld extenon over F of the ame order q. Then there an omorphm σ : E E uch that σ F = d. We can ue σ to nduce a map from the repectve projectve n-pace n a natural way by lettng σ : P n E P n E map [α 0,..., α n ] [σα 0,..., σα n ]. Th map welldefned nce f [α o,..., α n ] = [β 0,..., β n ], then by defnton there ome γ E uch that α = γβ for = 0, 1,..., n, thu [σα 0,..., σα n ] = [σγσβ 0,..., σγσβ n ] = [β 0,..., β n ]. By vrtue of the omorphc property of σ, σ a bjecton. Indeed, let α = [α 0,..., α n ] P n E be arbtrary, then σ map [σ 1 α 0,..., σ 1 α n ] α, hence σ urjectve. Alo, both P n E and P n E are fnte et, hence σ njectve. It moreover true that σ Hf E bjecton to H f E. We already know that the retrcton njectve. Let α = [α 0,..., α n ] H f E, then fα 0,..., α n = 0 and σ [σ 1 α 0,..., σ 1 α n ] = α. We need to how that [σ 1 α 0,..., σ 1 α n ] H f E. Now, σ act a the dentty on F, hence f σ 1 α 0,..., σ 1 α n = σ 1 f[α 0,..., α n ] = σ 1 0 = 0, hence [σ 1 α 0,..., σ 1 α n ] H f E. We conclude that H f E = H f E and hence N ndeed only depend on a clamed, nce th reult hold alo for the trval feld extenon F F and F F = F The Zeta Analogy. The congruence zeta functon and the Remann zeta functon atfy a hghly analogou relaton. In the cae of the Remann zeta functon we have ζ = n = n=1 p p, > 1 where the product over all prme p > 0. So far we have dealt wth the congruence zeta functon defned by one polynomal f. Let u now conder t over a et of polynomal {f 1,..., f m } F [x 1,..., x n ]. Let F be a fnte feld of order q and let V = {a A n F : f j a = 0 for j = 1, 2,..., m} be an algebrac et n A n F. Defnton 14. The functon =1 N u Z V u = exp where N the number of pont n A n F atfyng the equaton defned n V, called the zeta functon of V over F. 17 =1 By

20 Let K be the algebrac cloure of F. One can how that K the countable unon of omorphc cope to F of order q for all potve nteger. It natural to conder a feld contanng all feld related to the number N and K wll erve a that feld. We can then extend V o that V A n K and wth N pont whoe coordnate all le n F. Suppoe a = a 1,..., a n V and let L be the mallet feld contanng F {a 1,..., a n }. Suppoe L = q d, we then call a to be a pont of degree d. Lemma 4.3. The pont a, a q,..., a qd 1 all le n V and are parwe dtnct. Proof. Suppoe that a = a qj for ome j {1, 2..., d 1}, and let j be the mnmal element wth th property. Notce that th mple that a qkj = a for all k Z +. Addtonally, we have that a qd = a nce a L for = 1, 2,..., n. Thu, by the choce of j we have that j d. By bac fact about fnte feld, there ext a ubfeld L L of order q j and nce a = a qj equvalent to a L or = 1, 2..., n we have a contradcton by defnton of L. Th prove that the pont a, a q,..., a qd 1 are parwe dtnct. To how that the pont all le n V, we need to verfy that they atfy the polynomal equaton defnng V. Let ψ : L L be the automorphm whch map x x q, o that ψ F = d. Let a = a 1,..., a n V and let j {1,..., m} correpond to a polynomal defnng V. Then, by defnton, f j a 1,..., a n = 0. But then ψ k 0 = ψ k f j a 1,..., a n = f j ψ k a,..., ψ k a n = f j a q k,..., a qk = 0. Snce th hold for any j {1, 2,..., m} and k {1, 2,..., d 1}, we are done. Defnton 15. Let a V be a pont of degree d. A et of the form P = {a, a q,..., a qd 1 } called a prme dvor on V. The degree of P denoted δp and equal to d. Propoton 4.4. Let F be a fnte feld uch that [F : Z/pZ] = n. Then the ubfeld of F are n one-to-one correpondence wth the dvor of n. Lemma 4.5. N = d dn d where n d the number of prme dvor on V of degree d. Proof. The prme dvor partton V. Let α V be a pont uch that all coordnate le n F. Th F could be a larger feld than neceary.e. there ome d uch that F d F the mallet feld uch that α F d. We know from propoton 4.4 that d, o that α defne a unque prme dvor of degree d whch a dvor of. Th prove the lemma. Now we are n a poton to how how the congruence zeta functon and the Remann zeta functon bear conderable reemblance. Theorem 4.6. Z V u = 1 1 u δp. P Proof. By mergng the factor correpondng to prme dvor of the ame degree, the product can be rewrtten a 1 nk 24 1 u k. By takng the logarthmc dervatve of 24 we get d 1 nk ln du 1 u k = d n k ln1 u k du k=1 k=1 = 1 ku k n k u 1 u k. k=1 18 k=1

21 Expand the denomnator nto a geometrc ere and compute the coeffcent of u m to get 1 25 dn d u m = N m u m 1 u m=1 d m by lemma 4.5. Integratng the rght hand de of 25 and then takng the exponental conclude the proof The Ratonalty of the Zeta Functon. In th ecton we prove that the zeta functon Z f u of fx 0,..., x n = a 0 x m 0 + a 1x m a nx m n, where a 0,..., a n F and F a feld of order q 1 m, ratonal. In order to do th, we frt outlne a proof of the Hae-Davenport relaton, whch an nteretng reult n t own rght. Clearly f a homogeneou polynomal, t therefore defne a projectve hyperurface H f F where F F a feld extenon of degree. Theorem 3.11 gve n 1 N = q + 1 n 26 q χ a 1 gχ =0 m=1 χ 0,...,χ j=0 n where χ are character of F uch that χ m = ɛ, χ ɛ and χ 0 χ n = ɛ. Let χ be a character of F, compoe t wth N F/F to get χ = χ N F/F : F C. Th mappng n fact a character on F nce χ ab = χ N F/F ab = χ N F/F an F/F b = χ N F/F a χ N F/F b = χ aχ b by propoton Lemma 4.7. I χ ρ = χ ρ. II χ m = ɛ = χ m = ɛ. III χ a = χa for all a F. Proof. The frt follow from the lat part of propoton For the econd part, note that m χ NF/F a = χ m N F/F a = ɛ N F/F a = 1 for all a F, hence t the trval character. For the thrd part, mply note that χ a = χ N F/F a 1 = χa = χa, by propoton Th lemma how that by lettng χ vary over all character of F of order dvng m, the ame happen for the correpondng character χ of F. Thu equaton 26 can now be rewrtten a 27 n 1 N = q + 1 q =0 n χ 0,...,χ n j=0 χ a 1 gχ where χ are character of F uch that χ m = ɛ, χ ɛ and χ 0 χ n = ɛ. It turn out that gχ and gχ atfy a mple relatonhp. Theorem 4.8 Hae-Davenport Relaton. gχ = gχ Remark. We remnd that an nteger whch depend on χ, whch a character on F F defned above. In order to prove th, we frt need two lemma, the proof of whch wll be omtted ee [IR, Ch. 11.4]. Gven a monc polynomal fx = x n a 1 x n n a n F [x], we defne a mappng λ by λf = ψa 1 χa n and let λ1 = 1. Lemma 4.9. λfg = λfλg for all monc f, g F [x]. 19

22 Lemma gχ = δfλf /δf where the um over all rreducble monc polynomal of degree n F [x]. Gven th defnton of λ we have the dentty λft δf = f f 1 1 λft δf where the um over all monc polynomal, and the product over all monc rreducble polynomal n F [x]. Let u pend ome tme outlnng the legtmacy of th dentty. We expre the factor n the rght hand de a a geometrc ere 1 1 λft δf = λft δf k. k=0 Ung the fact that λ multplcatve by lemma 4.9 we get 1 1 λft δf = λf k t kδf f k=0 f = f 1 + λft δf λf k t kδf +... Now, the rng F [x] a UFD o n partcular each monc polynomal can be expreed a a product of rreducble monc polynomal n a unque way. Now, nce each power of each monc rreducble polynomal found n the um above, we ee, by multplyng the factor n the lat expreon, that each monc polynomal occur n the argument of λ n the reultng um. Inductvely applyng lemma 4.9 gve n =1 λf a = λ n =1 f a. Th together wth the bac fact that δ n = n =1 a δf how that the dentty hold. =1 f a Proof of the Hae-Davenport Relaton. By collectng term of equal degree, we have 28 λft δf = λf t. f =0 δf= where the left hand de a n 28. We proceed by analyzng the um δf= λf. If = 1, then 29 χaψa = gχ δf= λf = a F λx a = a F If > 1 t turn out that the um vanh, nce 30 λx a 1 x a. δf= λf = a F Now, λ only depend on the coeffcent a 1 and a n. By fxng thoe and lettng the other vary over F, we ee that 30 equal to q 2 χa ψa 1 = q 2 χa ψa 1 = 0 a 1, a n a a 1 by propoton 2.6 extended to character on arbtrary fnte feld. Thu we have λft δf = 1 + gχt = 1 1 λft δf. f f By takng logarthmc dervatve, we get d ln1 + gχt dt 20 = gχ 1 + gχt.

23 and d 1 ln dt 1 λft δf = d dt f f = f ln 1 λft δf λfδft δf 1 1 λft δf. By multplyng thee expreon by t, we have gχt 1 + gχt = f λfδft δf 1 λft δf. Expand the denomnator n geometrc ere to get =1 1 1 gχ t = λf k δft. kδf f k=1 By comparng coeffcent for t we get 1 1 gχ = δf δfλf /δf = gχ where the lat equalty the content of lemma Th conclude the proof of theorem 4.8. We now make ue of the Hae-Davenport relaton to conclude our analy of the number N aocated wth Z f u, where fx 0,..., x n = a 0 x m 0 + a 1x m a nx m n F [x 0,..., x n ]. Subttutng gχ = 1 +1 gχ nto equaton 27 we get 31 n 1 N = q + 1 n+1 =0 where χ are character of F uch that χ m and χ 0 χ n = ɛ. χ 0,...,χ n 1 n+1 q n j=0 χ a 1 gχ, = ɛ where m a n the polynomal f, χ ɛ Fnally, we are n a poton to tate the man theorem of th ecton, whch etablhe the ratonalty of Z f u under certan condton. Theorem Let a 0,..., a n F where F = q 1 m and fx 0,..., x n = a 0 x m 0 + a 1 x m a nx m n. Then the congruence zeta functon Z f u a ratonal functon of the form where P u = χ 0,...,χ n Z f u = P u 1n 1 u1 qu 1 q n 1 u, 1 1 n+1 1 q χ 0a 1 0 χ na 1 n gχ 0 gχ n u and the product over all n+1-tuple χ 0,..., χ n uch that χ m = ɛ, χ ɛ and χ 0 χ n = ɛ. Proof. Th a drect reult of our characterzaton of ratonal zeta functon.e. propoton 4.1 appled to the current tuaton. Note that f n even, then 31 can be wrtten a N = n 1 j=0 β j α, where β j = q j and α = 1 n+1 1 q χ 0a 1 0 χ na 1 n gχ 0 gχ n for ome n + 1-tuple χ 0,..., χ n ubject to the condton above. If n odd, then 31 can be wrtten a j β j α where β j = N and α = 0. Hence the exponent 1 n n the numerator of Z f u, nce the zeta functon of the form 1 α u j 1 β ju. 21