The exceptional set in the polynomial Goldbach problem

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1 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt International Journal of Number Theory c World Scientific Publishing Company The exceptional set in the polynomial Goldbach problem Paul Pollack Mathematics Department, niversity of Illinois, 409 West Green Street rbana, Illinois 680, S pppollac@illinois.edu Received Day Month Year ccepted Day Month Year Communicated by xxx For each natural number N, let RN denote the number of representations of N as a sum of two primes. Hardy and Littlewood proposed a plausible asymptotic formula for R2N and showed, under the assumption of the Riemann Hypothesis for Dirichlet L- functions, that the formula holds on average in a certain sense. From this they deduced under ERH that all but O ɛx /2+ɛ of the even natural numbers in [, x] can be written as a sum of two primes. We generalize their results to the setting of polynomials over a finite field. Owing to Weil s Riemann Hypothesis, our results are unconditional. Keywords: Goldbach conjecture, Hardy Littlewood conjectures, polynomials over finite fields, exceptional set Mathematics Subject Classificatio000: T55, P32. Introduction For each natural number N, write RN for the number of ordered representations of N as a sum of two primes. famous conjecture of Goldbach asserts that RN > 0 for each even N 4. The first substantial progress towards Goldbach s conjecture was made in the series of papers Some problems of partitio numerorum published in the early 920s by Hardy and Littlewood. In part III of this series, one finds the prediction [8, Conjecture ] that as N through even values, N RN SN log N 2, where SN := 2 p>2 p p 2 p 2.. p N p>2 To this day, it remains a famous open problem to prove that this asymptotic formula holds for all N. However, in part V of the same series, Hardy & Littlewood proved that the Riemann Hypothesis for Dirichlet L-functions ERH implies that

2 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 2 Paul Pollack as x, N x N even N RN SN log N 2 2 x 5/2+o..2 See [9, Theorem ]. It follows easily that. holds for almost all even N. Moreover, one can derive from.2 that there are x /2+o even values of N x for which RN = 0 [9, Theorem B]. In this paper we use the circle method to study an analogue of the Goldbach problem for F q [T ], the ring of one-variable polynomials over the finite field with q elements. Let α and β be nonzero elements of F q, and let γ := α + β. Let n be a positive integer. If γ 0, we suppose that is a univariate polynomial of degree n over F q with leading coefficient γ; otherwise we suppose is a nonzero polynomial of degree < n over F q. We define R = R α,β,n,fq by R :=, P,P 2 αp +βp 2= where the sum is over degree-n monic irreducibles P and P 2 in F q [T ]. Note when α + β = 0, it is natural to view R as counting twin irreducible pairs {P, P α } cf. [2]. What is the right analogue of.? If we recall that a polynomial of degree n over F q is irreducible with probability roughly /n, then standard probabilistic arguments cf. [5,.2.3] suggest that R Sq n /, where now S := P + P P P 2 with both products extended over monic irreducibles P. In order to make this prediction more precise, we introduce the notion of an even polynomial. Define the norm of a nonzero polynomial M F q [T ] as M = q deg M, so that M = #F q [T ]/M. We say that M is even if M is divisible by every irreducible of norm 2. If q > 2, then every element of F q [T ] is even, while when q = 2, the polynomial M is even precisely when it is divisible by T T +. s in the classical setting, it is easy to verify that if is not even, then R = O; also in this case, S = 0. So let us assume that is even. Then we conjecture that whenever q n,, R S qn..3 Note that this is a uniform conjecture, in the sense that we do not assume that any of the parameters q, n, α, β or is fixed. Our attack on this conjecture goes via the circle method, as developed in the polynomial setting by Hayes [0]. In this paper, Hayes proves a three-irreducible analogue of our conjecture.3, following the approach of [8]. Whereas Hardy & Littlewood needed an unproved hypothesis on the zeros of L-functions in their work,

3 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt The exceptional set in the polynomial Goldbach problem 3 Hayes s results and ours are unconditional, owing to Weil s proof of the geometric Riemann Hypothesis. Our principal result is an analogue of.2: Theorem.. Let α and β be nonzero elements of F q, and let R be defined as above. Then qn R S 2 q 5n+/2 n..4 Here the indicates that the sum is taken over degree-n polynomials with leading coefficient γ in the case γ 0, and over all nonzero polynomials of degree < n when γ = 0. The implied constant is absolute. From this, it is a simple matter see 4 to deduce the following estimate for the size of the exceptional set in Goldbach s problem: Theorem.2. Let α and β be nonzero elements of F q, and let γ := α+β. Suppose first that γ 0. Then the number of even polynomials of degree n and leading coefficient γ that cannot be written in the form αp + βp 2 for degree-n monic irreducible polynomials P, P 2 is q n+/2 n 3. If γ = 0, then the same bound holds for the number of even polynomials of degree < n that cannot be represented in this form. Here the implied constant is absolute. The bound of Theorem.2 is a bit more explicit than the x /2+o bound of Hardy and Littlewood quoted above. It may be compared with the result of Goldston see [7, bottom of p. 22] that on ERH, the number Ex of even N x lacking a representation as a sum of two primes is x /2 log x 4. The author has proposed a different attack on.3 in [3, Chapter 7]. That approach, which rests an explicit version of the Chebotarev density theorem for function fields, shows that.3 holds if q tends to infinity much faster than n and satisfies gcdq, 2n =. One consequence is that for an appropriate constant C, the exceptional set considered in Theorem.2 is empty if gcdq, 2n = and q > Cn! 4. We do not go into details here; related results will appear in joint work of the author with ndreas Bender [3]. See also [], [2]. We conclude this introduction by remarking that since the era of Hardy and Littlewood, there has been substantial progress towards estimating Ex unconditionally. In the late 30s, Chudakov [4], van der Corput [4], and Estermann [6] independently adapted methods of Vinogradov to show that Ex x/log x for each positive. The current record is due to Pintz see [], who has shown that Ex x θ for a certain θ < 2/3.

4 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 4 Paul Pollack 2. Preliminaries 2.. Notation and conventions We recall briefly the set-up of Hayes [0]. We write F q T for the completion of F q T at the prime associated to the /T -adic valuation, which we identify with the field of finite-tailed Laurent series in /T : { n } F q T = F q /T = a i T i : a i F q, n Z. i= We let denote the induced absolute value on F q T, so that n a i T i = q n if a n 0. i= Note that this agrees with the previous definition of M for M F q [T ]. The unit interval is defined as { } := a i T i : a i F q. i<0 Then is a compact abelian group; we use ν to denote the Haar measure on, normalized so that ν =. For notational simplicity, we always abbreviate fθ dνθ to fθ dθ. For θ and integers r, we define Bθ, r = {η : η θ < q r }. Then the ν-measure of Bθ, r is q r see [0, Corollary 3.2]. We write e: F q T S for the map defined by n 2πi e a i T i = exp p Tra, i= where the trace is from F q to its prime field F p Two lemmas on arithmetic functions complex-valued function f, defined on the multiplicative semigroup M of monic polynomials over F q, is said to be multiplicative if fb = ffb whenever and B are relatively prime. Two examples of multiplicative functions which appear repeatedly are the analogues of the Euler totient function and the Möbius function: Here φm = #F q [T ]/M, and { 0 if P 2 M for some irreducible P, µm = k if M is the product of k distinct monic irreducibles. The following crude lemma is often useful:

5 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt The exceptional set in the polynomial Goldbach problem 5 Lemma 2.. If G is a nonnegative multiplicative function, then deg d monic G q d deg P d where the implied constant is absolute. Proof. Define g : M C so that GP + + GP 2 GP P P , G = gd. D D monic By Möbius inversion, g = D,D monic GDµ/D. Since g is multiplicative and gp k = GP k GP k, we have that deg d monic G = deg d D monic D monic 2q d deg P d + gd q d + q d + + q deg D GP P Lemma 2.2. For every real i, we have deg =d monic φ i = Oq id, where the implied constant depends only on i. deg D d D monic + GP 2 GP P Proof. Define a multiplicative function G on M by setting i G := = i. φ P P. gd D Since = q d when deg = d, to prove the lemma it is enough to show that G = Oq d. 2. deg =d monic For each monic irreducible P we have GP i / P. Moreover, since G depends only on the irreducibles dividing, every difference GP k GP k with k > vanishes. By Lemma 2., deg =d monic G q d deg P d + O P 2 q d exp O P Now 2. follows since P 2 M M M 2 = n q n 2. P 2.

6 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 6 Paul Pollack 2.3. The fundamental approximation Let n be a positive integer. To study additive problems concerning degree-n irreducibles, one is led to investigate the behavior of the function f : C defined by fθ := ep θ, deg P =n where the sum is over monic irreducibles of degree n. We introduce the decomposition = I G/H, deg H n/2 gcdg,h= where I G/H = { } η : η G/H <. q deg H q n/2 Thus I G/H = BG/H, n/2 + deg H. The sets I G/H, with G and H as above, form a disjoint open cover of [0, Theorem 4.3]. We define the major arcs as the union of those intervals I G/H with deg H n/4, and we take 2 := \ the minor arcs. The function f can be well-approximated on each I G/H by a simpler function g. For θ I G/H, set { µh q n n gθ := e T n θ G/H if θ G/H < /q n, 0 otherwise. The following fundamental estimate is proved by Hayes as a consequence of Weil s Riemann Hypothesis see [0, Theorem 5.3 and Lemma 7.]: Lemma 2.3. For all θ, we have fθ gθ < 2q 3n+/4. 3. Proof of Theorem. For distinct polynomials, the functions eθ define orthonormal elements of L 2 see [0, Theorem 3.5]. Thus fαθfβθe θ dθ = eαp + βp 2 θe θ dθ = R. P,P 2 We decompose R = R + R 2, where in R the integration is taken over and in R 2 the integration is taken over 2. Then 2 qn R S R R S q. 3. Lemma 3.. We have fθ 2 dθ q n /n and gθ 2 dθ q n /n.

7 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt The exceptional set in the polynomial Goldbach problem 7 Proof. The first estimate is almost immediate. Writing πq; n for the number of monic irreducible polynomials of degree n over F q, we have by a well-known theorem of Gauss that πq; n q n /n. Thus fθ 2 dθ = eθp P 2 dθ = = πq; n qn n. P P,P 2 To handle the second estimate, observe that gθ 2 dθ = deg H n/2 qn deg H n/2 G,H= µh BG/H,n 2 G,H= and that the final sum here is n by Lemma µh q 2n dθ = qn Now recall the following elementary result from linear algebra: deg H n/2, squarefree Lemma 3.2 Bessel s inequality. Let e,..., e n be a finite collection of orthonormal vectors in a complex inner product space V. Then for any x V, Lemma 3.3. We have n x, e k 2 x 2. k= R 2 2 q 5n+/2 n. Proof. We view R 2 as the -th Fourier coefficient of the function fαθfβθ 2, where 2 is the indicator function of the set 2. So by Bessel s inequality, with the functions eθ playing the role of the e i, we see that R 2 2 fαθfβθ 2 dθ, 2 which, by the Schwarz inequality, is bounded by /2 /2 fαθ 4 dθ fβθ 4 dθ. 2 2 Since multiplication by elements of F q preserves the ν-measure of Borel subsets of, both of the above integrals coincide with 2 fθ 4 dθ. Now fθ 4 dθ 2 gθ 4 dθ + 2 fθ gθ 4 dθ. 2

8 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 8 Paul Pollack By Lemmas 2.3 and 3., fθ gθ 4 dθ sup fθ gθ 2 2 By Lemma 2.2, 2 gθ 4 dθ = q4n n 4 = q3n n 4 2 fθ 2 + gθ 2 dθ q 3n+/2 q n /n + q n /n q 5n+/2 n. n/4<deg H n/2 n/4<deg H n/2, squarefree 4 µh dθ BG/H,n G,H= 3 q3n n 4 Lemma 3.3 follows upon collecting the estimates. n/4<r n/2 q5n/2 q2r n 4. For H a monic polynomial over F q and any element of F q [T ], define c H by c H := eg/h. G mod H G,H= Here G runs over a reduced residue system modulo H, which will usually be chosen as the set of polynomials of degree < deg H and coprime to H. Then c H is a polynomial analogue of the usual Ramanujan sum. It is multiplicative in H for fixed and satisfies µh/h, c H =. 3.2 φh/h, Compare [0, Theorem 6.]. Lemma 3.4. We have where and R = S qn + E, S := deg H n/4 2 µh c H E := fαθfβθ gαθgβθe θ dθ. Proof. We have R = gαθgβθe θ dθ + fαθfβθ gαθgβθe θ dθ = gαθgβθe θ dθ + E,

9 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt The exceptional set in the polynomial Goldbach problem 9 and we need to show that the remaining integral is S q n /. Inserting the definition of g, we can rewrite this integral as q 2n 2 µh eα + βt n θ G/He θ dθ. deg H n/4 G,H= BG/H,n Write e θ = e G/He θ G/H and make the change of variables θ θ + G/H, so that the integration takes place over B0, n. This transforms the expression into q 2n deg H n/4 2 µh e G/H eα + βt n θ dθ. B0,n G,H= By the choice of, the polynomial α + βt n has degree < n; it follows that α + βt n θ < q for each θ B0, n. Recalling the definition of e, we see that the integrand here is identically. Since the measure of B0, n is q n, the above simplifies to q n 2 µh e G/H. deg H n/4 G,H= But the rightmost sum here is precisely c H = c H. Lemma 3.5. We have S S 2 q n/2 n 3. Proof. Since c H is multiplicative in H, we have 2 µh c H = + P 2 c P deg H>n/4 P = S. The factorization here is justified by the absolute convergence of the left-hand sum, which follows from 3.2. Hence S S = 2 µh µh/h, φh/h, = = D deg H>n/4 D squarefree, monic D H, H/D,= D D monic, squarefree φd µh 2 deg E>n/4 deg D E monic, E,= µh/d φh/d µe φe 2.

10 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 0 Paul Pollack ppealing to Lemma 2.2, this last double sum is q n/4 Thus D D monic, squarefree S S 2 q n/2 K, where K := pplying Lemma 2., Now KP P K q n = deg P n + D φd. D D monic, squarefree 2 D φd. KP. 3.3 P 2 P + P P 2 = 3 P + O P 2, and so the product on the right-hand side of 3.3 is exp 3 P + O exp 3 q r q r = exp3 log n + O n 3. r r n deg P n Piecing everything together, as desired. S S 2 q n/2 q n n 3 = q n/2 n 3, Lemma 3.6. We have E 2 q 5n+/2 n. Proof. By another application of Bessel s inequality, E 2 fαθfβθ gαθgβθ 2 dθ. Since fαθfβθ gαθgβθ 2 fαθ gαθ 2 fβθ 2 + fβθ gβθ 2 gαθ 2, we have by Lemmas 2.3 and 3., fαθfβθ gαθgβθ 2 sup f g 2 fβθ 2 + gαθ 2 q 3n+/2 fθ 2 + gθ 2 q 3n+/2 q n n = q 5n+/2 n,

11 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt The exceptional set in the polynomial Goldbach problem as desired. Lemma 3.7. We have R Sq n / 2 q 5n+/2 n. Proof. Observe that the sum to be estimated is R S q + S qn qn S n2 By Lemmas 3.5 and 3.6, this is as claimed. = E 2 + q2n n 4 q 5n+/2 n + q2n n 4 qn/2 n 3 q 5n+/2 n, 2 S S 2. Theorem. follows immediately upon combining 3. with the results of Lemmas 3.3 and Proof of Theorem.2 Lemma 4.. If is even, then S, where the implied constant is absolute. Proof. Since is even, P 2 P 2. s S P P P 2 P 4 P 2 d P >2 4 q d q 2d d < 4 q d = 4 q, 4. d it follows that S is bounded below by a positive constant S q say depending only on q. Moreover, for q 9, S P 2 P 2 4 q P P >2 Since S q > 0 for each of the finitely many q < 9, the lemma follows. Suppose now that is exceptional, so that is included in the sum.4 but R = 0. Then contributes S 2 q 2n /n 4 q 2n /n 4 to.4. So the number of such must be q5n+/2 n q 2n /n 4 = q n+/2 n 3, which is the assertion of Theorem.2.

12 May 3, 200 2:55 WSPC/INSTRCTION FILE gcremarks-ijnt 2 Paul Pollack cknowledgements The author is supported by an NSF postdoctoral research fellowship. This work is adapted from the author s Ph.D. thesis at Dartmouth College [3]. He thanks his advisor, Carl Pomerance, for generosity with both time and ideas. He also thanks the referee for a careful reading of the manuscript. References []. O. Bender, Decompositions into sums of two irreducibles in F q[t], C. R. Math. cad. Sci. Paris , no. 7-8, [2], Representing an element in F q[t] as the sum of two irreducibles in F q s[t], submitted, arxiv: v [math.nt], [3]. O. Bender and P. Pollack, On quantitative analogues of the Goldbach and twin prime conjectures over F q[t], submitted. [4] N. G. Chudakov, On the density of the set of even numbers which are not representable as a sum of two primes, Izv. kad. Nauk. SSSR 2 938, [5] R. Crandall and C. Pomerance, Prime numbers: computational perspective, second ed., Springer, New York, [6] T. Estermann, On Goldbach s problem: Proof that almost all even positive integers are sums of two primes, Proc. London Math. Soc , [7] D.. Goldston, On Hardy and Littlewood s contribution to the Goldbach conjecture, Proceedings of the malfi Conference on nalytic Number Theory Maiori, 989 Salerno, niv. Salerno, 992, pp [8] G. H. Hardy and J. E. Littlewood, Some problems of Partitio numerorum ; III: On the expression of a number as a sum of primes, cta Math , no., 70. [9], Some problems of Partitio numerorum ; V: further contribution to the study of Goldbach s problem, Proc. London Math. Soc , [0] D. R. Hayes, The expression of a polynomial as a sum of three irreducibles, cta rith. 966, [] J. Pintz, Landau s problems on primes, J. Théor. Nombres Bordeaux , no. 2, [2] P. Pollack, polynomial analogue of the twin prime conjecture, Proc. mer. Math. Soc , no., [3], Prime polynomials over finite fields, Ph.D. thesis, Dartmouth College, [4] J. G. van der Corput, Sur l hypothése de Goldbach pour presque tous les nombres pairs, cta rith 2 937,