For each number field k, the Bombieri-Lang conjecture for k-rational points
on surfaces of general type implies the existence of a polynomial f(x,y) in
k[x,y] inducing an injection k x k --> k.Comment: 4 page

Poonen, Bjorn

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For each number  field k, the Bombieri-Lang conjecture for k-rational points on
surfaces of general type implies the existence of a polynomial f(x, y) [element of] k[x, y] inducing an injection k  x k-->k

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MULTIVARIABLE POLYNOMIAL INJECTIONS ON RATIONALNUMBERSBJORN POONENAbstract. For each number field k, the Bombieri-Lang conjecture for k-rational points onsurfaces of general type implies the existence of a polynomial f(x, y) ∈ k[x, y] inducing aninjection k × k → k.1. IntroductionHarvey Friedman asked whether there exists a polynomial f(x, y) ∈ Q[x, y] such that theinduced map Q×Q→ Q is injective. Heuristics suggest that most sufficiently complicatedpolynomials should do the trick. Don Zagier has speculated that a polynomial as simpleas x7 + 3y7 might already be an example. But it seems very difficult to prove that anypolynomial works. Both Friedman’s question and Zagier’s speculation are at least a decadeold (see [Cor99, Remarque 10]), but it seems that there has been essentially no progress onthe question so far.Our theorem gives a positive answer conditional on a small part of a well-known conjecture.Theorem 1.1. Let k be a number field. Suppose that there exists a homogeneous polynomialF (x, y) ∈ k[x, y] such that the k-rational points on the surface X in P3 defined by F (x, y) =F (z, w) are not Zariski dense in X. Then there exists a polynomial f(x, y) ∈ k[x, y] inducingan injection k × k → k.Remark 1.2. If F (x, y) is separable (or equivalently, squarefree) and homogeneous of degreeat least 5, then X is of general type. So the hypothesis in Theorem 1.1 would follow fromthe Bombieri-Lang conjecture that k-rational points on a surface of general type are neverZariski dense.Remark 1.3. As the proof of Theorem 1.1 will show, if we have an algorithm for determiningthe Zariski closure of the set of k-rational points on each curve or surface of general type,then we can construct f(x, y) explicitly.Remark 1.4. To prove that a nonzero homogeneous polynomial F (x, y) defines an injectionk×k → k is to prove that X(k) is contained in the line x− z = y−w = 0. If F is separable,then X is a smooth projective hypersurface in P3, so it is simply connected. But as far as weknow, there is not a single simply connected smooth algebraic surface X with X(k) 6= ∅ suchthat X(k) is known to be not Zariski dense in X! If one uses nonhomogeneous polynomials,Date: September 23, 2010.2000 Mathematics Subject Classification. Primary 11C08; Secondary 11G30, 11G35.Key words and phrases. Bombieri-Lang conjecture, polynomial injection.This article is to appear in Acta Arith. The research was supported by NSF grant DMS-0841321.1one must instead understand rational points on affine 3-folds; this seems unlikely to improvethe situation. All this suggests that Friedman’s question cannot be answered unconditionallywithout a major advance in arithmetic geometry.Remark 1.5. One cannot hope to answer the question using local methods alone. Moreprecisely, if L is any local field of characteristic 0, and f(x, y) ∈ L[x, y] is nonconstant, thenthe induced map L× L→ L is not injective. To prove this, choose a point (x0, y0) ∈ L× Lwhere ∂f/∂x or ∂f/∂y is nonvanishing, and let c = f(x0, y0); then the affine curve f(x, y) = cis smooth at (x0, y0), so by the implicit function theorem it contains infinitely many L-points,each of which has the same image under f as (x0, y0).Remark 1.6. If k is any imperfect field, then there exists a polynomial injection k × k → k,by a construction that can be found in the proof of Proposition 8 in [Cor99]. Namely, letp = char k, choose t ∈ k − kp, and use f(x, y) = xp + typ. This applies in particular to anyglobal function field.Remark 1.7. The generalized abc-conjecture of [BB94] (more specifically, the 4-variable ana-logue) would imply that f(x, y) := xn + 3yn defines a polynomial injection Q×Q → Q forsufficiently large odd integers n: this was observed in [Cor99, Remarque 10].Remark 1.8. For the function field K of an irreducible curve over a base field k of character-istic 0, an analogue of the generalized abc-conjecture is known [Mas86, Lemma 2]. This ana-logue can be used to show that for some t ∈ K and m ≥ 1, the polynomial f(x, y) = xm+tymdefines an injection, under certain technical hypotheses. These hypotheses can be satisfiedwhen k is a number field, for instance. See [Cor99, Proposition 8] for details and for otherrelated results.2. Proof of theoremLet k, F , and X be as in Theorem 1.1. Let d = degF . Call a line in P3 trivial if it is givenby x− ζz = y− ζw = 0 for some ζ ∈ k with ζd = 1. Each trivial line is contained in X. Letw be the number of roots of 1 in k, and let p be a prime number such that p > 3 and p - w.When we speak of the genus of a geometrically irreducible curve, we mean the genus of itssmooth projective model. When we say that something holds for “most” elements of k orof kn, we mean that it holds outside a thin set in the sense of [Ser97, §9.1]. Such sets arisein the context of the Hilbert irreducibility theorem, which shows that a finite union of thinsets cannot cover all of kn.Lemma 2.1. Fix an integral closed subscheme Z of X. For most ( a bc d ) ∈ GL2(k) ⊂ k4, theinverse image Y of Z under the finite morphismP3 → P3(x : y : z : w) 7→ (axp + byp : cxp + dyp : azp + bwp : czp + dwp)satisfies:(i) If dimZ = 0, then Y (k) = ∅.(ii) If Z is a trivial line, then Y (k) is contained in a trivial line.(iii) If Z is any other curve in X, then Y (k) is finite.2Proof. We can compute Y in stages, by first taking the forward image of Z under the auto-morphismP3 α→ P3(x : y : z : w) 7→ (ax+ by : cx+ dy : az + bw : cz + dw)(technically, ( a bc d ) here should be the inverse of what it was before, but this does not matter),and then pulling back byP3 β→ P3(x : y : z : w) 7→ (xp : yp : zp : wp).(i) Here dimZ = 0. If Z is not a k-rational point, then Z(k) = ∅, so Y (k) = ∅. If Z is ak-rational point, then for most ( a bc d ) the value of (ax+ by)/(cx+dy) on Z is not a pth powerin k, so Y (k) = ∅.(ii) Here Z is x− ζz = y − ζw = 0 for some ζ ∈ k with ζd = 1. Then α(Z) = Z, so Y isxp − ζzp = yp − ζwp = 0. By choice of p, the pth-power map on k is injective, and moreoverζ = ηp for some η ∈ k with ηd = 1. So all points in Y (k) satisfy x− ηz = y − ηw = 0.(iii) Here Z is an irreducible curve in X that is not a trivial line. If Z is geometricallyreducible, then Z(k) is not Zariski dense in Z, so Z(k) is finite, and Y (k) is finite too. Soassume that Z is geometrically irreducible.If y = 0 on Z or if x/y defines a constant rational function on Z, then as in (i), for most( a bc d ) the value of x/y on α(Z) is not a pth power in k, so Y has no k-rational points exceptpossibly those where x = y = 0, so Y (k) is finite.Suppose that x/y defines a rational function of degree m > 1 on Z. By Bertini’s theorem([Har77, Corollary III.10.9]), ax+ by has distinct zeros on the normalization Z ′ of Z, outsidethe base locus of the linear system given by 〈x, y〉, for most a and b. The same applies tocx+dy, so for most ( a bc d ), the rational function (ax+ by)/(cx+dy) on Z′ has m simple zerosand m simple poles on Z ′. Adjoining the pth root of this function to the function field of Z ′yields the function field of a geometrically irreducible curve C of genus greater than 1, bythe Hurwitz formula. By [Fal83], C(k) is finite. Since Y admits a dominant rational map toC, the set Y (k) is finite too.Thus we may assume that x/y is of degree 1 on Z; in particular, Z is a rational curve.Similarly, we may assume that z/w is of degree 1 on Z. If the rational functions x/y and z/won Z were different, then for most ( a bc d ), the supports of the divisors of (ax+ by)/(cx+ dy)and (az + bw)/(cz + dw) on the normalization of Z would not coincide. Adjoining the pthroots of these functions would lead to a geometrically irreducible curve of genus greater than1, by the Hurwitz formula again. So Y (k) would be finite as before.Thus we may assume that x/y = z/w as rational functions on Z. So on Z, we havexdF (x, y) = xdF (z, w) = F (xz, xw) = F (xz, yz) = zdF (x, y).But F (x, y) does not vanish on Z (since x/y is nonconstant), so xd − zd vanishes on Z.Since Z is geometrically irreducible, x− ζz vanishes on Z for some ζ ∈ k with ζd = 1. Butx/y = z/w on Z, so y − ζw vanishes on Z too. Thus Z is a trivial line, a contradiction. Let W be the Zariski closure of X(k). By assumption, dimW ≤ 1. Applying Lemma 2.1 toeach irreducible component of W shows that by replacing F (x, y) with F (axp+byp, cxp+dyp)for suitable ( a bc d ), we may reduce to the case that W (k) contains at most finitely many points3outside the trivial lines. Repeating this construction lets us reduce to the case that W (k) iscontained in the trivial lines.Lemma 2.2. For G(x, y) := F (xp+1, yp+1) for the new F as above, the equation G(x, y) =G(z, w) has only finitely many solutions over k with (x, y) 6= (z, w).Proof. Suppose that x, y, z, w ∈ k are such that G(x, y) = G(z, w). Then there exists ζ ∈ kwith ζd = 1 such that xp + 1 = ζ(zp + 1) and yp + 1 = ζ(wp + 1). If ζ = 1, then this implies(x, y) = (z, w), by the choice of p. If ζ 6= 1, then xp + 1 = ζ(zp + 1) defines a geometricallyirreducible curve whose projective closure is smooth, and hence of genus p(p− 1)/2 > 1, soby [Fal83] it has at most finitely many solutions in k. The same applies to yp+1 = ζ(wp+1),and there are only finitely many ζ. Lemma 2.3. If G(x, y) ∈ k[x, y] is such that the equation G(x, y) = G(z, w) has onlyfinitely many solutions over k with (x, y) 6= (z, w), then for most (a, b) ∈ k2 the polynomialf(x, y) := G(axp + b, ayp + b) defines an injection k × k → k.Proof. By choice of p, if a 6= 0 and b is arbitrary, the polynomial axp + b defines an injectionk → k. Also, for most (a, b), the coordinates of the finitely many exceptional solutions toG(x, y) = G(z, w) are not in the range of this injection. Theorem 1.1 follows from Lemmas 2.2 and 2.3.AcknowledgementsThe idea for this article arose during a discussion session at the Hausdorff Institute inBonn. I thank Burt Totaro for a comment on a first draft of this article, and I thank thereferee for a few suggestions.References[BB94] J. Browkin and J. Brzezin´ski, Some remarks on the abc-conjecture, Math. Comp. 62 (1994), no. 206,931–939. MR 1218341 (94g:11021) ↑1.7[Cor99] Gunther Cornelissen, Stockage diophantien et hypothe`se abc ge´ne´ralise´e, C. R. Acad. Sci. ParisSe´r. I Math. 328 (1999), no. 1, 3–8 (French, with English and French summaries). MR 1674417(99i:03036) ↑1, 1.6, 1.7, 1.8[Fal83] G. Faltings, Endlichkeitssa¨tze fu¨r abelsche Varieta¨ten u¨ber Zahlko¨rpern, Invent. Math. 73 (1983),no. 3, 349–366 (German); English transl., Finiteness theorems for abelian varieties over numberfields (1986), 9–27. Erratum in: Invent. Math. 75 (1984), 381. MR 718935 (85g:11026a) ↑2, 2[Har77] Robin Hartshorne, Algebraic geometry, Springer-Verlag, New York, 1977. Graduate Texts in Math-ematics, No. 52. MR 0463157 (57 #3116) ↑2[Mas86] R. C. Mason, Norm form equations. I, J. Number Theory 22 (1986), no. 2, 190–207, DOI10.1016/0022-314X(86)90069-7. MR 826951 (87i:11040) ↑1.8[Ser97] Jean-Pierre Serre, Lectures on the Mordell-Weil theorem, 3rd ed., Aspects of Mathematics, Friedr.Vieweg & Sohn, Braunschweig, 1997. Translated from the French and edited by Martin Brown fromnotes by Michel Waldschmidt; With a foreword by Brown and Serre. MR 1757192 (2000m:11049)↑2Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA02139-4307, USAE-mail address: poonen@math.mit.eduURL: http://math.mit.edu/~poonen4

Acta Arithmetica

Multivariable polynomial injections on rational numbers

Bjorn Poonen

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Multivariable polynomial injections on rational numbers

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