UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Points of small height on plane curves Radzimski, Vanessa Elena 2014

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata

Download

Media
24-ubc_2014_spring_radzimski_vanessa.pdf [ 284.24kB ]
Metadata
JSON: 24-1.0103396.json
JSON-LD: 24-1.0103396-ld.json
RDF/XML (Pretty): 24-1.0103396-rdf.xml
RDF/JSON: 24-1.0103396-rdf.json
Turtle: 24-1.0103396-turtle.txt
N-Triples: 24-1.0103396-rdf-ntriples.txt
Original Record: 24-1.0103396-source.json
Full Text
24-1.0103396-fulltext.txt
Citation
24-1.0103396.ris

Full Text

Points of Small Height on Plane CurvesbyVanessa Elena RadzimskiB.Sc., Florida State University, 2012A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)April 2014c? Vanessa Elena Radzimski 2014AbstractLet K be an algebraically closed field, and let C be an irreducible plane curve,defined over the algebraic closure of K(t), which is not defined over K. Weshow that there exists a positive real number c0 such that if P is any point onthe curve C whose Weil height is bounded above by c0, then the coordinatesof P belong to K.iiPrefaceThis thesis was written by myself. The topic of this thesis and the methodol-ogy for proving the results were suggested by my supervisor, Professor DragosGhioca.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . v1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Generalities and preliminaries . . . . . . . . . . . . . . . . . . 11.2 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Outline of study . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Valuations over K(t) and its extensions . . . . . . . . . . . . . 42.2 Residue fields for valuations of K(t) . . . . . . . . . . . . . . . 73 Heights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Curves in Affine Space . . . . . . . . . . . . . . . . . . . . . . . 154.1 The ?-endomorphism . . . . . . . . . . . . . . . . . . . . . . . 154.2 Curves over K(t) . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . 28Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31ivAcknowledgementsFirst, an enormous thank you goes to my supervisor, Dragos Ghioca. I amextremely grateful for all of the support he has given me during my work withhim; for his mathematical insight, advice on teaching, and his everlastingencouragement. A special thanks also goes to Zinovy Reichstein for servingas the second reader of this thesis.All of the faculty, staff, and students in the math department deserve around of applause. Thanks to my professors for their difficult but enlighteningcourses, to the office staff who always bring a smile to my face, and to myfellow graduate students who have made my experience even brighter.Finally, a great thanks to my friends and family. It?s amazing to knowthat there are so many people, near and far, who believe in me no matterwhat.vChapter 1Introduction1.1 Generalities and preliminariesThroughout our study in number theory, we would like to determine thedegree of complexity of numbers. The theory of heights provides us withthe tools to study numbers in this setting. In order to understand what aheight function is, we could consider the elementary exponential height onthe rationals. This type of height may be extended to finite extensions ofthe rationals and even further to function fields over number fields. In thework ahead, we consider heights for points on affine varieties. We denoteby k the algebraic closure of a field k. We assume that the reader has anunderstanding of general valuation theory [4], Galois theory [1], and basicalgebraic geometry [1]. We will review valuation theory over function fieldsK(t) and its finite extensions, as well as the basic definitions and propertiesof the height function in this setting. Throughout our study, we will assumethe base field K to be algebraically closed.11.2 MotivationThe main focus of study ahead regards that of plane curves. We definek[X] = k[x1, . . . , xn] for some field k and consider the following definition.Definition 1.2.1. An affine variety V over a field k is the set of common zerosof a collection of polynomials f1(X), . . . , fm(X).If n = 2 and m = 1, we define the variety V as a plane curve C ={(x, y) ? k2|f(x, y) = 0}. As long as the polynomial f(X, Y ) ? k[X, Y ] isnon-constant, C is non-empty. The choice for the field k could be any field,in particular the field of rational functions K(t) for some constant field K.We would like to study whether or not there exists a positive lower boundfor the height of a point on a plane curve defined over the algebraic closureof a function field K(t). For k = Fp(t), Ghioca proved in [3] the following:Theorem 1.2.2. Let X be an affine subvariety of An defined over Fp(t).Let Y be the Zariski closure of X(Fp). There exists a positive constant Cdepending only on X such that if P ? X(Fp(t)) and h(P ) < C, then P ?Y (Fp(t)).This theorem states that within the varieties over Fp(t) the only variatiesthat contain many point of small height are the varieties defined over theconstant field Fp. So, it is natural to ask whether the above result can beextended to function fields with constant fields, such as C(t) or Q(t). Inorder to do this, we define a new endomorphism to replace the role of theFrobenius in the results of [3], and obtain the main theorem of this paper.Theorem 1.2.3. Let K be an algebraically closed field and let C ? A2 bean irreducible plane curve defined over K(t) and not over K. For (x, y) ?C(K(t)), there exists a real number c0 > 0 such that if h(x, y) < c0, then(x, y) ? C(K).21.3 Outline of studyWe begin our study with valuation theory over function fields. We then definethe basic notion of heights over function fields and extend the definition tofinite extensions of the base field. We provide the reader with a brief overviewof the definitions and basic properties of heights over function fields, and thenmove on to studying plane curves defined over K(t). As our constant fieldsare arbitrary, we examine what we refer to as the ?-endomorphism on K[t]and its interplay with heights over finite extensions of K(t). With the use ofa lemma of Derksen and Masser [2], we prove preliminary lemmas that thenlead to a proof of Theorem 1.2.3 over K(t).Aiming to extend this result to curves C defined over K(t), we reduce ourproof to a curve C ? which is defined over K(t); C ? is the union of the Galoisconjugates of C over K(t). Using the fact that this curve C ? contains C, weconclude that Theorem 1.2.3 holds for curves defined over K(t).3Chapter 2ValuationsWe assume the reader is familiar with general valuation theory. For our caseof interest, we take F = K(t), where K is algebraically closed and obtainthe following discrete valuations where v(K(t)?) = Z.2.1 Valuations over K(t) and its extensionsDefinition 2.1.1. Let f, g,? K[t], so that F =fg? K(t). We define the placeat infinity as:v?(F ) = deg g ? deg f .For any ? ? K, define the valuation associated to t? ? ? K[t] as:vt??(F ) = vt??(f)? vt??(g)with vt??(f) = vt??((t? ?)df1) = d and (t? ?) - f1.Proposition 2.1.2. If v is a discrete valuation on a field F , then the subringOv = {x ? F|v(x) ? 0} is a valuation ring with unique maximal idealMv = {x ? F|v(x) ? 1}.4Proof. By definition, we have that Ov is a ring. Now, we wish to show thatfor any x ? F , either x or x?1 is in Ov. If x /? Ov, then v(x) < 0. So,v(x?1) = v(1) ? v(x) = 0 ? v(x) > 0. Hence, x?1 ? Ov, making Ov avaluation ring. Now, if x ? O?v , then we have that x, x?1 ? Ov, whichhappens if and only if v(x) = v(x?1) = 0. Hence,Mv consists of all non-unitelements of Ov, so that Ov/Mv is a field, making Mv the unique maximalideal of Ov.From now on, we denote by ?K(t) the set of all valuations over K(t).Claim 2.1.3. Modulo taking multiples, the valuations of Definition 2.1.1 arethe only valuations on K(t).Proof. Let v be a discrete valuation of K(t) and let p be its place. We weconsider the two cases v(t) ? 0 and v(t) < 0.Suppose v(t) ? 0. Then, we have that K[t] ? Ov and v(K(t)?) 6= {0}.For J = K[t] ?Mv, we have that J is a non-zero ideal of K[t]. Since 1 /? J ,we have that J 6= K[t]. As Mv is maximal in Op, we have that J is primein K[t]. Hence, J = (t? ?) for some ? ? K.If f(t) ? K[t] is not divisible by t? ?, then f(t) /? Mv and v(f(t)) = 0.For g(t) non-zero in K(t), we may write g(t) = (t? ?)?g1(t)g2(t), where ? ? Z,g1, g2 ? K[t], and t ? ? - g1, g2. So, v(g(t)) = ?v(t ? ?) = mvt??(g(t)).Hence, v ? vt??.Now, suppose that v(t) > 0. Then, n = v(t?1) > 0 for t?1 ? Mv. Takef ? K[t] with deg (f) = d. Then,f(t) =d?i=0?iti = tdd?i=0?ixi?d= tdd?i=0?d?it?i5= td(?d +d?i=1?d?it?i)= td(?d + g(t)),where g(t) ? Mv. Since we have that ?d 6= 0 ? K, v(?d) = 0, so thatv(d?i=0?d?it?i) = 0. Finally, we have that v ? v? sincev(f(t)) = v(td) = ?dv(t?1) = kv?(f(t)).With this claim, we may consider the following familiar lemma.Lemma 2.1.4. For x ? K(t)?,?v??K(t)v(x) = 0.Proof. Let x ? K(t) with x = fg , f, g ? K[t]. By definiton of vt??, if P - fg,then vt??(x) = 0. Thus, the only valuations we need to consider are thoseassociated to t? ? which divide f or g, as well as the infinite place. Lettingt??1, . . . , t??n be all divisors of f and g, we have that f = (t??1)?1 . . . (t??n)?n and g = (t ? ?1)?1 . . . (t ? ?n)?n . The valuation of a polynomial att ? ?i corresponds to the order of vanishing at ?i ? K. So, we have thatdeg (f) =?i?i and deg (g) =?i?i. Hence,?v??K(t)v(x) =?t??vt??(x) + v?(x)=?t??vt??(x) + deg (g)? deg (f)Now, we have thatdeg (g)? deg (f) =?i(?i ? ?i)6and?t??vt??(x) =?i(vt??i(f)? vt??i(g)) =?i(?i ? ?i).Combining these results,?v??K(t)v(x) =?t??vt??(x) + deg (g)? deg (f)=?i(?i ? ?i) +?i(?i ? ?i)= 02.2 Residue fields for valuations of K(t)K(t) and its finite extensions L are fields satisfying Proposition 2.1.2 [4].Using the definitions from that proposition, we have the following:Ovt?? = {x =fg ? K(t)|f, g ? K[t], t? ? - g, gcd (f, g) = 1}Mvt?? = {x =fg ? K(t)|f, g ? K[t], t? ?|f, t? ? - g, gcd (f, g) = 1}andO? = {x =fg |f, g ? K[t], g 6= 0, deg (f)? deg (g) ? 0}M? = {x =fg |f, g ? K[t], g 6= 0, deg (f)? deg (g) < 0}We define the residue field ?v = Ov/Mv for each valuation v ? ?L.Proposition 2.2.1. For F = K(t) and place v ? ?F , for t? ? irreducible,the residue field ?vt?? is isomorphic to K[t]/(t? ?) and ?? ?= K.Proof. Let us first consider the place v = vt??, for ? ? K. Consider thecanonical projection Ovt?? ?? ?vt?? . We now define the map ? : K[t] ???vt?? as the restriction of the canonical map. The kernel of ? is (t ? ?), sowe must show that ? is surjective.7Let x =f(t)g(t)? Ovt?? , where (t ? ?) - g(t) and gcd (f, g) = 1. Sincet ? ? is irreducible, we must have that gcd (t? ?, g(t)) = 1. So, there existh(t), l(t) ? K[t] such that (t? ?)h(t) + g(t)l(t) = 1. Hence,x =f(t)g(t)(t? ?)h(t) + g(t)l(t) =(t? ?)h(t)f(t)g(t)+ f(t)l(t)We then have that ?(x) = f(t)l(t). So, since both f(t) and l(t) arein K[t] this proves that ? is surjective giving us the desired isomorphism?vt?? ?= K[t]/(t? ?).Now, let v = v?. First, we note that for any f(t) ? O?, f(t) may bewritten in the formf(t) =bntn + bn?1tn?1 + ? ? ?+ b0tn + cn?1tn?1 + ? ? ?+ c0Where n ? 0 and bi, ci ? K. Next, consider the map ? : O? ?? Kwhere ?(f(t)) := bn. The map ? is a ring homomorphism since for f(t) =n?i=1bitin?1?j=1cjtj + tnand g(t) =m?i=1ditim?1?j=1ejtj + tmwe have:?(f(t)g(t)) = ??????n?i=1bitin?1?j=1cjtj + tn?m?i=1ditim?1?j=1ejtj + tm?????= ???????n+m?i+j=kbidjtk(n?1?j=1cjtj + tn)(m?1?j=1ejtj + tm)??????= bndm= ?(f(t)) ? ?(g(t))8and?(f(t) + g(t)) = ???????n?i=1biti(m?1?j=1ejtj + tm)+m?i=1diti(n?1?j=1cjtj + tn)(m?1?j=1ejtj + tm)(n?1?j=1cjtj + tn)??????= ???????(bn + dm)tn+m +m+n?1?i+j=k?ktk(m?1?j=1ejtj + tm)(n?1?j=1cjtj + tn)??????= bn + dm= ?(f(t)) + ?(g(t)).Certainly, ? is surjective and ker (?) is the set of all f where bn is zero,i.e when the degree of the numerator is less than the denominator. Hence,we have that ker (?) =M?, proving that ?? ?= K as desired.Proposition 2.2.2. If the field K is algebraically closed, then for vt?? ??K(t), ?vt?? = K. Moreover, if L/K(t) is a finite extension with w|v, then?w = K.Proof. We must prove that the extension ?w/?v is algebraic.Given an algebraic extension L/K(t), consider some element x ? L ?Ow. Since L/K(t) is algebraic, x must be a root of some polynomial withcoefficients in K(t). That is, there exist a1, . . . , an ? K(t) whereanxn + an?1xn?1 + ? ? ?+ a0 = 0.Let v be a place of K(t) lying below w. Then, v corresponds to eithert? ? or v?.9Suppose v = vt??. Multiplying each ai by (t? ?)?mini v(ai), we have thateach of the ai now belongs to Ov. With ai =figi, at least one of the ai isa unit modulo Mv, that is, t ? ? does not divide fi or gi. Now, reducingmodulo Mw, we have:a?ix?i + ? ? ?+ a?0 = 0where i is the largest index such that ai is a unit moduloMv. As a remark, wenote that i 6= 0 since otherwise a?0 = 0 or a0 = 0, moduloMv, a contradiction.Now, suppose that v = v?. We still have that x satisfies an equationanxn+an?1xn?1+? ? ?+a0 = 0 for ai ? K(t). We want the ai in O?, so as eachai =figi, with fi, gi ? K[t], we will multiply each ai by tmini {deg(gi)?deg(fi)}.Now we have that deg (fi) ? deg (gi) for every i, so that v?(ai) ? Z and atleast one is a unit. This yields ai ? O?. Reducing modulo Mw, we havethata?ix?i + ? ? ?+ a?0 = 0Now in either case, we have that x is a root of an equation over ?v sinceeach ai ? Ov/(Mw ? Ov) = Ov/Mv = ?v. Hence, ?w/?v is algebraic.In the first case v = vP (t), with P (t) = t ? ?, and by Proposition 2.2.1,?v = K[t]/(t ? ?) = K. But as ?w is an extension of ?v, we have that?w = K as well.Similarly, by Proposition 2.2.1, ?? = K, yielding that ?w = K concludingthe proof.10Chapter 3HeightsNow that we have defined the valuations over K(t), we may move to definingheights.Definition 3.0.3. For x ? K(t), the local height of x at v ? ?K(t) ishv(x) = ?min {0, v(x)} = max {0,?v(x)}.The (global) height of x is defined by h(x) := hK(t)(x) =?v??K(t)hv(x).We may also define a multi-dimensional height, which will be useful inour study of plane curves. Let (x, y) ? K(t)2. Then, defineh(x, y) =?v??K(t)max {hv(x), hv(y)}.In the work ahead, we are concerned with the field K(t), so we may extendthe height to a intermediate field K(t) ? L ? K(t), where L/K(t) is finite.For any valuation v over K(t), v extends to a valuation w ? ?L [4]. We saythat w ? ?L lies above v if and only if Ov embeds into Ow. If this is the case,then we have that ?v ?? ?w as well. So, we define f(w|v) = [?w : ?v]. But,since our field K is algebraically closed, f(w|v) = 1, by Proposition 2.2.2.11Also, we define this valuation w ? ?L for x ? K(t) as w(x) = e(w|v)v(x),where e(w|v) = w(u) for u a uniformizer of v in K(t) (That is, v(u) = 1).We also have that the sum formula from Lemma 2.1.4 holds over L. Thegeneral sum formula for extensions L/K(t) is?w??Lnww(x) = 0, where x ? L?and nw = e(w|v)f(w|v) is the local degree. In our case though, f(w|v) = 1 asmentioned above, and e(w|v) has already been absorbed into the valuationw(x) since w(x) = e(w|v)v(x). Thus, our local degrees nw = 1 for everyw ? ?L and?w??Lw(x) = 0.For x ? L we define the local height of x at w ? ?L as hw(x) =1[L : K(t)]max {?w(x), 0} and the global height of x as h(x) =?w??Lhw(x).Similarly, we may extend the defintion to points (x, y) in A2 with hw(x, y) =1[L : K(t)]max {0,?w(x),?w(y)} and h(x, y) =?w??Lhw(x, y).Remark : In the proof of the following, Proposition 3.0.4, we use the factthat the sum of all the ramification indices equals the degree of the extension,i.e.?w|ve(w|v) = [M : L]. The proof of this fact can be found in [5].Proposition 3.0.4. The definition of the height is well defined.Proof. Let K(t) ? L ?M be a finite extension of fields. For x ? L, v ? ?L,and w ? ?M we have:?w|v1[M : K(t)]w(x) =?w|v1[M : K(t)]w(x)=?w|v1[M : K(t)]e(w|v)v(x)12=v(x)[M : K(t)]?w|ve(w|v)=v(x)[M : K(t)][M : L]=v(x)[L : K(t)].So, when considering the height and using the equality above, we haveh(x) =?v??Lhv(x) =?v??L1[L : K(t)]max {?v(x), 0}=?v??L?w|v1[M : K(t)]max {?w(x), 0}=?w??M1[M : K(t)]max {?w(x), 0}=?w??Mhw(x).Proposition 3.0.5. For any x ? L/K(t) and n ? Z, h(xn) = nh(x).Proof.h(xn) =?w??L1[L : K(t)]max {0,?w(xn)}=?w??L1[L : K(t)]max {0,?nw(x)}= n?w??L1[L : K(t)]max {0,?w(x)}= nh(x).13We conclude this section with a property of the two-dimensional height.Proposition 3.0.6. For x, y ? L/K(t), h(x, y) ? h(x) + h(y).Proof. Recalling that h(x, y) =?v??Lhv(x, y) =?v??Lmax {hv(x), hv(y)}, wenote that max {hv(x), hv(y)} = hv(x) + hv(y) if and only if hv(x) or hv(y)equals 0. So, hv(x, y) ? hv(x) + hv(y) for each place v ? ?L, and taking thesum over all places, we obtain the result h(x, y) ? h(x)+h(y) as desired.14Chapter 4Curves in Affine SpaceWhen working in A2, the varieties that are of interest to us are plane curves.In practice, we consider polynomials f(X, Y ) ? K(t)[X, Y ] and are inter-ested in the zero locus of this polynomial. For the equation f(X, Y ) = 0,the solutions of it will represent a curve C in A2. If we consider a subfieldL ? K(t), we set C(L) = {(x, y) ? L? L : f(x, y) = 0}.The main theorem for this paper is the following:Theorem 4.0.7. Let C ? A2 be an irreducible plane curve defined over K(t)and not over K. For (x, y) ? C(K(t)), there exists a real number c0 > 0 suchthat if h(x, y) < c0, then (x, y) ? C(K).4.1 The ?-endomorphismIn [3], the key endomorphism was the Frobenius, which one applies to thecoefficients of the curve. Since we are now extending the result of Theorem1.2.2 in A2 to arbitrary base fields, we need a different endomorphism.Define ? : K[t] ?? K[t], where ?(K) = idK and ?(t) = tm, for some fixedm ? N. The endomorphism ? then naturally extends to an endomorphism15K(t) ?? K(t). In fact, we claim the following regarding the extension of ?.Claim 4.1.1. ? : K(t) ?? K(t) is an automorphism.Proof. We know that ? : K(t) ?? K(t) is an injective endomorphism offields, so we wish to show that ? is surjective; that is, for every f ? K(t),there exists g ? K(t) such that ?(g) = f .Let us start with an arbitrary algebraic element f ? K(t). So, f satisfiesan equation anfn+an?1fn?1+. . .+a0 = 0 where P (X) = anXn+an?1Xn?1+. . .+ a0 is defined over K(t) and irreducible.We claim that ? : K(t1/m) ?? K(t) is a surjection, and in order to provethis all we need to show is that t has a pre-image in K(t1/m). For ai =m?j=0cijtjand ? ? K(t), define t = ?(?). Then,?(m?j=0cij?j) =m?j=0?(cij?j) =m?j=0cij?(?)j = aiWe note that ?(t1/m)m = ?(t) = tm, so that ?(t1/m) = ?mt, for someprimitive mth root of unity ?m ? K (which exists since K is algebraicallyclosed). Hence, we take ? = ??1m t1/m, which is indeed contained in K(t1/m).Now that we know that ? : K(t1/m) ?? K(t) is a surjection. If we definebi = ??1(ai) for i = 0, . . . , n, we claim that Q(X) = bnXn + ? ? ?+ b0 = 0 hasthe same number of solutions as P (X).We may consider a complete splitting of Q(X) over K(t) so thatbnXn + ? ? ?+ b0 = (X ? x1)e1 ? ? ? (X ? xm)emThen, since bi = ??1(ai), we have thatanXn + ? ? ?+ a0 = (X ? ?(x1))e1 ? ? ? (X ? ?(xm))emHence, P (X) and Q(X) have the same number of roots over K(t). But? maps a solution of Q(X) to a solution of P (X) and moreover, since16? : K(t) ?? K(t) is an endomorphism of fields, it must be injective. Inparticular, f ? K(t) has an image under ?, implying that ? : K(t) ?? K(t)is surjective.Alternatively, we note that ? induces an automorphism of K(t) since ?maps K to K identically, and then it sends one transcendental element ofK(t) (namely t) into a transcendental element (tm) of K(t). So essentially? maps an algebraic function f(t) ? K(t) into f(tm). This certainly meansthat ? is surjective (and thus an isomorphism) since ?(g) = f , where g(t) =f(t1/m).Definition 4.1.2. We say that a non-constant polynomial f ? K[t][X, Y ] isreduced if the coefficients ai of f do not share a non-constant common factorin K[t].If f1, . . . , fk ? K[t], we define the greatest common divisior (gcd) off1, . . . , fk as the unique monic polynomial of highest degree in K[t] whichdivides all of the fi. We define two different types of heights of non-zeroelements of algebraic extensions L/K(t).Definition 4.1.3. For a place w ? ?L and f ? L[X, Y ] we have the localheight hw(f) = maxi {hw(ai)} and the global height h(f) =?w??Lhw(f).When considering the height of a point x ? L/K(t), we would like aconnection between the height of x and the height of ?(x). The followingobservation, [2, Lemma 2.1], will be key in making this connection.Lemma 4.1.4. Let L be a finite extension of K(t) and suppose that x ? Lsatisfies f(x) = 0, for f(X) = anXn + . . . a0 reduced and irreducible overK(t). Then h(x) =h(f)n.Remark: If the reader refers to [2, Lemma 2.1], she will notice a factor of[L : K(t)]. This arises due to the height in [2] being a normalized height. For17our work, we do not need the normalized height and thus ignore the degreefactor.From this result, we may derive the following corollary:Corollary 4.1.5. For l ? N, define T = t1/l and let L be a finite extensionof K(t). Suppose that x ? L satisfies f(x) = 0, where f(X) ? K(T )[X] isreduced and irreducible over K(T). Then, h(x) =h(f)l ? deg (f)Proof. By Lemma 4.1.4, we have h(x) = hK(T ) =hK(T )(f)deg (f), where hK(T ) isthe height with respect to K(T ). But, we want the height h to be the usualheight that is with respect to K(t).If y ? K(T ), then we notehK(t)(y) =1[K(T ) : K(t)]?w??K(T )max {0,?w(y)}andhK(T )(y) =?w??K(T )max {0,?w(y)}.Combining these statements, we findh = hK(t) =1[K(T ) : K(t)]hK(T ) =1lhK(T ).By applying Lemma 4.1.4, we obtain our desired result ofh(x) =hK(T )(f)l ? deg (f)Claim 4.1.6. For x ? K(t), h(?(x)) = mh(x).Proof. Let ? : K(t) ?? K(t) be as we have defined. Then, for x ? K(t),let f = anXn + . . . + a0 be its minimal polynomial over L =??k=0K(t1/mk).Then, f is irreducible over L with ai ? K(t1/mki ), for some ki. But then,18ai ? K(t1/mk), for the maximum of the ki. We may clear denominators sothat we have ai ? K[t1/mk] = K[T ]. Thus,0 = ?(anxn + an?1xn?1 + . . .+ a0)= ?(an)?(x)n + ?(an?1)?(x)n?1 + . . .+ ?(a0)= ?(an)Yn + ?(an?1)Yn?1 + . . .+ ?(a0), for Y = ?(x).If we consider the restriction of ? as the map K(t1/mk) ?? K(t1/mk?1),this map is surjective. To see this, we note that ?(t1/mk)mk= ?(t) = tm.This yeilds ?(t1/mk) = ?t1/mk?1for an mk?1-th root of unity ?. So, we take?(??1t1/mk) = t1/mk?1, proving surjectivity as desired. As K(t1/mk) ??K(t1/mk?1) is an endomorphism of fields, it must be injective. By this, it iscertainly true that ? : L ?? L is an automorphism, which may be seen bythe following chain:K(t) ?? K(t1/m) ?? K(t1/m2) ?? ? ? ?We now want to show that ?(an)Y n + . . .+ ?(a0) = 0 is irreducible overL. Supoose on the contrary that this is not the case. Then,?(an)Y n + . . .+ ?(a0) = (bmY m + . . .+ b0)(ckY k + . . .+ c0) = 0,where each bi = ?(b?i), ci = ?(c?i), for some b?i, c?i ? L since ? is an automor-phism. Then, ?(f) = ?(g)?(h), yielding ?(f ? gh) = 0. But once again,using that ? is injective, we have that f = gh over L, a contradiction of theirreducibility of f over L.Now that we have the irreducibility of ?(an)Y n + . . . + ?(a0) , we mayapply Corollary 4.1.5 and obtain:h(?(x)) =max degi {?(ai)}mkn=mmax degi {ai}mkn=mh(f)mkn= mh(x).Hence, we have that h(?(x)) = mh(x). Additionally, in order to applyCorollary 4.1.5, we want that the gcd(?(ai)) = 1. By assumption, we have19that ai ? K[T ] (which is a PID) for each i and gcd(ai) = 1. Hence, we knowthat there exist ?i ? K[T ] so that1 =n?i=0?iai = ?(n?i=0?iai) =n?i=0?(?i)?(ai).Since ?(?i) = ci for some ci ? K[T ], we have that gcd(?(ai)) = 1, asdesired.Additionally, we have the following corollary which follows from Claim4.1.6.Corollary 4.1.7. For (x, y) ? K(t)2, h(?(x), ?(y)) ?m2h(x, y).Proof. We have that:h(?(x), ?(y)) ? max {h(?(x)), h(?(y))}= max {mh(x),mh(y)}, by Claim 4.1.6= mmax {h(x), h(y)}?m2(h(x) + h(y))?m2h(x, y)The last inequality follows by Proposition 3.0.6 since h(x, y) ? h(x) +h(y) ? 2 max {h(x), h(y)}.4.2 Curves over K(t)After defining and studying characteristics of the ?-endomorphism, we arenow ready to consider the proof of Theorem 1.2.3. In this section we provethe following special case of Theorem 1.2.3.Theorem 4.2.1. Let C ? A2 be an irreducible plane curve defined over K(t)and not over K. For (x, y) ? C(K(t)), there exists a real number c0 > 0 suchthat if h(x, y) < c0, then (x, y) ? C(K).20In order to prove Theorem 4.2.1, we begin with the following lemma.Lemma 4.2.2. Let f ? K[t][X, Y ] be a reduced polynomial of total degreed. Let m ? Z be such that m ? 2h(f), if (x, y) ? K(t)2satisfies f(x, y) = 0then either h(x, y) ?14dor f(?(x), ?(y)) = 0.Proof. Suppose f(X, Y ) =?i,jaijX iY j for i, j ? N? {0}. We choose an inte-germ with respect to f so thatm ? 2h(f) and construct the ?-endomorphismfrom this particular m. For each i, we will define mij = xiyj.Assume f(?(x), ?(y)) 6= 0. We wish to show that h(x, y) ?14d. Wedefine the extension L = K(t, x, y, ?(x), ?(y)). We note that for a givenf(x, y) ? K[t][X, Y ], ?(f(x, y)) = f(?(x), ?(y)) ? L. Define [L : K(t)] = Nfor some N ? N.If ? = f(?(x), ?(y)) 6= 0, then, by the sum formula,?w??L1Nw(?) = 0.Since f(x, y) = 0, we have:? = ? ? ?((f(x, y)))= f(?(x), ?(y))? ?(?i,jaijmij)=?i,jaij?(x)i?(y)j ??i,j?(aijmij)=?i,jaij?(mij)??i,j?(aij)?(mij)=?i,j(aij ? ?(aij))?(mij)Claim 4.2.3. tm ? t|?(g)? g, for any g ? K[t].Proof of Claim. Let g =N?i=0citi, so that ?(g) =N?i=0citim. Then, since tm ?t|tml ? tl, for any l ? N and ?(g) ? g =N?i=0citim ?N?i=0citi =N?i=0ci(tim ? ti),we have that tm ? t|?(g)? g as desired.21So, from Claim 4.2.3, ? = (tm ? t)?i,jbij?(mij), bij =aij ? ?(aij)tm ? t? K[t].Define the set S as follows:S = {w ? ?L : w lies above an irreducible factor in K[t] of tm ? t}.As we are working over an algebraically closed field K, we may write thepolynomial tm ? t in terms of m irreducible linear factors. For an m? 1throot of unity ? ? K, tm ? t = t??m?1=1(t? ?). So, the places w ? S lie aboveanm? 1th root of unity inK or zero. We now consider a series of inequalities.For each w ? S:1Nw(?) =1Nw((tm ? t)?i,jbij?(mij))=1N[w(tm ? t) + w(?i,jbij?(mij))]?1Nw(tm ? t) +1Nmini,j{w(bij?(mij))}=1Nw(tm ? t) +1Nmini,j{w(bij) + w(?(mij))}?1Nw(tm ? t) +1Nw(bIJ) +1Nw(?(mIJ)), for some I, J?1Nw(tm ? t) +1Nw(?(mIJ)), since w(bIJ) ? 0.Now, for any valuation w ? ?L,1Nw(?(mIJ)) =1Nw(?(xI)?(yJ))=1N(w(?(xI) + w(?(yJ)))=1N(I?(x) + J?(y))22?I + JNmin {w(?(x)), w(?(y)), 0}Since the degree of mIJ is at most d, we have:1Nw(?(mIJ)) ?dNmin {w(?(x)), w(?(y)), 0}.Now, recalling that hw(x, y) = ?1Nmin {w(x), w(y), 0}, we have that1Nw(?(mIJ)) ? ?dhw(?(x), ?(y)). (4.2.1)So,1Nw(?) ?1Nw(tm ? t)? dhw(?(x), ?(y)) (4.2.2)Next, taking ? =?i,jaij?(mij) and considering valuations w ? ?L \ S,1Nw(?) =1Nw(?i,jaij?(mij))?1Nmini,j{w(aij?(mij))}=1Nw(aIJ?(mIJ)), for some I, J=1Nw(aIJ) +1Nw(?(mIJ))Now, recall that hw(f) = ?1Nmin {w(aij), 0} for any w ? ?L. Sincew(aIJ) ? min {w(aIJ), 0}, we have:1Nw(?) ?1Nw(aIJ) +1Nw(?(mIJ))?1Nmin {w(aIJ), 0}+1Nw(?(mIJ))?1Nmin {w(aIJ), 0} ? dhw(?(x), ?(y)), by 4.2.123= ?hw(f)? dhw(?(x), ?(y)).Combining our results for all w ? ?L:?w??L1Nw(?) ??w(tm?t)>01Nw(tm ? t)? d?w??Lhw(?(x), ?(y))??w??L\Shw(f).Hence,0 =?w??L1Nw(?) ??w(tm?t)>01Nw(tm ? t)? dh(?(x), ?(y))? h(f) (4.2.3)Then, by our results from Proposition 3.0.4:?w(tm?t)>01Nw(tm ? t) =?v??K(t)v(tm ? t) = ?v?(tm ? t) = mBy Claim 4.1.6,0 ? m? dh(?(x), ?(y))? h(f)= m? d?v??Lhv(?(x), ?(y))? h(f)? m? d(h(?(x)) + h(?(y)))? h(f)= m? d(mh(x) +mh(y))? h(f)? m? d(mh(x, y) +mh(x, y))? h(f)= m? 2dmh(x, y)? h(f)This yields h(x, y) ?12d?h(f)2dm. But, as m ? 2h(f), we have thath(x, y) ?14d, as desired.24Now that we have proven the lemma, we may move to the proof of ourmain theorem for curves defined over K(t).Proof of Theorem 4.2.1. Let f(X, Y ) =?i,jaijX iY j = 0 be the irreduciblepolynomial which defines our curve C over K(t). As f(X, Y ) is irreducible,so is C. In order to make use of Lemma 4.2.2, we will consider f in its reducedform and make sure that at least one of the coefficients aij is non-constant sothat f is not defined over K. By Lemma 4.2.2, for (x0, y0) ? C and m suchthat m ? 2h(f), one of two cases must occur:(1) h(x0, y0) ?14d(2) f(?(x0), ?(y0)) = 0?? (?(x0), ?(y0)) ? CLet us take the first m we find such that m ? 2h(f) and assume that(x0, y0) 6? C(K). Our choice of m now determines our choice of ? correspond-ing to m. Also, denote by C? the zero locus of the curve f?(X, Y ), wheref?(X, Y ) =?i,j?(aij)X iY jClaim 4.2.4. C ? C? is finite.Proof of Claim. Suppose on the contrary that C ? C? is infinite. Then, wemust have that C is one of the irreducible components of C?.So, f? ? Id(C?) ? (f) = Id(C), since C is an irreducible component of C?.This implies that f |f?. Also, f and f? are of the same degree, so we musthave that there exists some ? ? K(t) such that f? = ?f . Now, ? =?(aij)aijfor every i, j. But since f is reduced, by the remark after Definition 4.1.2,gcd ({aij}i,j) = 1 ? K. Hence, there exist bij ? K[t] such that?i,jaijbij = 1.Applying ?, we have that ?(?i,jaijbij) = 1 =?i,j?(aij)?(bij). It follows thatgcd ({?(aij)}i,j) = 1. By this, we must have that ? ? K; otherwise, wehave a contradiction of the aij and ?(aij) being relatively prime. So, since25? =?(aij)aij? K for every i, j, we have that deg aij = deg ?(aij), so thataij ? K. Which then implies that f is defined over K. Contradiction.Similiarly, we define C??1as the zero locus of f??1(X, Y ) =?i,j??1(aij)X iY j.It is worth noting that C??1is not defined over K. If it were, C := ?(C??1)would be defineed over ?(K) = K, a contradiction. We claim that C?C??1isfinite. Suppose not, so that there exist infinitely many points (x, y) ? K(t)such that f(x, y) = 0 = f??1(x, y). Then, by applying ?, we have?(?i,jaijxiyj) = 0 = ?(?i,j??1(aij)xiyj)?i,j?(aij)?(x)i?(y)j = 0 =?i,jaij?(x)i?(y)jSo, there are infinitely many points in K(t) that satisfy f(x, y) = 0 =f?(x, y). This contradicts Claim 4.2.4, proving that C ? C??1is finite.Now, let {(x1, y1), . . . , (xN , yN)} ? C ? C??1, be all the points which arenot defined over K. For each j, we define hj = h(xj, yj). Note that hj > 0since the point is not defined over K. We let c0 = min { 14d ,mini hi}, while ifN = 0, i.e., all points in C?C??1are defined over K, then simply let c0 =14d.For (x0, y0) ? C(K(t)) such that h(x0, y0) < c0, we?ll show (x0, y0) ?C(K). Indeed by Lemma 4.2.2, we have that (?(x0), ?(y0)) ? C. This meansthat (x0, y0) ? C??1and so, (x0, y0) ? C ? C??1. But h(x0, y0) < c0 ? hi foreach i, and therefore (x0, y0) ? C(K).If N = 0, then automatically (x0, y0) ? C ? C??1yields that (x0, y0) ?C(K) since all those points in the intersection are defined over K.Remark: Theorem 1.2.3 requires that the curve is not defined over K, sowe will now see why this assumption is needed. Suppose our curve is definedby the polynomial f(X, Y ) = Y 2 ? X3 ? 1, which is defined over the base26field K, and let m ? 3.Now, suppose that we take (x, y) = (t,?t3 + 1) = p0 ? C(K(t)). Forp1 = (??1(x), ??1(y)), we have thatf(??1(x), ??1(y)) = (??1(?t3 ? 1))2 ? (??1(t))3 ? 1= ??1((?t3 ? 1)2 ? t3 ? 1)= ??1(0)= 0.Thus, p1 ? C(K(t)). Similarly for pn = (??n(x), ??n(y)), we havef(??n(x), ??n(y)) = (??n(?t3 ? 1))2 ? (??n(t))3 ? 1= ??n((?t3 ? 1)2 ? t3 ? 1)= ??n(0)= 0Hence, pn belongs to the curve C. Since ?n(pn) = p0, we may applyCorollary 4.1.7 to see that h(pn) ?(2m)nh(p0). If we let n go to infinity,then h(pn) must go to zero since m ? 3. Hence, for (x, y) ? C and againtaking n to infinity, we have that h(pn) goes to 0. We cannot find a boundc0 on the heights, and this shows that the conclusion of Theorem 1.2.3 canfail if C is defined over K.274.3 Proof of the main theoremAlthough we have been working with curves defined over K(t), we can extendthe result further to any curve defined over K(t). The fact that Theorem1.2.3 is true over K(t) will be a key tool in the extension to K(t).Let L/K be a function field of transcendence degree 1. We will takean element u ? L which is transcendental over K, so that L is finite overK(u). Suppose we take a curve C defined over L, which is not necessarilyirreducible. We define a new curve C ? =???GC? , where G is defined as theset {? ? Aut (L) : ? |K(u) = idK(u)}. From Galois theory, we know that forfinite extensions L/K(u), there exists some inseparable extension of degreepm,m ? 0 such that K(u) ? K(u1/pm) ? L, and L/K(u1/pm) is separable.We take t = u1/pmand obtain the following proposition.Proposition 4.3.1. C ? is defined and irreducible over K(t).Proof. As all elements in G fix K(u), they must fix K(t) as well since anypnth-root of unity in characteristic p is 1. Thus, the only elements of L fixedby all automorphisms in G are all the elements of K(t). So, C ? is defined overK(t) and is fixed by any automorphism of Gal (K(t)sep/K(t)).Now, we wish to show that C ? is irreducible over K(t). Suppose not: thenthere must exist distinct irreducible curves C1 and C2 defined over K(t) andproper inside C ? such that C1, C2 ? C ?. Since we have defined C ? =???GC? , wemust haveC1 =???S1C? and C1 =???S2C?for S1,S2 ? G. Since C1 is defined over K(t), it follows that C1 is fixed byany ? ? G. Then, C1 = C1? =???GC??.But this means that C?? ? C1 for any ? ? S1. If we then vary ? throughevery element in G, it follows that C? ? C1 for every ? ? G. Hence,???GC? =28C ? ? C1. Thus, C1 and C ? are equal; the same follows for C2, proving that C ?is irreducible as claimed.Finally, we show that Theorem 1.2.3 holds for curves defined over K(t).But first, as we will make use of the curve C ? above, we must verify that C ?is not defined over K.Suppose that C ? were defined overK. Let ?? be the automorphism ofK(t)such that it is the identity on K and ??(t) = t+?. Recalling that C ? =???GC? ,we must have that C?? conincides one of the C? , for some ? : L ?? L, fixingK(u). Since there exist infinitely many ? ? K, by the pigeonhole principlethere must be distinct ?, ? ? K such thatC?? = C? = C??This means that C is fixed by ??, which is the automorphism of K(t) suchthat ?? is the identity on K and ????(t) = t + ?, ? = ? ? ? 6= 0. For thecurve C being defined by f(X, Y ) =?i,jaijX iY j, the invariance of C under?? may be observed in the following.?i,j??(aij)X iY j =?i,j??(aij)X iY j???(?i,j??(aij)X iY j) = ???(?i,j??(aij)X iY j)?i,jaijX iY j =?i,j????(aij)X iY jBecause C is fixed by such a map, it must be defined over the fixed fieldfor such an automorphism, which in the case of ??, is K; but C has beentaken to not be defined over K. Contradiction.Now, let C be a plane curve defined over K(t), and C ? as in Proposition4.3.1. By this proposition and Lemma 4.2.2 we know that there exists a realnumber c0 > 0 such that if (x, y) ? C ?(K(t)) and h(x, y) ? c0, then (x, y) ?29C ?(K). So, let (x, y) ? C(K(t)) with h(x, y) ? c0. Since C(K(t)) ? C ?(K(t)),we may conclude that (x, y) ? C ?(K). Thus, we must have that (x, y) ? C(K)as well.30Bibliography[1] P. Aluffi, Algebra: chapter 0, Graduate Studies in Mathematics, vol. 104, AmericanMathematical Society, Providence, RI, 2009. MR2527940 (2010h:00001)[2] H. Derksen and D. Masser, Linear equations over multiplicative groups, recurrences,and mixing I, Proc. Lond. Math. Soc. (3) 104 (2012), no. 5, 1045?1083. MR2928336[3] D. Ghioca, Points of small height on varieties defined over a function field, Canad.Math. Bull. 52 (2009), no. 2, 237?244. MR2512312 (2010e:11061)[4] S. Lang, Algebra, third, Graduate Texts in Mathematics, vol. 211, Springer-Verlag,New York, 2002. MR1878556 (2003e:00003)[5] J.P. Serre, Local fields, Graduate Texts in Mathematics, vol. 67, Springer-Verlag,New York, 1979. Translated from the French by Marvin Jay Greenberg. MR554237(82e:12016)[6] , Lectures on the Mordell-Weil theorem, Third, Aspects of Mathematics, Friedr.Vieweg & Sohn, Braunschweig, 1997. Translated from the French and edited by MartinBrown from notes by Michel Waldschmidt, With a foreword by Brown and Serre.MR1757192 (2000m:11049)31

Cite

Citation Scheme:

        

Citations by CSL (citeproc-js)

Usage Statistics

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.24.1-0103396/manifest

Comment

Related Items