SOME SUMS OVER IRREDUCIBLE POLYNOMIALS

Transcription

1 SOME SUMS OVER IRREDUCIBLE POLYNOMIALS DAVID E SPEYER Abstract We rove a number of conjectures due to Dinesh Thakur concerning sums of the form P hp ) where the sum is over monic irreducible olynomials P in F q[t ], the function h is a rational function and the sum is considered in the T -adic toology As an examle of the simlest of our results, in F 2[T ], the sum P always converges P k + to a rational function, and is 0 for k = Introduction The oint of this document is to exlain some identities exerimentally found by Dinesh Thakur, involving sums over irreducible olynomials in finite fields We begin by stating the simlest of these identities: Let P be the set of irreducible olynomials in F 2 [T ] Then P = 0 P P Here the sum must be interreted as a sum of ower series in T examle, the first five summands are T = T +T 2 +T 3 + T +) = T T 2 +T +) = T 2 +T 3 + T 3 +T +) = T 3 + T 3 +T 2 +) = T 3 + As the reader can see, only finitely many terms contribute to the coefficient of each ower of T, and the coefficient of T j is 0 for each j We now introduce the notation necessary to state our general results To aid the reader s comrehension, we adot the following conventions: Integers will always be denoted by lower case Roman letters k,, q ); olynomials over finite fields will always be denoted by caital Roman letters A, F, P ), sets of such olynomials will always be denoted by calligrahic letters A, P, R, ), symmetric olynomials will be denoted by bold letters e k, k, ) Of course, there will be other sorts of mathematical objects as well, which we trust the reader to accommodate as they occur Let be a rime and q a ower of Let F q be the field with q elements Let R be the olynomial ring F q [T ] Let K be the fraction field F q T ) and For

2 2 DAVID E SPEYER let K be the T -adic comletion of K All infinite sums will be understood in the T -adic toology Let P be the set irreducible olynomials in R; let P be the set of monic irreducible olynomials Here is our main result for the case = 2 Theorem If = 2 then, for any ositive integer k 0 mod q, the sum P k P P is in K The reader may wonder what haens is we sum over all irreducible olynomials rather than monic ones; that is an easy corollary: Corollary 2 For any ositive integer k, the sum P k is in K P P Proof We rewrite the sum as P P follows from the identity in F q U) To rove this identity, write and recall that ax) k = X LCMk,q ) ax) k = /ax) kj a m = j= { m 0 mod q 0 otherwise The corollary then ap ) k We now discuss the case of a general rime Define the rational function G U) by G u) = U ) U) U) When = 2, we have G 2 U) = 2U+2U 2 = U/ U), so G 2 U) 2 2 /P ) = /P ) When is odd, we have the following alternate exressions for G : Proosition 3 If is odd, then, as rational functions in F U), we have j= U j j G U) = U) = U j j 0 j< j 0 mod

3 SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 3 Proof If is odd, then U ) U) = j= )j j) U j If = 2, there is also a 2U 2 term) We have ) j j ) = )j ) 2) j + ) 2 j )j This roves the first equality, and the second is immediate j Theorem 4 For any ositive integer k 0 mod q, the sum P P G /P k ) is in K mod As we noted, G 2 /X) = /X ), so Theorem 4 imlies Theorem Once again, we have a trivial variant where we sum over P: Corollary 5 For any ositive integer k, the sum G /P k ) is in K P P Proof If = 2, we roved this in Corollary 2, so we may and do) assume is odd As in the roof of Corollary 2, we rewrite the sum as P P a F G /ap ) k ) q We now need the identity G au) k ) = GCDq, k)g U LCMq,k) ) in F q U) To rove this identity, we use the formula G U) = j 0 mod U j /j and the identity { a m q m 0 mod q = 0 otherwise So G /au) k ) = j 0 mod Putting kj = LCMq, k)l, this is q ) l 0 mod as required = q ) jau) kj k LCMq,k)l U LCMq,k)l = j 0 mod kj 0 mod q ju kj kq ) LCMq,k) G U LCMq,k) ) = GCDq, k)g U LCMq,k) ) We also comute exlicit values for the sum when k is not too large

4 4 DAVID E SPEYER Theorem 6 Let k = q )l If l q/, then P P G /P k ) = 0 If q/ + < l 2q/, then G /P k T q T ) q+ ) = l T q2 T q )T q2 ) P P In rincile, our methods are caable of comuting P P G /P k ) for any k 0 mod q, but they become imractical beyond l = 2q/ History of the roblem Dinesh Thakur exerimentally discovered most of the relations described above in characteristic two, and susected there should be similar results in odd chracteristic Thakur ublished these comutations them in a rerint entitled Surrising symmetries in distribution of rime olynomials, arxiv: At Thakur s suggestion, Terence Tao romoted the roblem in osts on his blog and on the Polymath blog I am grateful to Thakur for finding such an elegant roblem and to Tao for bringing it to my attention My thanks also to all who articiated in the discussion on the Polymath blog Noam Elkies, Ian Finn, Ofir Gorodetsky, Jesse, Gil Kalai, David Lowry-Duda, Dustin G Mixon, John Nicol, Partha Solaurkar, John Voight, Victor Wang, Qiaochu Yuan, Joshua Zelinsky 2 The Carlitz exonential, and symmetric olynomials The main tool in our roofs is the theory of the Carlitz exonential Put Define D i = T qi T )T qi T q )T qi T q2 ) T qi T qi ) e C Z) = this sum is T -adically convergent for any Z K We will make use of the roduct identity: e C πz) = + Z ) πz A j=0 A R\{0} where π is the element of K q T ) such that the coefficients of Z q match This identity should be thought of as similar to Euler s identity: sinπz) = + z ) πz a a Z 0 We introduce the notations A for the nonzero olynomials of R, and A for the monic olynomials Writing e k for the elementary symmetric function of degree k, this imlies { π k /D j k = q j e k /A) A A = 0 otherwise T qj D j

5 SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 5 Since the ring of symmetric olynomials is generated by the e k, we deduce Proosition 2 If f is a homogenous symmetric olynomial of degree k, then f/a) A A is in π k K Here we note that f/a) A A is always defined, since only finitely many terms contribute to the coefficient of any articular ower of T The above considers symmetric olynomials in {/A} A A, but we would rather restrict to the case of A monic To this end, we have Proosition 22 e l /A q ) A A = { ) l π lq ) /D j l = qj q 0 otherwise Proof Grouing together scalar multiles of the same olynomial in the Carlitz roduct identity, we have e C πz) = ) Zq πz A q A A Equate coefficients of Z lq ) on both sides Corollary 23 If f is a homogenous symmetric olynomial of degree l, then f/a q ) A A is in π lq ) K 3 Proofs of rationality We now have enough background Theorem 4 and, hence, Theorem Throughout, let k 0 mod q Consider the symmetric olynomial g X, X 2,, ) := ) ) Xi X i The olynomial g has integer coefficients, so we may discuss lugging elements of K into it Let C be the cyclic grou of order, and let C act on A by rotating coordinates Let denote the small diagonal: := {A, A,, A)} A Then g /A k ) A A = A k A,,A ) A \ )/C Ak 2 Ak The sum is over cosets for the free action of C on A \ Let Φ = {A,, A ) A : GCDA,, A ) = } Any A,, A ) A can be uniquely factored as A i = DB i for some D A and B,, B ) Φ So we can factor the above sum as g /A k ) A A = D k B k D A Bk 2 Bk B,,B ) Φ\{,,)})/C

6 6 DAVID E SPEYER Now, from Proosition 23, g /A k ) A A, is in π k K Also from Proosition 23, D A /D k is in π k K, and a quick comutation shows that this sum is lus terms in t F q [[t ]], so it is not zero We deduce that B kbk 2 K Bk B,,B ) Φ\,,))/C For B A, let ΨB) be the set of -tules B, B 2,, B ) for which GCDB,, B ) = and B i = B Let ψb) = #ΨB) So we have shown that ψb)/ B k K B A \{} Here, to interret the numerator, we must divide ψb) by as integers and only then consider the quotient in F If B = P k P k 2 2 P r kr then there is an easy bijection between ΨB) and ΨP k ) ΨP k 2 kr 2 ) ΨPr ), so ψb) = ψp k i i ) If P is irreducible then ψp r ) is divisible by for any r > 0, since C acts freely on ΨP r ) So, if B is divisible by two different irreducible olynomials, then ψb) is divisible by 2 So we can rewrite the sum as ψp r )/ P rk P P r= We now comute ψp r ); which is the number of -tules P r,, P r ) with P r i = P r and GCDP r,, P r ) = In other words, we must count r,, r ) Z 0 with r i = r and minr,, r ) = 0 The number of r,, r ) Z 0 with r i = r is the coefficient of U r in / U) In order to imose minr,, r ) = 0, we subtract off the terms with minr,, r ) > 0 These are in bijection with s,, s ) Z 0 with + s i = r So ψ r ) is the coefficient of U r in / U) U / U) In other words, r=0 ψp r )U r = U )/ U) So ψp r ) U r = ) U U) = G U) r= ψp r )/ We deduce that r= = G P rk /P k ) We have now shown that P P G /P k ) K, as claimed We record the secific formula we have roved: Proosition 3 Let k be a ositive integer Then G /P k ) = g/ak ) A A P P A A /A k We will rewrite this formula in various ways in Section 5 We remark that this formula is correct even if k is not divisible by q, although we have only shown the ratio is in K when k 0 mod q

7 SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 7 4 Vanishing We will now rove the claim in Theorem 6 that the sum vanishes when k = q )l for l q/ From Proosition 3, it is equivalent to show that g /A lq ) ) A A = 0 To this end, we must exlicitly write g /A lq ) ) as a olynomial in the e k /A q ) The variables λ or µ will always denote artitions, meaning weakly decreasing sequences λ, λ 2,, λ r ) of ositive integers; sums over λ or µ imlicitly contain the condition that the summation variable is a artition We define e λ = s e λ s The symmetric olynomials e λ form an integer basis for the symmetric olynomials with integer coefficients Lemma 4 Write g X l, X l 2,, ) = for some integers c λ Then c = 0 λ =l c λ e λ X, X 2, ) Proof Note that e is in the only e λ with a nonzero coefficient of X l The coefficient of X l in g X l, Xl 2, ) is clearly 0 Now, suose that l q/, so we have l < q + So any artition λ,, λ r ) of l other than,,, ) contains a λ i between 2 and q By Lemma 22, e m /A q ) A A = 0 for 2 m q, so e λ /A q ) A A = 0 whenever λ is a artition of l other than,,, ) We deduce that g /A q ) A A = 0 as desired 5 Comutations for small k In this section, we will discuss the comutation of P P /P k for k 0 mod q and, in articular, rove the remaining half of Theorem 6 Our strategy is to combine Proositions 3 and 22 We must comute g /A lq ) ) A A and A A /A k Note the latter is l /A q ) ) A A We write k = q )l Put g X l, Xl 2, ) = λ =l c λe λ X, X 2, ) l X, X 2, ) = µ =l d µe µ X, X 2, ) Note that e m /A q ) A A = 0 unless m is of the form q j )/q ) So we only need to sum over artitions where all the arts of λ are of the form q j )/q ) From now on, we now imose that q/ + l 2q/ So l < q + Any artition of l cannot contain any arts of size q j )/q ), for j Similarly, l < 2q + 2, so a artition of l can contain at most one art of size q 2 )/q ) = q + and no arts of size q j )/q ) for j 2 We deduce that the only terms which contribute to our final answer come from λ =,,, ) or λ = q +,,,, ) when comuting g /A lq ) ) A A, and from µ =,,, ) in comuting

8 8 DAVID E SPEYER l /A q ) A A ) Moreover, from Lemma 4, the coefficient c,,,) is zero We deduce that P P /P k = c q+, l q ) e q+, l q ) /Aq ) A A d l e l/a q ) A A ) = c q+, l q ) e q+/a q ) A A e /A q ) A A ) l q d l e /A q ) A A ) l = c q+, l q ) e q+/a q ) A A d l e /A q ) A A ) q+ Here r is shorthand for r arts equal to We now use Proosition 22 The owers of π and ) cancel to give P P /P k = c q+, l q ) d l D q+ = c q+,l q ) T q T ) q+ D 2 d l T q2 T q )T q2 ) To finish the comutation, we must find c q+, s and d l The latter is easy: Comaring coefficients of X l on both sides of lx, X 2, ) = µ =l d µe µ X, X 2, ), we deduce that d l = To comute c q+, l q ), we begin with the formula g X l, X l 2, ) = l X, X 2,, ) l X, X 2, ) ) For brevity, we write fx) to indicate that the inuts to a symmetric olynomial are X, X 2, ) Note that we are working with symmetric olynomials with integer coefficients, so it makes sense to divide by We rewrite the right hand side of the revious equation as e X) l + ) e X) l + d q+, l q e q+ X)e X) l q + Here the ellises denote terms e λ where λ has some art not of the form q j )/q ) We deduce that Now, observe the identity c q+, l q = d q+, l q )) ) j j X)U j j j = log + X i U) i = log + X i U) = log i + ) e m X)U m m= The coefficient of U l on the left is )l l l Exanding the log on the right hand side as a Taylor series, only one term contributes to U l e q+ e l q

9 So we obtain ) l l SOME SUMS OVER IRREDUCIBLE POLYNOMIALS 9 l X) = )l q l q l q ) e q+ X)e l q X) + where the ellises denote a sum of e λ other than e q+ X)e l q X) So d q+, l q = ) q l and c q+, s = ) q l Plugging into our revious formula, and using that ) q mod, /P k T q T ) q+ = l T q2 T q )T q2 ) P P This concludes the roof of Theorem 6 We conclude by verifying one Thakur s conjectures which goes beyond the range l 2q/ Let = q = 2 Thakur conjectures P 3 = T 4 + T 2 P P We begin by comuting 3 X) = e X) 3 + 3e 3 X) 3 3e 2 X)e X) 3 X) 2 = e X) 6 + 6e X) 3 e 3 X) + 9e 3 X) 2 + Here and in the following equations, the ellises denote e λ terms where λ contains a art other than and 3 Note that 2 3 )/2 ) = 7, too large to contribute to a symmetric olynomial of degree 6) Similarly, So and 6 X) = e X) 6 + 6e X) 3 e 3 X) + 3e 3 X) 2 + g 2 X 3, X 3 2, ) = 2 g 2 /A 3 ) A A = e 3 /A) A A ) 2 = π6 D X) 2 6 X) ) = 3e 3 X) 2 + = π 6 T 4 T 2 ) 2 T 4 T ) 2 Recall that we are working modulo 2, so 3 = = and 2 = 0) Similarly, 6 /A) A A = e /A) A A 6 + e 3 /A) A A ) 2 ) 6 ) = π D + π 3 2 D 2 ) 6 ) ) 2 = π 6 T 2 T + T 4 T 2 )T 4 T ) We verify Thakur s claim: P 3 = /T 4 T 2 ) 2 T 4 T ) 2 ) /T 2 T ) 6 + /T 4 T 2 ) 2 T 4 T ) 2 ) = T 4 + T 2 P P