BOUNDS OF THE RANK OF THE MORDELLWEIL GROUP OF JACOBIANS OF HYPERELLIPTIC CURVES

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1 BOUNDS OF THE RANK OF THE MORDELLWEIL GROUP OF JACOBIANS OF HYPERELLIPTIC CURVES HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE Abstract. In this article we extend work of Lehmer, Shanks, and Washington on cyclic extensions, and elliptic curves associated to the simplest cubic elds. In particular, we give families of examples of hyperelliptic curves C : y 2 = f(x) dened over Q, with f(x) of degree p, where p is a Sophie Germain prime, such that the rank of the MordellWeil group of the jacobian J/Q of C is bounded by the genus of C and the 2-rank of the class group of the (cyclic) eld dened by f(x), and exhibit examples where this bound is sharp. 1. Introduction Let C/Q be a hyperelliptic curve, given by a model y 2 = f(x), with f(x) Q[x], and let J/Q be the jacobian variety attached to C. The MordellWeil theorem shows that J(Q) is a nitely generated abelian group and, therefore, J(Q) = J(Q) tors Z R J/Q, where J(Q) tors is the (nite) subgroup of torsion elements, and R J/Q = rank Z (J(Q)) 0 is the rank of J(Q). In this article we are interested in bounds of R J/Q in terms of invariants of C or f(x). In [20], Washington showed the following bound for the rank of certain elliptic curves, building on work of Shanks on the so-called simplest cubic elds (see [17]). Theorem 1.1 ([20, Theorem 1]). Let m 0 be an integer such that m 2 + 3m + 9 is square-free and not divisible by 3. Let E m be the elliptic curve given by the Weierstrass equation E m : y 2 = f m (x) = x 3 + mx 2 (m + 3)x + 1. Let L m be the number eld generated by a root of f m (x), let Cl(L m ) be its class group, and let Cl(L m )[2] be the 2-torsion subgroup of Cl(L m ). Then, rank Z (E m (Q)) 1 + dim F2 (Cl(L m )[2]). In this article, we extend Washington's result to curves of genus g 2. In order to nd other families of hyperelliptic curves of genus g 2 where a similar bound applies, we use a method of 2-descent for jacobians described by Cassels, Poonen, Schaefer, and Stoll (see Section 2; in particular, we follow the implementation described in [19]). The examples with genus g = 2 come from a family of cyclic quintic elds described by Lehmer ([10]). Theorem 1.2. Let t Q, let N(t) = t 4 + 5t t t + 25, let f t (x) = x 5 + t 2 x 4 2(t 3 + 3t 2 + 5t + 5)x 3 + (N(t) 4t 2 10t 20)x + (t 3 + 4t t + 10)x + 1, 1

2 2 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE and dene a hyperelliptic curve C : y 2 = f t (x) with jacobian J/Q. Let L/Q be extension dened by adjoining a root of f t (x). Then, there exists an explicit innite set T Q such that if t T, then rank Z (J(Q)) 2 + dim F2 (Cl(L)[2]) + 4 µ 5 (N(t)) where µ 5 (N(t)) is the number of prime divisors v of the numerator of N(t) such that ν v (N(t)) 0 mod 5. The set T of valid values of t mentioned in Theorem 1.2 is composed of x-coordinates of certain multiples of the generator of the MordellWeil group of a xed elliptic curve over Q (see the statement of Theorem 4.1 for more details). For instance, the rst such values of t ordered by height are t = 1, , , and Using Magma ([2]), we have veried that µ 5 (N(t)) = 0, for the rst 6 values of t T ordered by height (see Remark 6.6 in Section 6.2 for further details). It is also worth pointing out that Nakano ([13, Theorem 1]) has shown that Cl(L)[2] is always non-trivial for our choice of t. For higher genus g > 2, we construct hyperelliptic curves y 2 = f(x) such that f(x) denes the maximal real subeld of a cyclotomic extension of Q, and the degree of f(x) is p, a Sophie Germain prime. We obtain the following result. Theorem 1.3. Let q 7 be a prime such that p = (q 1)/2 is also prime, and let L = Q(ζ q ) + be the maximal totally real subeld of Q(ζ q ). Let f(x) be the minimal polynomial of ζ q +ζq 1 or (ζ q +ζq 1 ), let C/Q be the hyperelliptic curve y 2 = f(x), of genus g = (p 1)/2, and let J/Q be its jacobian. Then, there are constants ρ and j, that depend on q, such that rank Z (J(Q)) dim F2 Sel (2) (Q, J) ρ + j + dim F2 (Cl + (L)[2]), where ρ + j p 1. Further, if one of the following conditions is satised, (1) Taussky's Conjecture 2.17 holds, or (2) the order of 2 in (Z/pZ) is even, or (3) the order of 2 in (Z/pZ) is odd and equal to (p 1)/2, or (4) q 92459, then ρ = 0 and j = g = (p 1)/2, and dim F2 Sel (2) (Q, J) g + dim F2 (Cl(L)[2]). The constants ρ and j are, respectively, the dimensions of the kernel of signature map res : O L /(O L )2 {±1} p, and the dimension of the intersection of the image of res with a certain subgroup J (see Section 2.2 for more details). The organization of the paper is as follows. In Section 2, we review the method of 2-descent as implemented by Stoll in [19]. In Sections 2.1, 2.2, and 2.3, we specialize the 2-descent method for the situations we encounter in the rest of the paper, namely the case when f(x) denes a totally real extensions, or cyclic extension of Q, of prime degree. In Section 3, we prove Washington's theorem using the method of 2-descent. In Section 4, we prove Theorem 1.2, and in Section 5 we provide examples of hyperelliptic curves of genus g = (p 1)/2 where p is a Sophie Germain prime, and prove Theorem 1.3. Finally, in Section 6, we illustrate the previous sections with examples of curves, jacobians, and how their ranks compare to the bounds. Acknowledgements. The authors would like to thank Keith Conrad, Franz Lemmermeyer, Paul Pollack, and Barry Smith, for several helpful comments and suggestions.

3 RANK BOUNDS FOR JACOBIANS 3 2. Stoll's Implementation of 2-Descent In this section we summarize the method of 2-descent as implemented by Stoll in [19]. The method was rst described by Cassels [3], and by Schaefer [16], and Poonen-Schaefer [15] in more generality. Throughout the rest of this section we will focus on computing the dimension of the 2-Selmer group of the jacobian J of a hyperelliptic curve C, given by an ane equation of the form C : y 2 = f(x), where f Z[x] is square-free and deg(f) is odd (Stoll also treats the case when deg(f) is even, but we do not need it for our purposes). In this case, the curve C is of genus g = (deg(f) 1)/2 with a single point at innity in the projective closure. Before we can compute the dimension of the 2-Selmer group, we must dene a few objects of interest and examine some of their properties. We will follow the notation laid out in [19]. Let Sel (2) (Q, J) be the 2-Selmer group of J over Q, and let X(Q, J)[2] be the 2-torsion of the Tate-Shafarevich group of J (as dened, for instance, in Section 1 of [19]). Selmer and Sha t in the following fundamental short exact sequence: 0 J(Q)/2J(Q) Sel (2) (Q, J) X(Q, J)[2] 0. With this sequence in hand we get a relationship between for the rank of J(Q) and the F 2 -dimensions of the other groups that we dened. (1) rank Z J(Q) + dim F2 J(Q)[2] + dim F2 X(Q, J)[2] = dim F2 Sel (2) (Q, J). Using equation (1), we get our rst upper bound on the rank (2) rank Z J(Q) dim F2 Sel (2) (Q, J) dim F2 X(Q, J)[2]. This upper bound is computable, in the sense that J(Q)[2] and the Selmer group are computable, as we describe below. Denition 2.1. For any eld extension K of Q, let L K = K[T ]/(f(t )) denote the algebra dened by f and let N K denote the norm map from L K down to K. We denote L K = K[θ], where θ is the image of T under the reduction map K[T ] K[T ]/(f(t )), and L K is a product of nite extensions of K: L K = L K,1 L K,mK, where m K is the number of irreducible factors of f(x) in K[x]. Here, the elds L K,j correspond to the irreducible factors of f(x) in K[x], and the map N K : L K K is just the product of the norms on each component of L K. That is, if α = (α 1, α 2,..., α mk ), then N K (α) = m K i=1 N L K,i /K(α i ) where N LK,i /K : L K,i K is the usual norm map for the extension of elds L K,i /K. In order to ease notation, we establish the following notational conventions: when K = Q we will drop the eld from the subscripts altogether, and if K = Q v, we will just use the subscript v. This convention will apply to anything that has a eld as a subscript throughout the rest of the paper. As an example, L v = Q v [T ]/(f(t )) and L = Q[T ]/(f(t )). Following standard notational conventions, we let O K, I(K), and Cl(K) denote the ring of integers of K, the group of fractional ideals in K, and the ideal class group of K, respectively. We dene

4 4 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE analogous objects for the algebra L K as products of each component, as follows: O LK = O LK,1 O LK,mK, I(L K ) = I(L K,1 ) I(L K,mK ), Cl(L K ) = Cl(L K,1 ) Cl(L K,mK ). Denition 2.2. Let K be a eld extension of Q, and let L = Q[T ]/(f(t )) be as before. (1) Let I v (L) denote the subgroup of I(L) consisting of prime ideals in L with support above a prime v in Z. For a nite set S of places of Q, let I S (L) = I v (L). (2) For any eld extension K of Q, let v S\{ } H K = ker ( N K : L K /(L K )2 K /(K ) 2). For any place v of Q, we let res v : H H v be the canonical restriction map induced by the natural inclusion of elds Q Q v. (3) Let Div 0 (C) denote the group of degree-zero divisors on C with support disjoint from the principal divisor div(y). Remark 2.3. In our case, the curve is given by C : y 2 = f(x), and the support of div(y) is exactly the points with coordinates (α, 0), where α is a root of f, and the unique point at innity. Now for each K, there is a homomorphism F K : div(c)(k) L K, n P P (x(p ) θ) n P, P P and this homomorphism induces a homomorphism δ K : J(K) H K with kernel 2J(K) by [19, Lemma 4.1]. By abusing notation, we also use δ K to denote the induced map J(K)/2J(K) H K. All of these facts, together with some category theory, give us the following characterization of the 2-Selmer group of J over Q. Proposition 2.4 ([19, Prop. 4.2]). The 2-Selmer group of J over Q can be identied as follows: Sel (2) (Q, J) = {ξ H res v (ξ) δ v (J(Q v )) for all v}. In order to take advantage of this description of the Selmer group, we need some additional facts about the 2-torsion of J and the δ K maps. Lemma 2.5 ([19, Lemma 4.3]). Let K be a eld extension of Q. (1) For a point P C(K) not in the support of div(y), δ K (P ) = x(p ) θ mod (L K )2. (2) Let f = f 1 f mk be the factorization of f over K into monic irreducible factors. Then, to every factor f j, we can associate an element P j J(K)[2] such that: (i) The points {P j } generate J(K)[2] and satisfy m K j=1 P j = 0. (ii) Let h j be the polynomial such that f = f j h j. Then (3) dim J(K)[2] = m K 1. δ K (P j ) = ( 1) deg(f j) f j (θ) + ( 1) deg(h j) h j (θ) mod (L K )2.

5 RANK BOUNDS FOR JACOBIANS 5 Denition 2.6. Let I v = ker(n : I v (L)/I v (L) 2 I(Q)/I(Q) 2 ) and let val v : H v I v be the map induced by the valuations on each component of L v. Considering all primes at once, we get another map val: H I(L)/I(L) 2. More specically, the val map is the product of val v (res v ) over all places v. Next, the following lemma helps us compute the dimensions of various groups when K is a local eld or R. Lemma 2.7 ([19, Lemma 4.4]). Let K be a v-adic local eld, and let d K = [K : Q 2 ] if v = 2 and d K = 0 if v is odd. Then (1) dim J(K)/2J(K) = dim J(K)[2] + d K g = m K 1 + d K g. (2) dim H K = 2 dim J(K)/2J(K) = 2(m K 1 + d K g). (3) dim I K = m K 1. With all of this machinery the description of Sel (2) (Q, J) given in Proposition 2.4 can be rened as follows. Proposition 2.8 ([19, Cor. 4.7]). Let S = {, 2} {v : v 2 divides disc(f)}. Then Sel (2) (Q, J) = {ξ H val(ξ) I S (L)/I S (L) 2, res v (ξ) δ v (J(Q v )) for all v S}. This new characterization suggests the following method to compute Sel (2) (Q, J): S1: Find the set S. S2: For each v S, determine J v = δ v (J(Q v )) H v. S3: Find a basis for a suitable nite subgroup H L /(L ) 2 such that Sel (2) (Q, J) H. S4: Compute Sel (2) (Q, J) as the inverse image of v S J v under res v : H H v. v S Ignoring any complications that arise from computing and factoring the discriminant of f, we focus on steps 2 and 3. We omit the details of how to carry out step 4, since we are only interested in an upper bound for the F 2 -dimension of Sel (2) (J, Q). Step 2 can be broken down into three substeps: S2.1: For all v S \ { }, compute J v = δ v (J(Q v )) and its image G v = val v (J v ) in I v. S2.2: If G v = 0 for some v, with v odd, remove v from S. S2.3: Compute J. To complete step 2.3, we need the following lemma. Lemma 2.9 ([19, Lemma 4.8]). With notation as above: (1) dim J(R)/2J(R) = m 1 g. (2) J is generated by the δ (P ) for P C(R). (3) The value of δ (P ) only depends on the connected component of C(R) containing P. Next, for step 3, we see that if we let G = v S\{ } v S G v I(L)/I(L) 2, then the group {ξ H : val(ξ) G} contains Sel (2) (Q, J). In fact, the larger group H = {ξ L /(L ) 2 : val(ξ) G} also contains the 2-Selmer group and we can compute its basis using the following two steps.

6 6 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE S3.1: Find a basis of V = ker(val: L /(L ) 2 I(L)/I(L) 2 ). S3.2: Enlarge this basis to get a basis of H = val 1 (G). With notation as above, Stoll deduces an upper bound and a formula for the F 2 -dimension of the 2-Selmer group (see Lemma 4.10 and the discussion under Step 4), as follows. Proposition 2.10 ([19, Lemma 4.10]). With notation as above, dim F2 Sel (2) (Q, J) (m 1) + dim F2 (Cl(L)[2]) + dim F2 ker ( G Cl(L)/2Cl(L) ). The proof of the bound in Proposition 2.10 is explained at the beginning of the following section, before we specialize to totally real, and cyclic extensions Proof of Proposition The following commutative diagrams helps understand where the Selmer group ts: 0 J(Q)/2J(Q) 0 v J(Q v)/2j(q v ) δ v δv H v resv v H v The 2-Selmer group of J over Q is then given, as in Prop. 2.8, by val I(L)/I(L) 2 v valv v I v(l)/i v (L) 2 Sel (2) (Q, J) = {ξ H : val(ξ) I S (L)/I S (L) 2, res v (ξ) δ v (J(Q v )) for all v S}. The Selmer group is thus contained in H L /(L ) 2, and more precisely, Sel (2) (Q, J) {ξ H : val(ξ) G, res (ξ) J } where J = δ (J(R)), the group G is the product v S\{ } G v I(L)/I(L) 2, and recall that H is the kernel of the norm map from L /(L ) 2 down to Q /(Q ) 2. Thus, Sel (2) (Q, J) is contained in the larger group Ĥ = {ξ L /(L ) 2 : val(ξ) G, res (ξ) J }. We emphasize here that the denition of H in [19] does not impose a condition at, but the denition of Ĥ does to improve the bounds accuracy (thus Ĥ H). In an attempt to simplify notation, let L J be the subspace of L /(L ) 2 with a condition added at innite primes by L J = L /(L ) 2 res 1 (J ). Thus, Ĥ = {ξ L J : val(ξ) G}, and Sel (2) (Q, J) Ĥ. Note that Ĥ is the largest such subgroup L /(L ) 2 with val(ĥ) = G val (L J ). Let us dene subspaces V and W of L /(L ) 2 as follows: Let {ξ i } be generators of G val (L J ), and let W be the subspace generated by {val 1 (ξ i )}. Note that W L J L /(L ) 2. In particular, res (w) J for all w W. Moreover, W and val(w ) are isomorphic by construction, so W = val(w ) = G val (L J ) G val ( L /(L ) 2) = ker(g Cl(L)/2Cl(L)). Thus, dim(w ) dim ker(g Cl(L)/2Cl(L)). Next, let us write V = ker(val: L J I(L)/I(L) 2 ). It follows that Ĥ = V W (note that val(v ) is trivial, while val(w) is non-trivial for every w 0 in W ).

7 RANK BOUNDS FOR JACOBIANS 7 Lemma Let V = ker(val: L J I(L)/I(L) 2 ), let U = ( O L /(O L )2) L J, and let Cl(L J ) be the class group dened as follows Cl(L J ) = I(L)/P (L J ), where I(L) is the group of fractional ideals of L, and P (L J ) is the group of principal fractional ideals A = (α) with a generator such that res (α) J. Then, there is an exact sequence: where Proof. Consider 0 U V Cl(L J )[2] 0, Cl(L J )[2] = {[A] Cl(L J )[2] : A 2 = (α) for some α L J }. L res 1 (J ) 0 I(L) 2 L res 1 (J ) L J 2 I(L) I(L)/I(L) 2 0 and apply the snake lemma. Remark Let P (L) be the subgroup of all principal fractional ideals, let P + (L) be the subgroup of principal ideals generated by a totally positive element, and let P (L J ) be as above. Since the trivial signature (1, 1,..., 1) J, it implies that P + (L) P (L J ) P (L), and therefore there are inclusions Cl(L) Cl(L J ) Cl + (L), where Cl + (L) is the narrow class group of L. Putting all this together (and writing dim for dim F2 ), we obtain a bound dim Sel (2) (Q, J) dim(ĥ) = dim U + dim V + dim W = dim ( O L /(O L )2 res 1 (J ) ) + dim Cl(L J )[2] + dim G val (L J ) dim O L /(±(O L )2 ) + dim Cl + (L)[2] + dim G val ( L /(L ) 2) = (m 1) + dim Cl + (L)[2] + dim ker ( G Cl(L)/2Cl(L) ), where we have used the fact that dim(o L /(O L )2 ) = m, and the fact that res ( 1) is not in J (because J = δ (J(R)) H, which is the kernel of the norm map, so N(j) = 1 for j J, but the norm N( 1) = 1 because the degree of L is odd). We have shown Proposition We will improve on this bound by making certain assumptions about G and a more careful analysis of the dimension of the subgroup of totally positive units. Before we state our renements, we review some of the results on totally positive units that we shall need Totally Positive Units. Let L be a totally real Galois number eld of prime degree p > 2, with embeddings τ i : L R, for i = 1,..., p, and maximal order O L. Let Cl(L) = Cl(O L ) be the ideal class group of L, and let Cl + (L) be the narrow class group. Let V = {±1} p = (F 2 ) p and dene res : L /(L ) 2 V by res (α) = (τ 1 (α), τ 2 (α),..., τ p (α)). Let O L be the unit group of O L, and let O,+ be the subgroup of totally positive units. Thus, ker ( res O L /(O L )2 ) = O,+ L /(O L )2. val L 0

8 8 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE We refer the reader to [5] for heuristics and conjectures about the dimension of the totally positive units (in particular, the conjecture on page 4). In the following theorem, we use the notation of [5]. Theorem Let ρ, ρ +, and ρ be dened by ρ = dim F2 Cl(L)/2Cl(L), ρ + = dim F2 Cl + (L)/2Cl + (L), and ρ = dim F2 O,+ L /(O L )2 Then, (1) ρ = p dim F2 res (O L /(O L )2 ) = dim F2 {±1} p /res (O L /(O L )2 ). (2) We have 0 {±1} p /res (O L /(O L )2 ) Cl + (L) Cl(L) 0. In particular, max{ρ, ρ } ρ + ρ + ρ, and ρ + = ρ + ρ if and only if the exact sequence splits. (3) (Armitage-Fröhlich) ρ + ρ (p 1)/2. Proof. For part (1), note that ρ = dim F2 O,+ L /(O L )2 is the dimension of ker ( res O L /(O L )2 ), and the dimension of O L /(O L )2 is p. Thus, the dimension of the image of res O L /(O L )2 is p minus the dimension of the kernel. For part (2), see Section 2 of [5], and in particular Equation (2.9). Part (3) is shown in [1], where it is shown that ρ + ρ r 1 /2, where r 1 is the number of real embeddings of L. Here r 1 = p is an odd prime, so the proof is concluded. From the statement of the previous theorem, we see that ρ + ρ. However, ρ ρ is not necessarily true. In the following result, a condition is given that implies ρ ρ (see also [8], Section 3). Theorem 2.14 ([14, Corollaire 2c]). Let L/Q be a nite abelian extension with Galois group of odd exponent n, and suppose that 1 is congruent to a power of 2 modulo n. Then, in the notation of Theorem 2.13, we have ρ = ρ +. In particular, ρ ρ. We obtain the following corollary. Corollary Let L/Q be a cyclic extension of odd prime degree p, and suppose that the order of 2 in (Z/pZ) is even. Then, ρ = ρ + and, in particular, ρ = 0. Thus, res O is surjective. L /(O L )2 Proof. Suppose that Gal(L/Q) = Z/pZ for some prime p > 2, such that the order of 2 in (Z/pZ) is even (since Gal(L/Q) is cyclic of order p, this is equivalent to 1 being congruent to a power of 2 modulo p). Hence, Theorem 2.14 applies, and ρ = ρ +. By Theorem 2.13, part (2), we must have that {±1} p /res (O L /(O L )2 ) is trivial (so ρ = 0), which in turn implies that dim F2 res O = L /(O L p. )2 The odd primes below 100 such that the order of 2 is odd modulo p are 7, 23, 31, 47, 71, 73, 79, and 89, so the corollary applies to all other primes not in this list (i.e., 3, 5, 11, 13, 17, 19, etc.). The following result of Kim and Lim gives a specic criterion to check that ρ = ρ + for the maximal real subeld of a cyclotomic eld. Theorem 2.16 ([9, Theorems 3.1 and 3.3]). Let q and p be primes such that q = 2p + 1, and let L = Q(ζ q + ζ 1 q ), where ζ q is a primitive q-th root of unity. Further, assume either one of the following hypotheses:

9 RANK BOUNDS FOR JACOBIANS 9 (1) the order of 2 in (Z/pZ) is p 1, i.e., 2 is a primitive root modulo p, or (2) the order of 2 in (Z/pZ) is odd and equal to (p 1)/2. Then, ρ = dim F2 O,+ L /(O L )2 = 0. There is in fact a conjecture of Taussky that says that ρ = 0 in the case of L = Q(ζ q + ζ 1 q ), where p = (q 1)/2 is a Sophie Germain prime. For more on Taussky's conjecture, see [6], [9], and [18]. Conjecture 2.17 (Taussky's conjecture). Let q and p be primes such that q = 2p + 1, and let L = Q(ζ q + ζq 1 ), where ζ q is a primitive q-th root of unity. Then, ρ = dim F2 O,+ L /(O L )2 = 0. We conclude this section with some remarks about how to compute an upper bound of ρ in the cyclic case, working in coordinates over F 2. Let L be a cyclic extension of Q of degree p > 2, and let Gal(L/Q) = σ. Then, L is totally real (since L is Galois over Q, then it is totally real or totally imaginary, so [L : Q] = p > 2 is odd, and p = 2r 2 is impossible). Let u ±1 be a xed (known) unit in O L and let res (u) = (ε 1,..., ε p ) where τ 1,..., τ p are the real embeddings of L and ε i is the sign of τ i (u) R. We order our embeddings in the following manner. Let g u (x) be the minimal polynomial of u over Q, and let r 1,..., r p be the real roots of g u (x) ordered so that r 1 < r 2 < < r p. Then, {r i } = {τ j (u)}, and we choose τ i so that τ i (u) = r i for all 1 i p. With this notation, res (u) = ( 1, 1, 1,..., 1, 1, 1), i.e., it consists of a non-negative number of 1 signs followed by a non-negative number of +1 signs. Recall that u is in the kernel of res if and only if u is a totally positive unit. Attached to the generator σ G = Gal(L/Q), there is a permutation φ = φ σ S p, where φ is considered here as a permutation of {1, 2,..., n}, such that τ i (σ(u)) = r φ(i), and therefore res (σ(u)) = (ε φ(1),..., ε φ(p) ). Since τ is an embedding (injective), if τ i (α) = r i, then τ i (σ(α)) = r φ(i). It follows that τ i (σ n (u)) = r φ n (i), and res (σ n (u)) = (ε φ n (1),..., ε φ n (p)), for all n 1. Now, in addition, suppose that u is a unit of norm 1, and let G u be the subgroup of units generated by the conjugates of u, i.e., G u = u, σ(u), σ 2 (u),..., σ p 1 (u) O L generate a subgroup of O L. Note that the product p 1 n=0 σn (u) = 1, so Then, G u = u, σ(u), σ 2 (u),..., σ p 2 (u). res (G u) = (ε φ n (1),..., ε φ n (p)) : 0 n p 2 V, where we have dened V = {±1} p. If we x an isomorphism ψ : {±1} = F 2, and write f i = ψ(ε i ), then the map res : G u V can be written in F 2 -coordinates, and the corresponding p (p 1)

10 10 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE matrix over F 2 is given by M,u = ( ) f φ j (i) 1 i p 0 j p 2 f 1 f φ(1) f φ p 2 (1) f 2 f φ(2) f φ = p 2 (2) f p f φ(p) f φ p 2 (p) Lemma Let u be a unit of norm 1, and let d,u be the dimension of the column space of M,u or, equivalently, the dimension of res (G u). Then, ρ (p 1) d,u. In particular, if d,u = p 1, then ρ = 0. Proof. If u is of norm 1, then 1 G u, because the norm of 1 is 1, and the norm of every element in G u is 1. In particular, res ( 1), res (G u) is a space of dimension 1 + d,u. Hence, the kernel of res is at most of dimension p (1 + d,u ). It follows that ρ (p 1) d,u as desired Renements of the bound on the rank. Now we are ready to improve the bound in Proposition We will continue using the notation of Section 2.2. Proposition Let p be an odd prime, let C : y 2 = f(x) with f(x) of degree p (and genus g = (p 1)/2), such that L, the number eld dened by f(x), is totally real of degree p, and let J/Q be the jacobian of C/Q. Let ρ = dim F2 O,+ L /(O L )2, and let j = dim F2 (res (O L /(O L )2 ) J ). Then: In particular, dim Sel (2) (Q, J) j + ρ + dim Cl + (L)[2] + dim G val (L J ), (1) ρ + j p 1. (2) j dim J = (p 1)/2 = genus(c). (3) dim Cl + (L)[2] ρ + dim Cl(L)[2]. In particular, dim Sel (2) (Q, J) j + 2ρ + dim Cl(L)[2] + dim G val (L J ), (4) If G val (L J ) is trivial, then dim Sel (2) (Q, J) j + ρ + dim Cl + (L)[2] j + 2ρ + dim Cl(L)[2]. Proof. Let p be an odd prime, let C : y 2 = f(x) with f(x) of degree p, such that L, the number eld dened by f(x), is totally real of degree p, and let J/Q be the jacobian of C/Q. Recall that in Section 2.1 we showed Clearly, dim Sel (2) (Q, J) dim(ĥ) = dim U + dim V + dim W = dim ( O L /(O L )2 res 1 (J ) ) + dim Cl(L J )[2] + dim G val (L J ). dim ( O L /(O L )2 res 1 (J ) ) = dim res (O L /(O L )2 ) J + dim ker(res O L /(O L )2 ) = j + dim O,+ L /(O L )2 = j + ρ. Now, for part (1), notice that ( O L /(O L )2 res 1 (J ) ) O L /(O L )2 so the dimension as a F 2 - vector space is at most p. Moreover, res ( 1) is not in J (because J = δ (J(R)) H, which

11 RANK BOUNDS FOR JACOBIANS 11 is the kernel of the norm map, so NQ L(j) = 1 for j L such that res (j) J, but NQ L ( 1) = 1 since the degree of L is odd). Thus, j + ρ = dim ( O L /(O L )2 res 1 (J ) ) p 1, as claimed. For part (2), recall that we have dened J = δ (J(R)/2J(R)) H. By Lemma 2.9, and the fact that δ is injective (Lemma 4.1 in [19]), we have dim(j ) = m 1 g = p 1 p 1 = p 1 2 2, where we have used the fact that L is totally real to claim that m = p. Part (3) follows from Theorem 2.13, which shows that ρ + ρ + ρ. And part (4) is immediate from (3), so the proof is complete. Now we can put together Corollary 2.15 and Proposition 2.19 to give a bound in the cases when the multiplicative order of 2 mod p is even. Theorem Suppose L is a cyclic number eld of degree p > 2, such that the order of 2 in (Z/pZ) is even. Then, in the notation of Proposition 2.19, we have dim Sel (2) (Q, J) g + dim Cl(L)[2] + dim ker ( G Cl(L)/2Cl(L) ). Proof. Suppose L is a cyclic number eld of degree p > 2, such that the order of 2 in (Z/pZ) is even. Then, Corollary 2.15 implies that ρ = ρ + and ρ = 0. In particular, res (O L /(O L )2 ) = V, and in particular j = dim J = (p 1)/2 = g(c). Thus, the bound follows from Proposition It follows, as explained after Corollary 2.15, that Theorem 2.20 applies for p = 3, 5, 11, 13, 17, 19, etc. 3. Genus 1 The goal of this section is to show an alternative proof of Theorem 1.1, using Stoll's implementation of the 2-descent algorithm and the results we showed in the previous section. Let m be an integer such that D = m 2 + 3m + 9 is square-free and not divisible by 3. Let C be the (hyper)elliptic curve given by the Weierstrass equation C : y 2 = f m (x) = x 3 + mx 2 (m + 3)x + 1. Since C is elliptic, the jacobian J is isomorphic to C, so we will identify C with J. We will conclude the bound stated in the theorem as a consequence of Theorem Since the order of 2 1 mod 3 is 2, Corollary 2.15 shows that O,+ L = (O L )2 (a fact that is directly shown in [20]). Thus, in order to prove the theorem, it is enough to show that G is trivial. Lemma 3.1. Let m 0 be an integer such that D = m 2 + 3m + 9 is square-free and not divisible by 3. Let v = 2, or let v be a prime divisor of D, and let f m (x) = x 3 + mx 2 (m + 3)x + 1. Then, f m (x) is irreducible as a polynomial in Q v [x]. Proof. Let v = 2. Then, if we consider f m (x) as a polynomial in F 2 [x], we have { x 3 + x if m 1 mod 2, f m (x) = x 3 + x + 1 if m 0 mod 2. In both cases, f m is irreducible over F 2, hence it is irreducible over Q 2.

12 12 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE Now, let v > 2 be a prime divisor of D. Then, by assumption, v > 3 and f m (x m/3) = x 3 D D (2m + 3) x is integral over Z v. Since D = m 2 + 3m + 9, it follows that 4D (2m + 3) 2 = 27 and therefore the greatest common divisor of D and 2m + 3 divides 27. Since by assumption D is not divisible by 3, then gcd(d, 2m + 3) = 1; this together with the fact that D is square-free implies that f m (x m/3) is Eisenstein over Z v. Hence, f m is irreducible over Q v, as claimed. We are now ready to prove Theorem 1.1. Theorem 3.2 ([20, Theorem 1]). Let m 0 be an integer such that m 2 + 3m + 9 is square-free and not divisible by 3. Let E m be the elliptic curve given by the Weierstrass equation E m : y 2 = f m (x) = x 3 + mx 2 (m + 3)x + 1. Let L m be the number eld generated by a root of f m (x), and let Cl(L m ) be its class group. Then, rank Z (E m (Q)) 1 + dim F2 (Cl(L m )[2]). Proof. We shall use Theorem The order of 2 1 mod 3 is 2, even, so it remains to compute G m = v S m\{ } G m,v. The discriminant of f m is D 2 = (m 2 + 3m + 9) 2, so we have S m = {, 2} {v D}. However, by Lemma 3.1, the polynomial f m (x) is irreducible over Q v for any nite prime v S m. It follows that the number of irreducible factors of f m (x) over Q v is 1, and therefore I m,v is zerodimensional by Lemma 2.7. Since G m,v I m,v, we conclude that G m,v is always trivial. Hence, G m is trivial, and Theorem 2.20 implies that dim Sel (2) (Q, J m ) g + dim Cl(L m )[2] + dim ker ( G m Cl(L m )/2Cl(L m ) ) = g + dim Cl(L m )[2], where J m is the jacobian of the elliptic curve E m. Since the genus of E m is 1, then E m = Jm over Q. Moreover, f m (x) is irreducible over Q, and therefore the 2-torsion E(Q)[2] is trivial. Hence, as desired. rank Z (E m (Q)) dim Sel (2) (Q, E m ) 1 + dim Cl(L m )[2], 4. Genus 2 In this section we apply Stoll's implementation of 2-descent as described in Section 2 to prove that the jacobians of a certain family of hyperelliptic curves of genus 2 have Selmer rank bounded only in terms of the class group of the eld of denition of the 2-torsion points. The family of hyperelliptic curves that we discuss here arises from a family of degree 5 cyclic extensions of Q described by Lehmer in [10], in section 5, which we describe next. For every n N, we let f n (x) = x 5 + n 2 x 4 2(n 3 + 3n 2 + 5n + 5)x 3 + (N 4n 2 10n 20)x + (n 3 + 4n n + 10)x + 1 with N(n) = n 4 + 5n n n Then, f n denes a cyclic extension of Q. Further, the discriminant of f n is given by disc(f n ) = h(n) 2 N(n) 4,

13 RANK BOUNDS FOR JACOBIANS 13 where h(n) = n 3 + 5n n + 7. In fact, if t is a rational number such that f t (x) is irreducible, then we still have that f t (x) denes a cyclic extensions of Q of degree 5. Thus, thinking about f t (x) as taking a rational parameter gives us a full 1-parameter family of cyclic extensions of Q. We will consider the hyperelliptic curve C t : y 2 = f t (x) for some well chosen t Q so that disc(f t ) is exactly a rational 4th power. However, since f t (x) is not integral for t Q, we nd an integral model for C t as follows. Let t = a/b with relatively prime integers a, b and consider f a/b (x). The least common denominator of the coecients of f a/b (x) is b 4, therefore the polynomial ( x ) g a,b (x) = f a/b b 4 b 20 = x 5 + a 2 b 2 x 4 + ( 2a 3 b 5 6a 2 b 6 10ab 7 10b 8 )x 3 + (a 4 b 8 + 5a 3 b a 2 b ab b 12 )x 2 + (a 3 b a 2 b ab b 16 )x + b 20 is integral and monic. Moreover, note that C t : y 2 = f t (x) and C a,b : y 2 = g a,b (x) are isomorphic over Q via the map (x, y) (b 4 x, b 10 y). Further, since disc(f t ) = (t 3 + 5t t + 7) 2 N 4, then disc(g a,b (x)) = b 80 disc(f a/b (x)) = b 80 h(a/b) 2 N(a/b) 4. Hence, disc(g a,b ) is a 4th power exactly when t is a rational number such that h(t) = t 3 +5t 2 +10t+7 is a square in Q. In other words, disc(g a,b ) is a 4th power if there is a point P = (t, s) on the elliptic curve E : Y 2 = X 3 + 5X X + 7. The elliptic curve E/Q has Cremona label 368e1, rank 1, and trivial torsion subgroup over Q. Since the rank of E over Q is positive, we know that there are innitely many specializations of t = a/b Q such that the discriminant of g a,b (x) is a 4th power, as desired. With this choice of t values, our goal is to prove the following theorem. Theorem 4.1. Let E : Y 2 = X 3 + 5X X + 7, let P = ( 1, 1) E(Q) be the generator of the Mordell-Weil group of E, let n 1 be an integer congruent to 1 or 5 mod 6, and let t = t n = x(np ) be the x-coordinate of np. Let f t (x) be as above and let L t /Q be the cyclic degree 5 Galois extension dened by adjoining a root of f t (x). Let C t be the hyperelliptic curve cut out by the equation y 2 = f t (x) with jacobian J t (equivalently, C t is dened by y 2 = g a,b (x) where t = a/b in lowest terms). Then, rank Z (J t ) 2 + dim F2 Cl(L t )[2] + 4 µ 5 (N(t)) where µ 5 (N(t)) is the number of prime divisors v of the numerator of N(t) such that ν v (N(t)) 0 mod 5. To prove the theorem, we apply the process of 2-descent described in Section 2 for C t, but instead we use the isomorphic model C a,b : y 2 = g a,b (x) since g a,b has been constructed specically to be monic and integral so that [19] applies. We will only need to carry out steps 1 and 2 of the descent in order to compute G, and then we will apply Theorem 2.20, to conclude the theorem. The proof will be put together in Section 4.3. Remark 4.2. As one would expect, it is rare to encounter a value of t such that µ 5 (N(t)) > 0. In fact, if n 1 or 5 mod 6 with 1 n 17, and t = x(np ) as in statement of Theorem 4.1, then a Magma computation shows that µ 5 (N(t)) = 0. Note, however, in general there are values of t such that N(t) has a prime factor p such that ν p (N(t)) 0 mod 5. For instance, N(7721) = , so µ 5 (N(7721)) = 1, but 7721 is not the x-coordinate of a point np as above. Remark 4.3. Since g a,b (x) = f a/b (x/b 4 )b 20, both polynomials f t and g a,b dene the same extension L t /Q. We will continue to denote by L t,p the algebra over Q p as in Stoll notation, that is dened

14 14 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE by either of the polynomials f t or g a,b. Notice that L t /Q is a cyclic Galois extension of degree 5, so L t,p is either an extension of degree 5 of Q p, or the product of ve copies of Q p Step 1. Recall that we have computed where now and disc(g a,b (x)) = b 80 disc(f a/b (x)) = b 80 h(a/b) 2 N(a/b) 4, = b 58 (b 3 h(a/b)) 2 (b 4 N(a/b)) 4, H(a, b) = b 3 h(a/b) = a 3 + 5a 2 b + 10ab 2 + 7b 3 N (a, b) = b 4 N(a/b) = a 4 + 5a 3 b + 15a 2 b ab b 4 are integers relatively prime to b, since gcd(a, b) = 1. Thus, S t = {, 2} { v prime : v disc(g a,b ) } = {, 2, v b} {v H(a, b)} {v N (a, b)}, and continue on to step Step 2. The goal of this step is to compute the group G t = v S t\{ } G t,v I(L t )/I(L t ) 2 and, in fact, prove that the component G t,v is trivial for most v S t \ { }. Recall that m t,v is the number of factors in the factorization of g a,b (x) over Q v. In particular, if m t,v = 1, then I t,v is trivial by Lemma 2.7, and therefore G t,v = val p (J t,v ) I t,v must be trivial as well. Since S t = {, 2, v b} {v H(a, b)} {v N (a, b)}, we will consider several cases according to what factor of the discriminant v divides (and we will treat v = 2 and 5 separately) Case of v = 2. Recall that we have chosen t to be the x-coordinate of a point Q = np, where P = ( 1, 1) is a generator of E(Q) = Z, and n 1 or 5 mod 6. Since ( 1, 1) (1, 1) mod 2 is non-singular on Ẽ/F 2, and Ẽns (F 2 ), the group of non-singular points over F 2, contains two points ((1, 1) and O), it follows that if n is odd, then x(np ) 1 mod 2, and in particular, the denominator of x(np ) is not even. Thus, if t = x(np ) = a/b written as a fraction in reduced terms, then b is odd. If we reduce g a,b (x) = x 5 + a 2 b 2 x 4 + ( 2a 3 b 5 6a 2 b 6 10ab 7 10b 8 )x 3 + (a 4 b 8 + 5a 3 b a 2 b 10 modulo 2, then we obtain either + 15ab b 12 )x 2 + (a 3 b a 2 b ab b 16 )x + b 20 g a,b (x) { x 5 + x if a 0 mod 2, x 5 + x 4 + x 2 + x + 1 if a 1 mod 2. Since both polynomials are irreducible over F 2 and g a,b (x) is integral and monic, it follows that g a,b (x) is irreducible over Q 2. Hence m t,2 = 1, and I 2 is trivial. Since G t,2 I 2, we conclude that G t,2 is trivial as well.

15 Case of v = 5. If we reduce RANK BOUNDS FOR JACOBIANS 15 g a,b (x) = x 5 + a 2 b 2 x 4 + ( 2a 3 b 5 6a 2 b 6 10ab 7 10b 8 )x 3 + (a 4 b 8 + 5a 3 b a 2 b ab b 12 )x 2 + (a 3 b a 2 b ab b 16 )x + b 20 modulo 5, for all possible values of a and b mod 5, we nd that g a,b (x) is irreducible, unless (a, b) (1, 3), (2, 1), (3, 4), or (4, 2) mod 5. However, notice that t = x(np ) is an x-coordinate of a rational point on an elliptic curve in Weierstrass form Y 2 = h(x), with h cubic and monic, therefore the denominator b of t must be a perfect square, and thus congruent to 0, 1, or 4 mod 5. If (a, b) (2, 1) or (3, 4) mod 5, then x(np ) = t = a/b 2 mod 5. Now, P (4, 4) mod 5 generates Ẽ(F 5), which is cyclic of order 6, and x(np ) 2 mod 5 if and only if n 3 mod 6. Since n was chosen to be 1 or 5 mod 6, it follows that x(np ) 2 mod 5, and therefore (a, b) (2, 1) nor (3, 4) mod 5, and it follows that g a,b (x) is irreducible over F 5 and Q 5. Hence, m t,5 = 1 and I 5 is trivial, which implies that G t,5 is trivial as well. Remark 4.4. One can go into more detail and show that, in fact, the pair (a, b) (3, 4) mod 5 does not happen for any x-coordinate x(np ) = a/b, but we do not need this here Primes that divide N (a, b). First suppose that v is an odd prime, dividing disc(g a,b ), and such that v is inert or ramied in L t /Q (note that if v ramies, then it is totally ramied because L t /Q is Galois of prime degree). In this case, the v-adic extension L t,v /Q v is unramied or totally ramied, respectively, of degree 5, and therefore g a,b (x) must be irreducible over Q v, because g a,b (x) and f t (x) both dene L t,v. In particular, m t,v = 1 and therefore G t,v is trivial as explained in the rst paragraph of Step 2. We need the following lemma. Lemma 4.5. Let v 2, 5 be a prime that divides N = N (a, b), such that ν v (N ) is not a multiple of 5. Then, v ramies in L t /Q, where t = a/b. Proof. Let v 2, 5 be a prime that divides N (a, b). Since N (a, b) and b are relatively prime (as noted above in Step 1), it follows that ν v (N(a/b)) > 0. Considering t Q v, the polynomial f t (x) (and also g a,b (x)) describes the local algebra L t,v over Q v, so it suces to show that the roots of f t (x) have fractional valuation to prove that L t,v /Q v denes a totally ramied extension of degree 5, and therefore v ramies in L t /Q. Let us dene f t (x) = f t (x t 2 /5). It follows that f t and f t dened the same algebra L t,v, and f t is still monic, and integral over Q v (recall that ν v (b) = 0 and v 5). Then, f t (x) is given by f t (x) = x 5 + a 3 (t)x 3 + a 2 (t)x 2 + a 1 (t)x + a 0 (t), where the a i (t) are polynomials in t that factor as follows: and a 3 = N(t), a 2 = N(t) (t 2 + 5/2t + 5/4), a 1 = N(t) (t 4 + 5t /3t 2 50/3) ( a 0 = N(t) t t t4 50t t We claim that the v-adic valuation of b 0 (t) = a 0 (t)/n(t) is 0. Indeed, Euclid's algorithm yields polynomials ). x(t) = t t t t t

16 16 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE and such that y(t) = t t t , x(t) N(t) + y(t) b 0 (t) = 1. Since v 2, 5, we have x(t), y(t), N(t), and b 0 (t) Z v, and ν v (N) > 0 implies that ν v (b 0 ) = 0, as claimed. Now, let m = ν v (N) > 0 and recall that we are assuming that m 0 mod 5. Then, the Newton polygon of f t (x) = f t (x) a 0 (t) = 1 + a 1(t) a 0 (t) x a 0 (t) x5 is given by a single line from (0, 0) to (5, m). Thus, f t (x) (and therefore f t (x)), has 5 roots of valuation m/5 which is not an integer by our hypothesis. We conclude that the extension L t,v over Q v dened by f t (x) (and also by f t (x)) is totally ramied of degree 5, as we needed. In particular, by Lemma 4.5, if v divides N (a, b), and the valuation is not a multiple of 5, then G t,v is trivial. Otherwise, for each of those primes v such that the valuation of N at v is a positive multiple of 5, the group G t,v may be at most (m t,v 1)-dimensional (where m t,v 5), and therefore they may contribute up to 4µ 5 (N (a, b)) to the dimension of the kernel of G Cl(L)/2Cl(L) Primes that divide H(a, b). Now suppose that v is an odd prime divisor of H(a, b). In this case, we shall see that J v = δ v (J(Q v )) is non-trivial, but G v = val v (J v ) is trivial. We shall use Lemma 2.5 to describe J v and the denition of the val v map to show that G v vanishes. Note that dim J(Q v )/2J(Q v ) = dim J(Q v )[2] by Lemma 2.7 (because v 2). Thus, δ v (J(Q v )), is generated by δ v (P j ) where P 1,..., P mt,v 1 are generators of the 2-torsion J(Q v )[2]. The values δ v (P j ) are given by Lemma 2.5, and in order to compute these values, we need a bit more information about the roots of f t (x) over Q v. If v is inert or ramied in L t /Q, then g a,b (x) is irreducible over Q v, and so m t,v = 1 and G v is trivial. Thus, we shall assume that v splits completely in L t /Q and therefore g a,b (x) splits completely in Q v [x]. Recall that b and H(a, b) = b 3 h(a/b) = a 3 +5a 2 b+10ab 2 +7b 3 are relatively prime, because gcd(a, b) = 1. Since we are assuming that v divides H(a, b), it follows that v does not divide b and v divides h(a/b). Further, g a,b (x) = f t (x/b 4 )b 20, hence the roots {α i } 5 i=1 of g a,b and the roots {β i } 5 i=1 of f t over Q v dier by a unit (namely α i = b 4 β i for all 1 i 5), so we will work with f t (x) instead, where t = a/b Z v. If we consider f t (x) as a polynomial in (Z[t])[x] for the time being, and reduce the coecients modulo h(t), then f t (x) factors as follows over (Z[t]/(h(t))[x]: f t (x) (x (t 1)) (x (t 2)) (x (t 2 + 4t + 5)) (x + (t 2 + 3t + 4)) 2 mod h(t)z[x]. Therefore, if β 1,..., β 5 are the roots of f t (x) in Q v, then there are two roots, say β 4 and β 5, such that β 4 β 5 (t 2 + 3t + 4) mod v, because ν v (h(t)) = ν v (H(a, b)) > 0. Further, we note that (t 3 + 5t t + 18)h(t) (t 2 + 5t + 5)N(t) = 1 is an identity in Z[t]. It follows that if ν v (h(t)) > 0, then ν v (N(t)) = 0. Recall that we have chosen t Q such that h(t) is a square. In other words, t is the X-coordinate of a rational point on the elliptic curve Y 2 = h(x). Thus, if ν v (h(t)) > 0, then the valuation is even.

17 RANK BOUNDS FOR JACOBIANS 17 Suppose ν v (h(t)) = 2k > 0. Then, the v-adic valuation of disc(f t (x)) is given by ν v (disc(f t (x))) = ν v (h(t) 2 N(t) 2 ) = ν v (h(t) 2 ) + ν v (N(t) 2 ) = 2ν v (h(t)) + 2ν v (N(t)) = 4k, since ν v (h(t)) = 2k and ν v (N(t)) = 0. Since β 4 and β 5 are congruent roots modulo h(t)z v, it follows that ν v (β 4 β 5 ) = 2k also. Moreover, no other roots can be congruent modulo v, because if another pair was congruent, say β i β j mod v, with {i, j} {4, 5}, then ν v (disc(f t (x))) = ν v (β i β j ) 2 4k + 2, 1 i<j 5 but we saw the valuation is 4k. Thus, the only roots of f t (x) that are congruent modulo v are β 4 and β 5. Since α i = b 4 β i, we conclude that the only roots of g a,b (x) that are congruent modulo v are α 4 and α 5, and ν v (α 4 α 5 ) = ν v (β 4 β 5 ) = 2k is even. Now we are ready to resume our computation of δ v (J t (Q v )). Let P j = (α j, 0), for 1 j 5 be the 2-torsion points on J(Q v ), and let L t,v = L t,v,1 L t,v,5 = (Qv ) 5 where L t,v,j is the copy of Q v corresponding to the factor (x α j ) of f t (x). Recall that we had set up notation L t,v = Q v [θ], where θ is the image of T under the reduction map Q v [T ] Q v [T ]/(f t (T )). In particular, θ = α i in L t,v,i. In the notation of Lemma 2.5, we have g a,b (x) = f 1 (x)h 1 (x) with f 1 (x) = x α 1 and h 1 (x) = (x α 2 ) (x α 5 ), and Thus, the image of P i via δ v is given by δ K (P j ) = ( 1) deg(f j) f j (θ) + ( 1) deg(h j) h j (θ) mod (L K )2. δ v (P 1 ) = ((α 1 α 2 )(α 1 α 3 )(α 1 α 4 )(α 1 α 5 ), α 1 α 2, α 1 α 3, α 1 α 4, α 1 α 5 ), δ v (P 2 ) = (α 2 α 1, (α 2 α 1 )(α 2 α 3 )(α 2 α 4 )(α 2 α 5 ), α 2 α 3, α 2 α 4, α 2 α 5 ),. δ v (P 5 ) = (α 5 α 1, α 5 α 2, α 5 α 3, α 5 α 4, (α 5 α 1 )(α 5 α 2 )(α 5 α 3 )(α 5 α 4 )). Since G t,v = val p (J v ), and J v = δ v (J t (Q v )) is generated by the values δ v (P i ), we compute val v (δ v (P i )). Recall that I v = ker(n : I v (L t )/I v (L t ) 2 I(Q)/I(Q) 2 ) and val v : H v I v is the map induced by the valuations on each component of L t,v. Moreover, by our work above, the only roots that are congruent modulo v are α 4 and α 5, and ν v (α 4 α 5 ) = 2k. Thus, val p v(δ v (P 1 )) = val v ((α 1 α 2 )(α 1 α 3 )(α 1 α 4 )(α 1 α 5 ), α 1 α 2, α 1 α 3, α 1 α 4, α 1 α 5 ) = ((1), (1), (1), (1), (1)), val v (δ v (P 2 )) = val v (α 2 α 1, (α 2 α 1 )(α 2 α 3 )(α 2 α 4 )(α 2 α 5 ), α 2 α 3, α 2 α 4, α 2 α 5 ) = ((1), (1), (1), (1), (1)), val v (δ v (P 3 )) = ((1), (1), (1), (1), (1)), val v (δ v (P 4 )) = val v (α 4 α 1, α 4 α 2, α 4 α 3, (α 4 α 1 )(α 4 α 2 )(α 4 α 3 )(α 4 α 5 ), α 4 α 5 ) = ((1), (1), (1), (p) 2k, (p) 2k ) ((1), (1), (1), (1), (1)) mod I(Q) 2, val p (δ p (P 5 )) = val p (α 5 α 1, α 5 α 2, α 5 α 3, α 5 α 4, (α 5 α 1 )(α 5 α 2 )(α 5 α 3 )(α 5 α 4 )) = ((1), (1), (1), (p) 2k, (p) 2k ) ((1), (1), (1), (1), (1)) mod I(Q) 2. Hence, G t,v = val v (J v ) must be trivial.

18 18 HARRIS B. DANIELS, ÁLVARO LOZANO-ROBLEDO, AND ERIK WALLACE Primes that divide b. Suppose that v is an odd divisor of b, where t = a/b, and recall that we have chosen t to be the X-coordinate of a rational point (X 0, Y 0 ) E(Q) on the elliptic curve E/Q : Y 2 = h(x). The denominator of t = X 0 is therefore a perfect square (h(x) is a monic cubic polynomial), and it follows that ν v (b) = 2k > 0 for some k 1. The polynomial g a,b (x) is given by ( x ) g a,b (x) = f a/b b 4 b 20 = x 5 + a 2 b 2 x 4 + ( 2a 3 b 5 6a 2 b 6 10ab 7 10b 8 )x 3 + (a 4 b 8 + 5a 3 b a 2 b ab b 12 )x 2 + (a 3 b a 2 b ab b 16 )x + b 20. We dene g a,b (x) = b 20 g a,b (x) Q v [x]. Then, g a,b (x) = 1 + b 7 (a 3 + 4a 2 b + 10ab b 3 )x + b 12 (a 4 + 5a 3 b + 11a 2 b ab 3 + 5b 4 )x 2 + b 15 ( 2a 3 6a 2 b 10ab 2 10b 3 )x 3 + a 2 b 18 x 4 + b 20 x 5. Since a and b are relatively prime integers, and taking into account that the valuation of b is 2k, it follows that the Newton polygon of ga,b (x) has vertices (0, 0), (1, 14k), (2, 24k), (3, 30k), (4, 36k), (5, 40k), and slopes 14k, 10k, 6k (length of segment is 2), and 4k. Therefore, there are single roots of valuation ν v (α 1 ) = 4k, ν v (α 2 ) = 10k, and ν v (α 3 ) = 14k, and two roots of valuation ν v (α 4 ) = ν v (α 5 ) = 6k. In order to investigate the valuation of α 4 α 5 further, let us put α j = b 3 α j and g (x) = g a,b (b3 x) so that α j for j = 4, 5 are roots of g of valuation 0. Thus, g (x) = 1 + b 4 (a 3 + 4a 2 b + 10ab b 3 )x + b 6 (a 4 + 5a 3 b + 11a 2 b ab 3 + 5b 4 )x 2 + b 6 ( 2a 3 6a 2 b 10ab 2 10b 3 )x 3 + a 2 b 6 x 4 + b 5 x 5. The vertices are now (0, 0), (1, 4), (2, 6), (3, 6), (4, 6), (5, 5) so, indeed, there exactly two roots of zero valuation, namely α 4 and α 5. By considering b6 g (x), we see that b 6 g (x) = (a 4 + 5a 3 b + 11a 2 b ab 3 + 5b 4 )x 2 + ( 2a 3 6a 2 b 10ab 2 10b 3 )x 3 + a 2 x 4 + b (b 5 + b(a 3 + 4a 2 b + 10ab b 3 )x + x 5 ), and since g (α j ) = 0, we discover that α j is a root of x 2 a 2 (x a) 2 + bĝ (x) = 0, for some polynomial ĝ (x) Z v [x]. Therefore α j a mod vk for j = 4, 5. In fact, let us now consider α j = b 3 α j = b3 (a + α j b) and g (x) = g (a + bx). Then, b 4 g (x) is of the form a 4 (2 3x + x 2 ) + bĝ (x), for some polynomial ĝ (x) Z v [x]. It follows that α j 1 or 2 mod v2k. Hence, we conclude that α j b 3 (a + b(1 + κ 1 b)) or b 3 (a + b(2 + κ 2 b), for some κ i Z v. Thus, α 5 α 4 = b 4 (1 + (κ 2 κ 1 )b) is of valuation ν v (α 5 α 4 ) = ν v (b 4 ) = 8k, even. Hence, the valuation of all dierences α i α j for 1 i < j 5 is even.

19 RANK BOUNDS FOR JACOBIANS 19 Therefore, we can compute val v (δ v (P i )) as in Section and conclude that these values are trivial in I v, for all 1 i 5, because we are working modulo squares of ideals. Hence, G t,v is trivial Proof of Theorem 4.1. By our work on Steps 1 and 2 above, we have shown that G t,v is trivial for all primes v 2, except, perhaps, for those primes v such that v 5 divides N (a, b). If G t,v is non-trivial, then its dimension is at most that of I v, which is m t,v 1 4. Therefore, the dimension of G = v S t\{ } G t,v is at most 4 µ 5 (N (a, b)), where µ 5 (m) is the number of distinct prime divisors of the numerator of m Q such that ν v (m) 0 mod 5. Since v cannot divide N (a, b) and v simultaneously, and N (a, b) = b 4 N(t), then µ 5 (N (a, b)) = µ 5 (N(t)). In particular, dim(ker(g Cl(L t )/2Cl(L t ))) dim G 4 µ 5 (N(t)). Since the extension L/Q is cyclic of degree 5, and the order of 2 mod 5 is 4, even, we can apply Theorem 2.20 to show dim Sel (2) (Q, J) g + dim Cl(L)[2] + dim ker ( G Cl(L)/2Cl(L) ) g + dim Cl(L)[2] + 4 µ 5 (N(t)), as claimed. This concludes the proof of Theorem Genus g = (p 1)/2, where p is a Sophie Germain prime The goal of this section is to nd examples of hyperelliptic curves of genus g > 2 where the dimension of the Selmer group can be bounded in terms of a class group, as in Theorem 1.1 (genus 1 case) or Theorem 4.1 (genus 2 case). We begin by looking at polynomials f(x) that cut out extensions of degree p, contained inside a q-th cyclotomic extension, where q is another prime. Theorem 5.1. Let q > 2 be a prime such that the multiplicative order of 2 mod q is either q 1 or (q 1)/2, and let p > 2 be a prime dividing q 1. Let Q(ζ q ) be the q-th cyclotomic eld, and let L be the unique extension of degree p contained in Q(ζ q ). Further, suppose that O L = Z[α] for some algebraic integer α L, let f(x) be the minimal polynomial of α, and let J/Q be the jacobian variety associated to the hyperelliptic curve C : y 2 = f(x). Then, dim Sel (2) (Q, J) ρ + g + dim Cl + (L)[2]. If in addition the multiplicative order of 2 mod p is even, then dim Sel (2) (Q, J) g + dim Cl(L)[2]. Proof. We apply Stoll's algorithm to y 2 = f(x), in order to compute G. Note that f(x) is monic, integral, and irreducible over Q, since α generates O L as a ring. Moreover, since O L = Z[α], it also follows that disc(f(x)) = disc(o L ), and since L Q(ζ q ), the only prime dividing disc(o L ) is q. Hence, the set S = {, 2, q}. We will show that G 2 and G q are trivial, and therefore G is trivial as well. Indeed: Let v = 2. Since the order of 2 modulo q is q 1 or (q 1)/2 by hypothesis, it follows from [12, Theorem 26] that 2 splits into 1 or 2 prime ideals in Z(ζ q )/Z, and therefore 2 must be inert in the intermediary extension L/Q of degree p. In particular, the polynomial f(x) is irreducible over Q 2, since it denes an unramied extension L 2 /Q 2 of degree p. Hence,