aa r X i v : . [ m a t h . N T ] M a y ELLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELDTOWERS
LISA BERGER Introduction
We study the arithmetic structure of elliptic curves over k ( t ), where k is an algebraically closedfield. In [Shi86] Shioda shows how one may determine rank of the N´eron-Severi group of a Delsartesurface–a surface that may be defined by four monomial terms. To this end, he describes an explicitmethod of computing the Lefschetz number of a Delsarte surface. He proves the universal boundof 56 on the rank of an elliptic curve defined by an equation of the form y = x + at n x + bt m over k ( t ), where k is an algebraically closed field of characteristic zero. In [Shi92] Shioda shows that therank of 68 is obtained for the curve y = x + t + 1 over C ( t ). In recent work, Heinje [Hei11]characterizes all Delsarte elliptic surfaces. He determines 42 families of Delsarte elliptic curves andshows, through explicit computation, that 68 is the maximal rank over k ( t ), k algebraically closedof characteristic zero. By relating a Delsarte surface to a Fermat surface, Shioda is able to exploitthe relationship between divisor classes on his surface and the Mordell-Weil group of its genericfiber. In [Ber08] the author describes a more flexible construction of elliptic surfaces. We explicitlyconstruct families of surfaces, dominated by products of curves, with the additional property thatthey retain this DPC property under base extension. The N´eron Severi group of a product ofcurves may be expressed in terms of divisorial correspondences on the product, and Ulmer [Ulm11]utilizes this relationship to prove an explicit formula for the ranks of the Jacobians of the curvesconstructed in [Ber08]. He produces elliptic curves with rank at least 13 over C ( t ), and Occhipinti[Occ10] produces an elliptic curve over ¯ F p ( t ) whose ranks over the fields ¯ F p ( t /d ) grow at leastlinearly with d prime to p . The goal of this note is to show that the large rank examples obtainedvia our construction are rare. We determine all elliptic curves obtained via the construction in[Ber08], and we find that, for all but finitely many families, the Mordell-Weil group of E/k ( t /d )has rank zero, for each d prime to the characteristic of K = k ( t ), k an algebraically closed field ofarbitrary characteristic.To state the main theorems, we first recall the construction and notation in [Ber08]. Let C and D denote smooth, projective curves over a field k , and let f and g denote separable rational functionsin k ( C ) and k ( D ), respectively. We have a canonically defined rational map: C × k D P k , P [ f ( P ) : g ( P )] , defined away from the locus of points f = g = 0 and f = g = ∞ . A blow-up ofthis locus resolves the map to a morphism from the often singular surface in C × D × P , defined bythe vanishing of tf − g , where t = TS , T and S coordinates on P . Let S denote a smooth, properminimal model of this surface, with generic fiber X f,g , a curve over K = k ( t ). By construction, S is DPCT: it is dominated by a product of curves in towers of non-constant field extensions of theform t t d , d prime to the characteristic of k . That the surface is DPC is clear; it is birationalto C × D . That this property is retained in towers is detailed in [Ber08]. Let m := deg( f ) and n := deg( g ), m i , m ′ i ′ the orders of the zeroes and poles of f , n j , n ′ j ′ the orders of the zeroes andpoles of g . Date : October 21, 2018.2000
Mathematics Subject Classification.
Primary 14G05, 14H52; Secondary 11C08, 14K15.
Theorem. ([Ber08], [Ulm11])
Assume that the orders of zeros and poles of f and g haveno common divisor and that they are relatively prime to the characteristic of K = k ( t ) . Then thegeneric fiber X of a smooth projective model S of the surface defined by the vanishing of tf ( x ) − g ( y ) is an absolutely irreducible curve of geometric genus: g = mg D + ng C + ( m − n − − X ( i,j ) δ ( m i , n j ) − X ( i ′ ,j ′ ) δ ( m ′ i ′ , n ′ j ′ ) , where g D and g C denote the genera of the curves D and C , respectively, and δ ( a, b ) = ( a − b − + (( a,b ) − , and the sums are taken over all pairs ( i, j ) , ( i ′ , j ′ ) . Let C = D = P . Take rational functions f and g with div( f ) = P ki =1 m i a i − P k ′ i ′ =1 m ′ i ′ a ′ i ′ anddiv( g ) = P ℓj =1 n j b j − P ℓ ′ j ′ =1 n ′ j ′ b ′ j ′ , with all a i , b j , a i ′ , and a j ′ ∈ k , and with a i , a i ′ all distinct and b j , b ′ j ′ all distinct. Assume ( m, n ) = 1, and write rm = P m i = P m ′ i ′ and rn = P n j = P n j ′ .Then the generic fiber, X f,g , of the surface constructed above is a bidegree ( rm, rn ) curve birationalto the curve defined by the equation: tf ( x ) − g ( y ) = 0. The main work we present in this noteis an analysis of those partitions of ( rm, rn ), the multiplicities of the zeros and poles of f and g ,for which our construction yields an absolutely irreducible curve with geometric genus one, and weobtain the following:1.2. Theorem. (1)
Let E f,g denote an elliptic curve over k ( t ) , constructed as above: the generic fiber of asmooth, proper model of the surface tf − g ∈ C × D × P . Assume also that rm := deg( f ) ≤ deg( g ) =: rn . Then, for all but finitely many bidegrees ( rm, rn ) , with (m,n)=1, f has ex-actly one zero and one pole. (2) Let K = k ( t ) , k = ¯ k , and let E f,g denote an elliptic curve over K , with defining equationas in the preceding statement: C = D = P , f has exactly one zero and one pole. Let d range over non-negative integers, prime to the characteristic of K . Then the rank of theMordell-Weil group of E/k ( t /d ) is zero. The proof of part one is computational and consists of an analysis of our genus formula, in the caseof genus one. Along the way we give explicit models for the finitely many families of curves thatare not of this form. Part two is a corollary to this classification theorem and to an explicit rankformula in [Ulm11].It is a pleasure to acknowledge the work of Erick Galinkin, a former Stony Brook undergradu-ate, who carried out some initial computations for this project. Thanks are also due to TommyOcchipinti and Doug Ulmer for comments, suggestions and encouragement.2.
Genus one partitions C = D = P , construct the curve defined by tf ( x ) − g ( y ) as above, and continueto assume in what follows that m ≤ n . Set δ := P i,j δ ( m i , n j ), δ ∞ := P i ′ ,j ′ δ ( m ′ i ′ , n ′ j ′ ), and δ := δ + δ ∞ . Our goal is to impose singularities with multiplicities to ensure that the smoothmodel X f,g has geometric genus one. We first explicitly determine the maximum obtainable valuefor δ and for δ ∞ ; we denote by δ max this maximum value, and we show, without loss of generality,that a genus one curve may only be obtained when δ = δ max or when δ = δ max − r . Finally, we de-scribe the defining equations of all families of genus one curves obtained through our construction.Let k and k ′ denote the numbers of zeros and poles of f , ℓ and ℓ ′ the numbers of zeros and poles of g . LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 3
Lemma.
Given positive integers r , m and n and a partition ([ { m i } ] , [ { n j } ]) of the bidegree ( rm, rn ) . The maximum possible value for δ is δ max := r mn − rm − rn + r .Proof. We have δ = k X i =1 ℓ X j =1 δ ( m i , n j ) = X i,j ( m i − n j −
1) + ( m i , n j ) − , and δ = r mn − ℓrm − krn + P i,j ( m i , n j )2 , so, for fixed r , m and n , we find the maximum possible value of D := k X i =1 ℓ X j =1 ( m i , n j ) − ℓrm − krn. When ℓ = k = 1, we have P i,j ( m i , n j ) = ( rm, rn ) = r , and D = r − rm − rn . We show thatno larger value of D may be obtained by increasing ℓ or k , the numbers of parts of our partitions.In what follows we suppose an increase in k . The argument is identical if we instead assume anincrease in ℓ .Re-ordering terms if needed, consider a partition: m ′ k + m ′′ k = m k of m k . We show that, foreach j , ( m k , n j ) + n j ≥ ( m ′ k , n j ) + ( m ′′ k , n j ). First suppose n j | m ′ k . If n j also divides m ′′ k then wehave equality. Otherwise, since ( m ′′ k , n j ) divides m k , the inequality follows. If both ( m ′ k , n j ) and( m ′′ k , n j ) < n j then their sum is bounded by n j , and the strict inequality holds.From this we obtain, P ℓj =1 ( m k , n j ) ≥ P ℓj =1 (( m ′ k , n j ) + ( m ′′ k , n j )) − rn .This yields k X i =1 ℓ X j =1 ( m i , n j ) ≥ k − X i =1 ℓ X j =1 ( m k , n j ) + ℓ X j =1 (( m ′ k , n j ) + ( m ′′ k , n j )) − rn, and k X i =1 ℓ X j =1 ( m i , n j ) − ℓrm − krn ≥ k − X i =1 ℓ X j =1 ( m i , n j ) + ℓ X j =1 (( m ′ k , n j ) + ( m ′′ k , n j )) − ℓrm − ( k + 1) rn. No larger value for D may be obtained by increasing the number of elements in the partitions;the maximum value for D is r − rm − rn , and the maximum value for δ and for δ ∞ is as claimed. (cid:3) rm and rn so that δ = 2 δ max − r . Indeed,letting g a denote the arithmetic genus, we have g a − δ max + r = ( rm − rn − − ( r mn − rm − rn + r ) + r = 1. Assume without loss of generality that δ ≥ δ ∞ . In the remainder of this sectionwe find that a genus one partition is obtained only when δ = δ ∞ = δ max − r and when δ = δ max , δ ∞ = δ max − r . We show that, for all but finitely many bidegrees ( rm, rn ), we require k = k ′ = 1to obtain genus one, and we determine all partitions that yield genus one. LISA BERGER
We have(2.1) δ = r mn − ℓrm − krn − P i,j ( m i , n j )2 , and if we assume δ = δ max − r then we obtain the relation(2.2) ( ℓ − rm + ( k − rn = X ( m i , n j ) ≤ min { ℓrm, krn } . We use the upper bound in 2.2 to prove the following:2.4.
Proposition.
Suppose δ = δ ∞ = δ max − r . Then ( m, n ) = (1 , n ) and r = 2 .Proof. • We assume first that ℓ , k = 1. From the upper bound in 2.2 we obtain ( ℓ − rm ≤ rn and ( k − rn ≤ rm . Combining these yields ( k − ℓ − rm ≤ rm , and this implies that ℓ = k = 2. Making this substitution in 2.2 we have(2.3) rm + rn = ( m , n ) + ( m , n ) + ( m , n ) + ( m , n ) ≤ min { rm, rn } . The upper bound in 2.3 now implies that m = n and, since ( m, n ) = 1, our bidegree is( r, r ). The equality in 2.3 becomes: 2 r = ( m , n ) + ( m , n ) + ( m , n ) + ( m , n ). Since,for j = 1 ,
2, we have P i ( m i , n j ) ≤ r , each sum is exactly r . Hence, ( m i , n j ) = m i = n j , andall summands are equal. When each summand is 1, so that the common divisor is one, weobtain an irreducible (2 ,
2) curve. Several families of (2 ,
2) curves are analyzed in [Ber08],[Occ10] and [Ulm11]. Otherwise, for all i and j , we have ( m i , n j ) = r >
1. Hence, in order toobtain an irreducible curve we now determine the complementary partitions [ { m ′ i ′ } ], [ { n ′ j ′ } ]of ( rm, rn ) = ( r, r ) which yield δ ∞ = δ max − r , satisfying ( m ′ , · · · m ′ i ′ , n ′ , · · · n ′ j ′ , r ) = 1.From the upper bound in 2.2, assuming an ( r, r ) curve, we find that the only possiblepartitions are of the form ℓ ′ = k ′ = 2 and ℓ ′ = 2, k ′ = 1. (Since m = n = 1, we need not con-sider the symmetric case ℓ ′ = 1, k ′ = 2.) In the first case, as above, r = ( m ′ , m ′ , n ′ , n ′ , r ).Hence, our bidegree is (2 , r = ( r, n ′ ) + ( r, n ′ ). It follows,since n ′ + n ′ = r , that ( r, n ′ j ) = n ′ j , for each j . If n ′ = 1 then n ′ = 1, since n ′ | r and r = 1 + n ′ . So we have a (2 ,
2) curve. Otherwise, assume ( n ′ , n ′ ) = 1 but suppose, forsome positive integers k and k , that r = n ′ k = n ′ k = n ′ + n ′ . If r = n ′ n ′ = n ′ + n ′ ,then r = 4, and 2 = ( n ′ , n ′ , r ). Otherwise we must have r >
4, and we have r > n ′ n ′ ,since each n ′ i divides r , and since ( n ′ , n ′ ) = 1. However, for n ′ + n ′ = r , n ′ i = 1, we have n ′ n ′ > r , a contradiction. So we obtain only bidegree (2 ,
2) curves when ℓ , k = 1. • Assuming now that k = 1 for the first partition, again setting δ = δ max − r yields:( ℓ − rm = ( rm, n ) + ( rm, n ) + · · · + ( rm, n ℓ ) ≤ min { ℓrm, rn } . The only possible set of summands is rm + rm + · · · + rm + rm + rm . To ensure thatthe common divisor of the summands is one, we assume rm = 2. Since r is even, r = 2 and m = 1, and we obtain families of (2 , n ) curves. (cid:3) We note that, except for the (2 ,
2) case described above, we have proved that, whenever δ = δ ∞ = δ max − r our genus one (2 , n ) models are determined by partitions of the form:( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 , n ) [2][2 r , · · · r ℓ − , r ℓ − + 1 , r ℓ + 1] , [2][2 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 1 , r ′ ℓ ′ + 1] LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 5
We next show that the only other way to obtain a genus one curve is by imposing singularitiesso that, without loss of generality, δ = δ max .2.5. Proposition.
Suppose a < r and let δ = δ max − a . Then a = 0 , k = 1 , and ( rm, n , · · · , n ℓ , r ) = r Proof.
Substituting δ = δ max − a into equation 2.1 gives:(2.4) ( ℓ − rm + ( k − rn + r − a = X ( m i , n j ) ≤ min { ℓrm, krn } . We first note that either ℓ or k must be equal to one: Since a < r , we have r − a >
0. Hence,if both ℓ and k were greater than one, we would have ( ℓ − rm + ( k − rn + r − a exceeding theupper bound in 2.4.Assuming k = 1 and ℓ ≥ ℓ − rm +( r − a ) = ( rm, n )+( rm, n )+ · · · ( rm, n ℓ ).One possible solution is ( rm, n i ) = rm , for i = 1 , · · · ( ℓ − rm, n ℓ ) = r − a . With thissolution rm divides n i , for i = 1 , · · · , ( ℓ − r divides rn = P ℓi =1 n i , it follows that r divides n ℓ . Since r also divides rm , r | ( r − a ), which is ( rm, n ℓ ). Since r − a is positive, it followsthat a = 0, and δ = δ max . Hence, r divides each element of { rm, n , · · · , n ℓ } We also observe that there is no other set { ( rm, n j ) } satisfying ( rm, n ) + · · · ( rm, n ℓ ) = ( ℓ − rm + r . Indeed, suppose for some j that ( rm, n j ) < rm . We then have ( rm, n j ) ≤ rm . Hence, iftwo or more terms in our sum are each less than rm , we cannot sum to ( ℓ −
1) + r .Finally, since we assume m ≤ n , the equality 2.4 is not satisfied for ℓ = 1, k ≥ (cid:3) rm, rn ) yielding δ ∞ = δ c max := δ max − r . Further,we are only interested in those partitions that satisfy ( m ′ , · · · , m ′ k ′ , n ′ , · · · n ′ ℓ ′ , r ) = 1. Assuming δ ∞ = δ c max , (now writing m i , n j , k and ℓ for m ′ i ′ , n ′ ℓ ′ , k ′ and ℓ ′ ), we have:(2.5) ( ℓ − rm + ( k − rn − r = X ( m i , n j ) ≤ min { ℓrm, krn } . Except for the case where k = 1, there exist finitely many values of ℓ and k that satisfy thisrelation. We will consider each of these cases and determine all corresponding bidegrees. Towardthis end, we have the following:2.7. Proposition.
Suppose δ ∞ = δ c max . Then: (1) k = 1 and ℓ > or (2) k = 2 and ℓ = 2 , , or or (3) k = ℓ = 3 or (4) ℓ = 1 and k = 2 or .Proof. When ℓ = k = 1 we have δ ∞ = δ max , so k = 1 implies ℓ > ℓ = 1 implies k >
1. Wenext show that, for k > ℓ is bounded above by 4. From the upper bound in 2.5 we obtain therelations ( k − n − ≤ m and ( ℓ − m − ≤ n . Combining these we obtain:( ℓ − m − ≤ m + 1 k − , and LISA BERGER ℓ ≤ km ( k −
1) + 1 k − . That ℓ ≤ k = 2and m = 1.Beginning again with the bound in 2.5, we have(2.6) ( ℓ − k − n − ( ℓ − ≤ n + 1 . From this we obtain ( ℓk − ℓ − k ) n ≤ ℓ , and we consider three cases. Case 1 : When ℓk − ℓ − k = 0, we have ℓk = ℓ + k , so ℓ = k = 2. Case 2 : When ℓk − ℓ − k < ℓ ( k − < k . Either k = 1 or ℓ < kk − <
2, so ℓ = 1. Case 3 : Last, take ℓk − ℓ − k >
0. Then, from 2.6, we obtain n ≤ ℓℓk − ℓ − k , so we determine those ℓ and k for which ℓℓk − ℓ − k ≥
1. Setting ℓ ≥ ℓk − ℓ − k we obtain k ≤ ℓℓ − = 2 + ℓ − . From thisinequality it follows that ℓ = k = 3 or k ≤ ℓ = 1 implies k = 2 or k = 3. Substituting ℓ = 1 into 2.5 we have:(2.7) ( k − rn − r = X ( m i , n j ) ≤ min { rm, krn } . From the upper-bound in 2.7 we have ( k − rn − r ≤ rm , so n ≤ m +1 k − . Since we also assume m ≤ n , we have m ≤ n ≤ m +1 k − . From this it follows that k ≤
3. When k = 3 we find that m = n = 1, so we obtain bidegree ( r, r ). This case is identical to the case where ℓ = 3 and k = 1,so we will not consider this case below. (cid:3) We next examine each of the cases (1)-(4) in Proposition 2.7, and we determine all correspondingbidegrees. We show that, except for the cases where k = 1, there are finitely many bidegreessatisfying δ ∞ = δ c max , under the additional assumption that ( m i , · · · , m k , n , · · · n ℓ , r ) = 1. Wecontinue with our assumption that m ≤ n .2.8. Proposition.
Suppose δ ∞ = δ c max that ℓ = k = 2 , and that ( m , m , n , n , r ) = 1 . Then theonly possible bidegrees are: (2 , , (2 , , (3 , , (3 , , (4 , , and (4 , .Proof. When ℓ = k = 2, the formula in 2.5 becomes:(2.8) rm + rn − r = X i,j ( m i , n j ) ≤ min { rm, rn } . The upper bound on the sum in 2.8 limits the possible values for m and n . Indeed, first set rm + rn − r ≤ rm . Then rn − r ≤ rm , so n − ≤ m . Then, taking rm + rn − r ≤ rn , we have m − ≤ n . Combining these, taking m ≤ n , we have m = n or m + 1 = n . LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 7
Case : n = m + 1 . We first assume n = m + 1 in 2.8 and obtain 2 rm = P ( m i , n j ). Toattain this sum we must have, for each j , ( m , n j ) + ( m , n j ) = rm , since rm is an upper bound onthe sum of these two terms. Further, since, for each i and j , ( m i , n j ) is bounded by m i , and since m + m = rm , we conclude for each i and j that ( m i , n j ) is exactly m i . This means that m i | n j for each i and j , so m i divides n + n = rn for each i . We show next that each m i divides n .Since ( m i , r ) divides m i , ( m i , r ) | n j for each i and j , so ( m i , r ) divides each partition summand.Hence we must take ( m i , r ) = 1. Combined with the fact that each m i divides rn , it follows thateach m i divides n = m + 1. When m = m we find that this common value divides rm and rn ,hence it divides r . From this we must assume either m = m , or m = m = 1. If m = m = 1then rm = 2. When m = 2 and r = 1 we obtain bidegree (2 , m = 1 and r = 2 we obtainbidegree (2 , m = m , and since m and m divide n = m + 1, we have rm = m + m < m + 2 ≤ m + 1. From this it follows that m = 1 and r = 3, or r = 2, or r = 1. • When m = 1 and r = 3 we obtain bidegree (3 , • When r = 2 the bidegree is (2 m, m + 2), and we have m + m + 2 = 2 n . Since m | ( m + 1)and m | ( m + 1), and since we assume m = m , we have, without loss of generality, m ≤ ( m + 1) and m ≤ m +12 . This yields m ≤
3. When m = 3 we obtain bidegree (6 , δ ∞ = δ c max leaves a common divisor in the summands. When m = 1we obtain bidegree (2 , m = 2 we have bidegree (4 , • Finally assume r = 1, so we have an ( m, m + 1) curve, and formula 2.8 becomes 2 m = P ( m i , n j ). Since, for each i and j , ( m i , n j ) = m i , m i | ( n + n ). So m i | ( m + 1), which isequivalent to m i | ( m + m + 1). From this we obtain m | ( m + 1). It follows that either m = m = 1 or that m = 1 and m = 2. Since we assume m = m , we are in the lattercase, and we obtain bidegree (3 , Case : m = n = 1 . When m = n = 1 we have an ( r, r ) curve, and P ( m i , n j ) = r .When the two partitions of r are identical, so that m = n and m = n , we have P ( m i , n j ) = m + m + ( m , n ) + ( m , n ) > r , a contradiction. So we assume distinct partitions { [ m i ] } , { [ n j ] } of r , and we may also assume without loss of generality that n < m ≤ m < n . This gives us theinequalities ( m , n ) ≤ n , ( m , n ) ≤ n , ( m , n ) ≤ m , and ( m , n ) ≤ m . We set d := ( m , n )and consider four cases. • Suppose first that d ≥ r . Since m ≤ m , we have m ≥ r . Since m is a multiple of d satisfying m ≥ r , we take m = 2 d ≥ r . Then n ≥ r = r , a contradiction, so d < r . • Suppose next that r < d < r . Since m ≥ r , d = m , and we assume m = 2 d and that n = 3 d . Then m = r − d and n = r − d . We have r < d , so n < r . It followsthat ( n , m ) < r . Then ( n , m ) < r ; this is because ( n , m ) = ( n , m ), and because( n , m ) | n < r . Since r < d = m , we have m < r . Note also that ( m , n ) = m ;otherwise we would have ( m , n ) dividing ( m , m ). So ( m , n ) would divide m , n and n , and hence also m , since m + m = n + n = r . So we would have a common divisor inthe partition. (If the common divisor is one, then r = 2, and the (2 ,
2) curves have alreadybeen considered.) Since ( m , n ) = m , we have ( m , n ) ≤ r . Then, since ( m , n ) < r ,we have P ( m i , n j ) < r + r + r + r = r < r , a contradiction. LISA BERGER • In the case where d = r , we obtain δ c max via an ℓ = 2 = k partition of a (4 ,
4) curve withthe partition [2 , , • Last, take ( m , n ) < r , and suppose that ( m , n ) = n Then ( m , n ) divides m , m ,and n . Hence, it divides r and also n , and there is a common divisor in the partition.We assume the common divisor is one, so ( m , n ) = 1, and each other term is strictly lessthan r . It follows that r <
4, and this is not possible, since we assume here that the parti-tions are distinct. Otherwise, for each i , we have ( m i , n ) < n < r , so each ( m i , n ) is lessthan r . Similarly, we have ( m , n ) strictly less than r , and P ( m i , n j ) < r , a contradiction.Tracing through this proof of Proposition 2 .
8, we obtain the following genus one partitions:( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 ,
3) [2][3], [1,1][2,1](2 ,
3) [2][2,1], [1,1][2,1](2 .
4) [2][4], [1,1][2,2](2 .
4) [2][4], [1,1][3,1](2 .
4) [2][2,2], [1,1][2,2](2 .
4) [2][2,2], [1,1][3,1](3 ,
6) [3][6], [2,1][2,4](3 ,
6) [3][3,3], [2,1][2,4](4 ,
6) [4][6], [3,1][3,3](4 ,
6) [4][4,2], [3,1][3,3](3 ,
4) [3][4], [2,1][2,2](3 ,
4) [3][3,1], [2,1][2,2](4 ,
4) [4][4], [3,1][2,2] (cid:3)
Proposition.
Suppose δ ∞ = δ c max , ℓ = k = 3 and that ( m , m , m , n , n , n , r ) = 1 . Thenthe bidegree is (3 , .Proof. Substituting ℓ = k = 3 into 2.5 we have(2.9) 2 rm + 2 rn − r = X i,j ( m i , n j ) ≤ min { rm, rn } . From the upper bound in 2.9 we obtain the inequalities 2 rn − r ≤ rm and 2 rm − r ≤ rn .Combining these we have n = m = 1, so we obtain an ( r, r ) model and determine that 3 r = P ( m i , n j ). We have, for each i , P j ( m i , n j ) ≤ r and, for each j , P i ( m i , n j ) ≤ r . Hence, since thetotal sum of terms is 3 r , we have equalities: P j ( m i , n j ) = r and P i ( m i , n j ) = r . For any fixed j ,consider the sum P i ( m i , n j ). Each term is bounded by m i , and P i m i = r = P i ( m i , n j ). Hence,each term ( m i , n j ) is exactly m i . So, for each i , ( m i , n j ) = m i . Analogously, for fixed i , for each j , ( m i , n j ) = n j . From this it follows that ( m i , n j ) = m i = n j for all i and j . Since we assume( m , m , m , n , n , n , r ) = 1, we obtain ( m i , n j ) = m i = n j = 1. The bidegree is (3 ,
3) and wehave the partition: ( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](3 ,
3) [3][3], [1,1,1][1,1,1] (cid:3)
LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 9
Proposition.
Suppose δ ∞ = δ c max , ℓ = 3 and k = 2 and that ( m , m , m , n , n , r ) = 1 .Then the bidegrees are: (2 , , (2 , , (3 , , and (4 , .Proof. Substituting ℓ = 3 and k = 2 into 2.5 we have(2.10) 2 rm + rn − r = X ( m i , n j ) ≤ min { rm, rn } From the upper bound in 2.10 we obtain the inequalities rn − r ≤ rm and 2 rm − r ≤ rn .Combining these we find that m ≤ n ≤
3. In particular, m = 2 and n = 3 or m = 1 and n = 1 or m = 1 and n = 2. We analyze each of these cases.2.10.1. Case : m = 2 , n = 3 . Assuming m = 2 and n = 3 in 2.10 we obtain the relation6 r = P ( m i , n j ). Combining this with the fact that, for each i , P j ( m i , n j ) ≤ r , we have theequality P j ( m i , n j ) = 3 r . Analogously, we have, for each j , P i ( m i , n j ) = 2 r . It follows that,for each i and j , ( m i , n j ) = r . Indeed, suppose without loss of generality that the term ( m , n )is less than r . Then ( m , n ) = 2 r − ( m , n ) > r . Since our summands are positive integers,( m , n ) | m implies m > r . Since m + m = 2 r , m < r . Then ( m , n ) , ( m , n ) < r , contradict-ing P j ( m , n j ) = 3 r . Hence each summand is exactly r , so the common divisor of our partition is r . Since we assume ( m , m , m , n , n , r ) = 1, we find that r = 1 and the bidegree is (2 , Case : m = n = 1 . Substituting m = n = 1 into 2.10 we obtain bidegree ( r, r ), and P ( m i , n j ) = 2 r . It follows, for each i and j , that ( m i , n j ) = n j . Since n j divides m and m , n j divides m + m = r . Hence, for each j , n j is an integer of the form ra , and we determine those a , b , c ∈ Z + satisfying ra + rb + rc = r . The only solutions are ( a, b, c ) = (2 , , , ,
3) and(2 , , , r = m + m . From the second, the common divisor of the partitionsummands is 2. From the last triple we obtain bidegree (4 , Case : m = 1 , n = 2 . Assuming m = 1 and n = 2 in 2.10 we obtain 3 r = P ( m i , n j ).Hence, for each j , P i ( m i , n j ) = r . It follows that each ( m i , n j ) = m i . This is because m + m = rm = r and because no divisor can exceed m i . Hence, for each i and j , m i | n j , and it follows thateach m i divides P n j = 2 r . So, we have divisor sums of the form ra + rb = r , where a and b arenon-negative integers. The only solutions are a = 3, b = 6 and a = b = 4. The former correspondsto bidegree (3 ,
6) and the latter to bidegree (2 , rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 ,
3) [2][3], [1,1][1,1,1](2 ,
3) [2][2,1], [1,1][1,1,1](4 ,
4) [4][4], [2,2][2,1,1](2 ,
4) [2][4], [1,1][2,1,1](2 ,
4) [2][2,2], [1,1][2,1,1](3 ,
6) [3][6], [2,1][2,2,2](3 ,
6) [3][3,3], [2,1][2,2,2] (cid:3)
Proposition.
Suppose δ = δ c max , ℓ = 4 and k = 2 , and that ( m , m , n , n , n , n , r ) = 1 .Then the bidegree is (2 , .Proof. Substituting k = 2 and ℓ = 4 into 2.5 we have(2.11) 3 rm + rn − r = X ( m i , n j ) ≤ min { rm, rn } . From the upper bound in 2.11 we obtain rn − r ≤ rm and 3 rm − r ≤ rn . Combining theseinequalities we find that n = 2 or n = 1. When n = 1 we have m < , so there is no correspondingbidegree. Setting n = 2 in 2.11 we obtain m = 1, and the only possible bidegree has the form ( r, r ).Making this substitution in 2.11 yields 4 r = P ( m i n j ). Hence, for each j , ( m , n j ) + ( m , n j ) = r .From this it follows that ( m i , n j ) = m i for each i and j . So, since m i | n j , it follows that m i | P n j ,so m i | r . We determine positive integers a and b with ra + rb = r . The only solutions ( a, b ), up toreordering, are (3 ,
6) and (4 , ,
4) and (3 , ,
4) yields genus one. The partitions are:( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 ,
4) [2][4], [1,1][1,1,1,1](2 ,
4) [2][2,2], [1,1][1,1,1,1] (cid:3)
We next determine those bidegrees corresponding to δ ∞ = δ c max when ℓ = 1 and k = 2. Stillassuming m ≤ n , we prove:2.12. Proposition.
Suppose δ ∞ = δ c max , ℓ = 1 , k = 2 and ( m , m , rn, r ) = 1 . Then, the bidegreesare: (2 , , (2 , , (3 , , (3 , , (4 , , and (5 , .Proof. When k = 2 the relation 2.7 is(2.12) rn − r = ( m , rn ) + ( m , rn ) ≤ min { rm, rn } , so n ≤ m +1. Since the case m = n has already been considered above, and since we assume m ≤ n ,we take n = m + 1. Making this substitution in 2.12 yields rm = ( m , rn ) + ( m , rn ), so m and m each divide rn = r ( m + 1). As in the proofs of the preceding two propositions, we determinepositive integer solutions to: m +1 a + m +1 b = m , and we obtain the triples ( m, a, b ): (1 , , , , , , , , , ,
4) and (5 , , ,
4) and the second bidegree (3 , ,
3) and the fourth to bidegree (4 , , , LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 11 ( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 ,
3) [2][3], [1,1][3](2 ,
3) [2][2,1], [1,1][3](2 ,
4) [2][4], [1,1][4](2 ,
4) [2][2,2], [1,1][4](3 ,
4) [3][4], [1,2][4](3 ,
4) [3][3,1], [1,2][4](3 ,
6) [3][6], [1,2][6](3 ,
6) [3][3,3], [1,2][6](4 ,
6) [4][6], [1,3][6](4 ,
6) [4][4,2], [1,3][6](5 ,
6) [5][6], [2,3][6](5 ,
6) [5][5,1], [2,3][6] (cid:3) k = 1, k ′ >
1. We next determine all those obtained by setting k = k ′ = 1. Still writing ℓ and k in place of ℓ ′ and k ′ , substituting k = 1 into 2.5, we have:(2.13) ( ℓ − rm − r = ( rm, n ) + · · · + ( rm, n ℓ ) . For simplicity, we first consider the case where r = 1, and prove the following:2.14. Proposition.
Suppose δ ∞ = δ c max , k = 1 and r = 1 . Then the bidegrees are (2 , n ) , (3 , n ) , (4 , n ) and (6 , n ) , where n may be any integer satisfying n ≥ m and ( m, n ) = 1 .Proof. Assuming r = 1 in 2.13 we obtain ( ℓ − m − m, n ) + · · · ( m, n ℓ ), and we first notethat all but three of the terms ( m, n j ) must be equal to m . Indeed, supposing there are four termsless than m , we have ( ℓ − m + 4( m ) ≥ ( ℓ − m −
1, so m ≤
1, and our curve X f,g would not havegenus one. Hence, the partition is of the form:( ℓ − m − m + · · · + m + ( m, n ℓ − ) + ( m, n ℓ − ) + ( m, n ℓ ) , so we determine restrictions on the last three terms, and we need(2.14) 2 m − m, n ℓ − ) + ( m, n ℓ − ) + ( m, n ℓ ) . We first assume that each term on the right hand side in 2.14 is less than m . From this weobtain m ≥ m −
1, which implies that m ≤
2. When m = 2 we obtain (2 , n ) curves, 2 ∤ n . Sincewe assume r = 1, we cannot have m = 1, since in this case X f,g would be a rational curve.We next assume the first term is m , leaving m − m, n ℓ − ) + ( m, n ℓ ). One possible sum is m +( m − m − | m . We have implicitly assumed m is even,and the only solutions to the divisibility condition are m = 4, and m = 6. When m = 4 we obtain(4 , n ) curves, n odd. When m = 6 we obtain (6 , n ) curves, n ≡ m + ( m − m ≡ m − | m . These conditions implythat m = 3 or that m = 6. The case m = 6 is identical to, (with partition symmetric to), the m = 6case above. When m = 3 we obtain (3 , n ) curves, n ≥
2. Setting the sum as m + ( m −
1) we obtain m = 4, and this case has also been completed. There are no other partitions mt +( ( t − mt −
1) of m − (cid:3) rm, rn ), curves, under theassumption that r = 1 and k = k ′ = 1:( rm, rn ) [ { m i } ] [ { n j } ], [ { m ′ i ′ } ] [ { n ′ j ′ } ](2 , n ) n ≡ r , · · · r ℓ − , r ℓ + 1] , [2][2 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 1 , r ′ ℓ ′ − + 1 , r ′ ℓ ′ + 1](3 , n ) n ≡ r , r , · · · r ℓ − , r ℓ + 1] , [3][3 r ′ , r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 2 , r ′ ℓ ′ + 2] n ≡ r , r , · · · r ℓ − , r ℓ + 2] , [3][3 r ′ , r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 1 , r ′ ℓ ′ + 1](4 , n ) n ≡ r , r , · · · , r ℓ − , r ℓ + 1] , [4][4 r ′ , r ′ , · · · , r ′ ℓ ′ − + 2 , r ′ ℓ ′ + 3] n ≡ r , r , · · · , r ℓ − , r ℓ + 3] , [4][4 r ′ , r ′ , · · · , r ′ ℓ ′ − + 2 , r ′ ℓ ′ + 1](6 , n ) n ≡ r , r , · · · r ℓ − , r ℓ + 1] , [6][6 r ′ , r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 3 , r ′ ℓ ′ + 4] n ≡ r , r , · · · r ℓ − , r ℓ + 5] , [6][6 r ′ , r ′ , · · · r ′ ℓ ′ − , r ′ ℓ ′ − + 3 , r ′ ℓ ′ + 2]We next assume r >
1, and we have:2.16.
Proposition.
Suppose δ ∞ = δ c max , k = 1 , r > and ( rm, n , · · · n ℓ , r ) = 1 . Then the bidegreeis of the form: (2 , n ) , (3 , n ) , (4 , n ) , (4 , s + 2) , (6 , n ) , (6 , s + 2) , (6 , s + 4) or (6 , s + 3) ,where s is a positive integer . Proof.
Substituting k = 1 into 2.5 gives ( ℓ − rm − r = P ℓj =1 ( rm, n j ), and we first note that wemay not have more than four terms in the sum less than rm . This would give ( ℓ − rm + rm ≥ ( ℓ − rm − r , and we would have m ≤ . • In the case where we have exactly 4 terms less than rm , we have 3 rm − r ≤ rm , whichimplies m ≤
1. Setting m = 1 in 2.5 we have 2 r = P j =1 ( r, n j ). Since we assume( rm, n , · · · n ℓ , r ) = 1, it follows that r = 2. We obtain a family of (2 , n ) models, n odd, with the partitions:(2 , n ) [2][2 r , · · · r ℓ − , r ℓ ] , [2][2 r ′ , · · · , r ′ ℓ ′ − , r ′ ℓ ′ − + 1 , r ′ ℓ ′ − + 1 , r ′ ℓ ′ − + 1 , r ′ ℓ ′ + 1] • When exactly three terms are not equal to rm it is sufficient to consider partitions thatsatisfy: 2 rm − r = ( rm, n ) + ( rm, n ) + ( rm, n ) . Since each term is less than rm we have rm ≥ rm − r . Then 3 m ≥ m −
2, so m ≤ m = 2 we have:3 r = (2 r, n ) + (2 r, n ) + (2 r, n ) , and the only solution,comes from a (2 , n ) curve, r = 1, since ( rm, n , · · · , n ℓ , r ) = r . Thiscase has been completed above. Setting m = 1 we consider partitions that satisfy: r = ( r, n ) + ( r, n ) + ( r, n ) , The only possible partitions are ( r + r + r ), ( r + r + r ) and ( r + r + r ); the only possiblebidegrees are (3 , n ), (4 , n ), and (6 , n ), with the partitions:(3 , n ) [3][3 r , · · · , r ℓ ] , [3][3 r ′ , · · · r ′ ℓ − , r ′ ℓ − + 1 , r ′ ℓ − + 1 , r ′ ℓ + 1](4 , n ) [4][4 r , · · · , r ℓ ] , [4][4 r ′ , · · · r ′ ℓ − , r ′ ℓ − + 2 , r ′ ℓ − + 1 , r ′ ℓ + 1](6 , n ) [6][6 r , · · · , r ℓ ] , [6][6 r ′ , · · · r ′ ℓ − , r ′ ℓ − + 3 , r ′ ℓ − + 2 , r ′ ℓ + 1] LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 13 • Finally, take the case where exactly two terms are less than rm , so we consider partitionsthat satisfy: rm − r = ( rm, n ℓ − ) + ( rm, n ℓ ) , Reasoning as in the proof of the preceding proposition, we obtain, for the partition rm + ( rm − r ), the restriction that m = 3, m = 4, or m = 6. The first possibility, m = 3, isonly possible in the case where r is even. In fact, we obtain new partitions only for m = 3;when m = 4 and m = 6, we have r = 1, which has been considered above. Further, when m = 3 we have the restriction r = 2, and we have bidegrees (6 , s + 2) and (6 , s + 4).For the partition rm + ( rm − r ), we obtain the restriction m = 2, m = 3, or m = 6, andwe may have m = 2 only if r ≡ m = 2 and r = 3, so we have the bidegrees (6 , s + 3).For the partition rm + ( rm − r ) we have m = 2 or m = 4, and m = 2 is possible only if r is even. Further, we obtain new partitions only when m = 2 and r = 2, and we have thebidegrees (4 , s + 2). The partitions are:(6 , s + 2) [6][6 r , · · · r ℓ − , r ℓ + 2] , [6][6 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ − + 3 , r ′ ℓ + 5](6 , s + 4) [6][6 r , · · · r ℓ − , r ℓ + 4] , [6][6 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ − + 3 , r ′ ℓ + 1](6 , s + 3) [6][6 r , · · · r ℓ − , r ℓ + 3] , [6][6 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ − + 2 , r ′ ℓ + 1](4 , n + 2) [4][4 r , · · · r ℓ − , r ℓ + 2] , [4][4 r ′ , · · · r ′ ℓ ′ − , r ′ ℓ − + 1 , r ′ ℓ + 1] (cid:3) , N ), (3 , N ), (4 , N ), (5 ,
6) and (6 , N ). Further, all but finitely many of familiescome from those partitions that satisfy k = k ′ = 1. The table below describes these exceptionalfamilies, those for which the defining function f ( x ) does not have a unique zero and a unique pole.Bidegree Families1 (2 , tx ( x − y + 1)( y − b ) = ( x + 1)( x − a ) y ( y − a , b = 0, 12 (2 , tx ( y − y − a )( y − b )( y − c ) = y ( y − d ) ( x − x + 1) a , b , c = 0, d = 1, a , b , c , tx ( y − y − a )( y − b ) = y ( y − d )( x − x + 1) a , b = 0 d = a , b , 14 (3 , tx ( y − y + 1)( y − a ) = y ( x − x + 1)( x − b ) a , b = 0, 1, −
15 (3 , tx ( y − ( y − a ) = y ( y − b )( x − x + 1) a = 0, bb = 16 (3 , t ( y − a ) ( y − b ) ( y − x = y ( y − d ) ( x − ( x + 1) a , b = 0, d = a , b , 17 (4 , t ( y − ( y + 1)( y − a ) x = y ( x − ( x + 1) a = −
1, 08 (4 , tx ( y − ( y − a ) = y ( y − b ) ( x − ( x + 1) a = 0, bb = 19 (5 , tx ( y − = ( x − ( x + 1) y ( y − a ) a = 1Note that, for each bidegree, a displayed family may admit degeneration and hence correspondto more than one of the partitions we determined in this section.2.18. Other Products
C × D . Above we restrict to the case considered in [Ber08], setting C = D = P . Here we show that there are no other curves C , D for which our construction yields agenus one curve at the base of the tower. When C and D are elliptic curves the genus is: g = rm + rn + ( rm − rn − − X i,j δ ( m i , n j ) − X i ′ ,j ′ δ ( m ′ i , n ′ j ) . Setting g = 1, simplifying, we obtain:( ℓ + ℓ ′ ) rm + ( k + k ′ ) rn = X i,j ( m i , n j ) + X i ′ ,j ′ ( m ′ i , n ′ j ) , where ℓ , ℓ ′ , k and k ′ are defined as above. We have already noted that the sum P ( m i , n j ) + P ( m ′ i , n ′ j ) ≤ min { krn, ℓrm } + min { k ′ rn, ℓ ′ rm } . It follows that we cannot obtain a genus one modelvia our construction in this case, and an analogous argument shows that we cannot consider curves C , D of higher genus. The only other case where a genus one curve could be obtained at the baseof our construction would be for C = P and D = E , an elliptic curve. In that case we obtain therestriction:(2.15) ( ℓ + ℓ ′ ) rm + ( k + k ′ − rn = X i,j ( m i , n j ) + X i ′ ,j ′ ( m ′ i , n ′ j ) . From the upper bounds on each of the sums on the right hand side of equation 2.15 one showsthat k = k ′ = 1, and we have:(2.16) ℓrm = ( rm, n i ) + · · · + ( rm, n ℓ ) and ℓ ′ rm = ( rm, n ′ i ′ ) + · · · + ( rm, n ′ ℓ ′ ) . Each summand is rm , and this is possible only when rm = 1; otherwise we would not have anabsolutely irreducible generic fiber. But when rm = 1 we have a rational curve. We have proved,and now restate, Theorem 1 . Theorem.
Let E f,g denote the elliptic curve over k ( t ) , constructed as above, the generic fiberof a smooth, proper model of the surface tf − g ⊆ C × D × P . Assume deg( f ) = rm ≤ rn = deg( g ) .Then, for all but finitely many bidegrees ( rm, rn ) , with ( m, n ) = 1 , f has exactly one zero and onepole. Bounded Ranks f ( x ) defined with exactly one zero and one pole. Assume as stated in theintroduction that k is an algebraically closed field. In this section we study our ( rm, rn ) genus onecurves over the fields K = k ( t ). The main theorem is: except for the exceptional families in thetable above, all of our elliptic curves E/k ( t /d ) have Mordell-Weil groups with rank zero, for all d prime to the characteristic of K .3.2. For a global field K , by the Mordell-Weil theorem, the group E ( K ) is a finitely generatedabelian group. One may also consider curves over the fields k ( t ) where k is an arbitrary field, andthe group of k ( t ) points of the Jacobian variety J K := J ( X f,g ) need not be finitely generated. Let A denote an abelian variety over the field K . One defines the K/k -trace of A to be an abelian variety B/k with a K -homomorphism τ : B ⊗ k K → A , satisfying the following universal property: If C/k is another abelian variety with homomorphism ψ : C k ⊗ k K → A , then we have a homomorphism τ ′ : C ⊗ k K → B ⊗ k K , and the following commutative diagram: LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 15 C k ⊗ k K τ ′ (cid:15) (cid:15) ψ % % JJJJJJJJJ B k ⊗ k K τ / / A K . That is, B is the largest abelian variety, defined over k , with B × k K mapping to A as above. Inthis work we are interested in the case where A K is the Jacobian variety J K , as above. We definethe Mordell-Weil group M W ( J K ) := J ( K ) /τ ( B ( k )), and by the Lang-N´eron Theorem [LN59], thisquotient is a finitely generated abelian group. See [Con06] for a complete discussion of the K/k trace of an abelian variety over K and the Lang-N´eron Theorem.3.3. Let S d denote the base change t t d of the surface S described in the introduction. In arecent preprint Ulmer shows that the surface S d is a birational model of the quotient ( C d × D d ) /µ d ,where the curves C d , D d are smooth, projective models of: w d = f ( x ) and v d = g ( y ), and where µ d acts via: ( x, w ) ( x, ζ d w ), ( y, v ) ( y, ζ d v ), [Ulm11]. He considers a birational model X d of S d ,and a smooth, proper morphism π d : X d → P , which factors through ( C d × D d ) /µ d , and he uses thegeometry of this construction to determine an explicit formula for the ranks of the Mordell-Weilgroups, as defined above, of the Jacobians of our curves over k ( t ). Key to the rank formula is ourconstruction of an elliptic surface as a birational model of a product variety, so that the N´eron-Severi group of the surface may be expressed in terms of divisorial correspondences on C × D . Therank formula follows from this, combined with the Shioda-Tate formula and a thorough analysis ofthe geometry in our construction. In this section we use Ulmer’s rank formula to bound the ranksof the elliptic curves described in the preceding section. Combined with the classification theorem,we find that the large rank examples in [Ber08], [Occ10] and [Ulm11] are rare: there are finitelymany bidegrees ( rm, rn ) for which our construction yields elliptic curves with non-zero rank overthe fields k ( t (1 /d ) ) when k is algebraically closed.3.4. We briefly discuss Ulmer’s rank formula and refer the reader to [Ulm11] for further details.Let f d,v denote the number of irreducible components in the fiber of π d : X d → P , over the closedpoint v . Define c ( d ) := P v =0 , ∞ ( f d,v − . When k is algebraically closed this becomes c ( d ) = d X v =0 , ∞ ( f ,v − . Let P i and P ′ i ′ denote the zeros and poles of f , Q j , Q ′ j ′ the zeros and poles of g , and let t d,i,j and t ′ d,i ′ ,j ′ denote the numbers of closed points of the surface ( C d × D d ) /µ d over the points ( P i , Q j ) and( P ′ i , Q ′ j ), respectively. Let f ′ d, and f ′ d, ∞ denote the number of irreducible components in the fibersof ( C d × D d ) /µ d P lying over 0 and ∞ , respectively.Define c ( d ) := X i,j t d,i,j + X i,j t ′ d,i ′ ,j ′ − f ′ d, − f ′ d, ∞ + 2 . Note also that, in our construction, the covers C d and D d are often reducible. In the case wherethe base curves C and D are both rational, we let e d,f and e d,g denote the number of irreduciblecomponents of C d and D d , respectively. We write C ′ d and D ′ d for the smooth, proper models of w d/e d,f = ( f ( x )) /e d,f and v d/e d,g = ( g ( y )) /e d,g , respectively. When we take for our constant fieldan algebraic closure of k , we have t d,i,j = gcd( m i , n j , d ) and t ′ d,i ′ ,j ′ = gcd( m ′ i , n ′ j , d ). Our formulabecomes: c ( d ) := X i,j gcd( m i , n j , d )+ X i,j gcd( m ′ i ′ , n ′ j ′ , d ) − X i ( m i , e d,g ) − X j ( n j , e d,f ) − X i ′ ( m ′ i ′ , e d,g ) − X j ′ ( n ′ j ′ , e d,g )+2 . The constant c ( d ) varies with d ; it is clearly periodic, hence bounded. We have the following:3.5. Theorem. ([Ulm11]).
Let C and D denote smooth projective curves over and algebraicallyclosed field k , f ∈ k ( C ) , and g ∈ k ( D ) , separable rational functions. Let X f,g denote a smoothmodel of the curve tf − g , constructed as above. Write J C d and J D d for the Jacobians of the curves C d and D d , and write J = Jac ( X f,g ) and J d = J/k ( t /d ) . With notation as above, the rank over k ( t /d ) of the Mordell-Weil group of the Jacobian of X f,g is:Rank M W ( J d ) = RankHom k − av ( J C ′ d , J D ′ d ) µ d/ ( ed,f · ed,g ) − c ( d ) + c ( d ) , where Hom k − av ( J C ′ d , J D ′ d ) µ d/ ( ed,f · ed,g ) denotes those homomorphisms commuting with the actionof the group µ d/ ( e d,f · e d,g ) .In the remainder of this section we show that, for “most” elliptic curves arising in our construc-tion, the Rank Hom ab − v ( J C , J D ) µ d and c ( d ) terms in Ulmer’s formula are both zero, and that theranks are bounded in towers of function field extensions. To complete the proof of Theorem 1 . k − av ( J C ′ d , J D ′ d ) µ d/ ( ed,f · ed,g ) , writing µ ′ d for µ d/ ( e d,f · e d,g ) Lemma.
With notation as in the statement of the theorem, let X f,g denote the curve over k ( t ) ,constructed as above: the generic fiber of a smooth, proper model of the surface tf − g ∈ C × D × P ,and assume also that f has exactly one zero and one pole. Then RankHom ab-v ( J C ′ d , J D ′ d ) µ ′ d = 0,and the invariant c ( d ) = 0 for all d . Proof.
In this case the curve C ′ d is a smooth, projective model of: w d/e d,f = x rm/e d,f , a rationalcurve. The Jacobian J C ′ d is trivial; RankHom ab-v ( J C ′ d , J D ′ d ) µ ′ d = 0; and there is no contributionfrom this term to the rank of E f,g ( k ( t /d )).To prove the second part of the Lemma, we suppose first that d = 1. Since Rank Hom ab − v ( J C , J D ) µ d =0. The rank formula reduces to Rank( X f,g ( K )) = − c (1) + c (1). But c (1) = ( ℓ − k −
1) +( ℓ ′ − k ′ − k = k ′ = 1, so c (1) = 0. Since Rank( X f,g ( K )) ≥ c (1) = 0, andsince c ( d ) is linear in d , the Lemma follows. (cid:3) Hence, the Mordell-Weil rank of the Jacobian of our curve, in the case where f has exactly onezero and pole, is determined by the function c ( d ). To bound this rank over the fields K d := k ( t /d ),completing the proof of Theorem 1 . Lemma.
Let E f,g denote the elliptic curve over k ( t ) , constructed as above: the generic fiberof a smooth, proper model of the surface tf − g ∈ C × D × P , C = D = P , where f has exactly onezero and one pole. Then the invariant c ( d ) = 0 for all d .Proof. Recall that we consider the curves over k ( t /d ), k = ¯ k . In this case, the invariant c ( d ) is: c ( d ) = X i,j ( m i , n j , d ) + X i ′ ,j ′ ( m i ′ , n j ′ , d ) − ( X j ( d, n j , rm ) + ( d, n , · · · , n j , n ′ , · · · n ′ j ′ , rm )) − ( X j ′ ( d, n ′ j ′ , rm ) + ( d, n ′ , · · · , n ′ j ′ , n , · · · n j , rm )) + 2 , LLIPTIC CURVES WITH BOUNDED RANKS IN FUNCTION FIELD TOWERS 17 and this simplifies to: c ( d ) = X i,j ( m i , n j , d ) + X i ′ ,j ′ ( m i ′ , n j ′ , d ) − X j ( d, n j , rm ) − X j ′ ( d, n ′ j ′ , rm ) . But P i,j ( m i , n j , d ) = P j ( rm, n j , d ), and P i ′ ,j ′ ( m ′ i ′ , n ′ j ′ , d ) = P j ′ ( rm, n ′ j ′ , d ),so the invariant c ( d ) = 0. (cid:3) This proves the second part of Theorem 1 .
2, which we restate here:3.8.
Theorem.
Let K = k ( t ) , k an algebraically closed field, and let E f,g denote an elliptic curveover K , the generic fiber of the surface tf − g ∈ C × D × P , and assume that f has exactly onezero and one pole. Let d range over non-negative integers, prime to the characteristic of K . Thenthe rank of the Mordell-Weil group of E/k ( t /d ) is zero. Remarks E ( k ( t /d )). It is clear that the combinatorial argument could beextended to classify our Jacobians of higher dimension.4.2. In [AZ01], Avanzi and Zannier give a complete classification of genus one curves defined byequations of the form f ( x ) = g ( y ), f ( x ), g ( x ) ∈ K [ x ], where K is a field of characteristic zero,under the assumption that gcd(deg f, deg g ) = 1. In our classification of genus one curves we repeatsome of the results of Avanzi-Zannier, but for our construction we are able to say more. First, weconsider rational functions f ( x ) and g ( x ). Second, we have a stronger irreducibility result for ourcurves, allowing us to remove the assumption that deg f and deg g are relatively prime. References [AZ01] Roberto M. Avanzi and Umberto M. Zannier. Genus one curves defined by separated variable polynomialsand a polynomial Pell equation.
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