Root numbers and parity phenomena

The parity conjecture has a long and distinguished history. It gives a way of predicting the existence of points of infinite order on elliptic curves without having to construct them, and is responsible for a wide range of unexplained arithmetic phenomena. It is one of the main consequences of the Birch and Swinnerton‐Dyer conjecture and lets one calculate the parity of the rank of an elliptic curve using root numbers. In this handbook, we explain how to use local root numbers of elliptic curves to realise some of these phenomena, with an emphasis on explicit calculations. The text is aimed at a ‘user’ and, as such, we will not be concerned with the proofs of known cases of the parity conjecture, but instead, we will demonstrate the use of the theory by means of examples.

1. Introduction 1.1.The parity conjecture.The Mordell-Weil rank of an elliptic curve E over a number field K remains mysterious.Calculating it is generally either hard or impossible, and there is currently no method that will provably work for every curve.All we have is ways of finding upper bounds by calculating the rank of Selmer groups, lower bounds by searching for K-rational points, and fingers to cross that these coincide.In general, it can be hard to find out if an elliptic curve has even one K-rational point of infinite order.
The parity conjecture predicts points of infinite order and most known phenomena about elliptic curves effortlessly, after a simple calculation.It quantifies how the behaviour of E over completions of K should control the parity of the rank of E/K.It does this using the global root number of E/K, w(E/K) ∈ {±1}, which is defined as the product of easily computable local factors at each place v of K, local root numbers.They are straightforward to compute since, despite having a non-constructive definition, they have been classified for all places (see §2).
Parity Conjecture.Let E be an elliptic curve over a number field K. Then (−1) rk(E/K) = w(E/K), where w(E/K) is the global root number of E/K.
Note that if w(E/K) = −1, the parity conjecture implies that the rank is odd and, in particular, greater than 0. The parity conjecture is often the only way of predicting that a point of infinite order exists on an arbitrary elliptic curve.For now, take for granted that if E is a semistable elliptic curve defined over K, then w(E/K) = (−1) m+u , where m is the number of primes of K where E has split multiplicative reduction and u is the number of infinite places of K (see Corollary 2.5).
Example 1.1.Let us take E : y 2 − 23y = x 3 − 99997x 2 − 17x + 42.Magma [4] tells us that the rank of E/Q is 0, so that E does not have a rational point of infinite order.However, it also returns false, indicating that it has not proved that 0 is the rank of the curve 1 .With a simple calculation, the parity conjecture tells us that this curve should have a point of infinite order.Indeed, we have ∆ E = 17 • 655943686625481101 and non-split multiplicative reduction at both primes.Hence, since Q has one infinite place, the above result tells us that w(E/Q) = (−1) 1 = −1 and so the parity conjecture implies that the rank of E/Q is odd and that E has a point of infinite order.
One could use the conjecture of Birch and Swinnerton-Dyer to calculate Mordell-Weil ranks of elliptic curves.However, utilising the Birch-Swinnerton-Dyer conjecture requires computing L-functions, and, not only is this a rather lengthy calculation, we cannot provably compute ord s=1 L(E/Q, s) unless it is 0, 1, 2 or 3, nor do we know the analytic continuation of L-functions of elliptic curves over general number fields.The beauty of the parity conjecture is that it is free of the conjectural machinery of L-functions and is an entirely arithmetical statement, yet it still allows us to predict the existence of points of infinite order.
In terms of applications, parity-related arguments have many notable consequences.For example, a known case of the p-parity conjecture 2 was used by Bhargava and Shankar to prove that a positive proportion of elliptic curves over Q satisfy the Birch-Swinnerton-Dyer conjecture [1].The parity conjecture settles most cases of the congruent number problem (see §3.12) and has been used to prove many results about the ranks of families of elliptic curves [16,38,53,55].A variant on the 2-parity conjecture was used to prove that Hilbert's Tenth Problem has a negative answer over rings of integers of number fields [46] and results on the parity of Selmer ranks were used to prove that certain Kummer surfaces satisfy the Hasse principle [63].Similar arguments have also been used to prove results on class numbers of e.g.simplest cubic fields [71].Root numbers also seem to have played a role in the early investigation of Heegner points, see e.g.[34] Chapter 5, Corollary 21.3 and Conjecture 23.2.
The parity conjecture is easily deduced from the Birch-Swinnerton-Dyer conjecture together with the Hasse-Weil conjecture.The latter states that the completed L-function has analytic continuation to the whole of the complex plane and satisfies the functional equation L(E/K, s) = w(E/K) L(E/K, 2 − s), where w(E/K) is the global root number.Assuming the Hasse-Weil conjecture, w(E/K) = +1 if and only if the completed L-function is symmetric around s = 1, which happens if and only if its order of vanishing at s = 1 is even.This tells us that w(E/K) = (−1) ords=1 L(E/K,s) , which, assuming the Birch-Swinnerton-Dyer conjecture, gives the parity conjecture.1.2.A historical note.The investigation of parity-type questions appears to have begun with calculations of the parity of ranks of Selmer groups.In his 1954 paper 'A conjecture concerning rational points on cubic curves' [59], Selmer gave an explicit definition of the first and second descent for E/Q with a rational 2-isogeny by writing out the equations they lead to.Selmer wrote that he had performed numerical calculations that suggest "when the first descent indicated at most three generators, then none or two of these seem to be excluded by the second descent" (at most three as this was the case for all curves he numerically tested).Based on this, he conjectured that 1. "The second descent excludes an even number of generators".In his 1962 paper 'Arithmetic on curves of genus 1, III, The Tate-Šafarevič and Selmer groups' [6], where he coined the term 'Selmer group', Cassels extended Selmer's conjecture and proved that the dimension of the image of Sel p n (E/K) under the map induced by multiplication by p n−1 differs from that of Sel p (E/K), considered as a vector space over F p , by an even integer.This also treats the case of third descent, fourth descent, and so on.In [59], Selmer also conjectured the stronger statement, that 2. "The number of generators indicated by a first descent differs from the true number of generators by an even number".Extending Selmer's second conjecture to general p, as Cassels did for Selmer's first conjecture, it says that the rank of the p-Selmer group determines the parity of the Mordell-Weil rank and is equivalent to saying that δ p is even, where δ p is the multiplicity of Q p /Z p in X(E/K)[p ∞ ] and is conjecturally 0. In light of the parity conjecture and Selmer's second conjecture, we would expect that (−1) rkp(E/K) = w(E/K), where rk p (E/K) = rk(E/K)+δ p is the Z p -corank of the p-infinity Selmer group lim n→∞ Sel p n (E/K).This is known as the p-parity conjecture.Although we will not discuss it or its applications in this handbook, tackling the p-parity conjecture has proved to be more fruitful than tackling the parity conjecture itself and there have been numerous works on the subject (see §1.3 for references).In this context, one can prove parity-related statements without assuming the finiteness of the Tate-Shafarevich group (the fabled Shafarevich-Tate conjecture), while the parity conjecture itself appears to be, as of now, unapproachable without this assumption.
1.3.Using this handbook.This exposition is designed to explain how to calculate local and global root numbers of elliptic curves, and give a taste of what sorts of phenomena are predicted by the parity conjecture.The latter are concerned with the existence of rational points on elliptic curves for which there is as yet no construction, and that remain open problems without assuming the parity conjecture.As of now, the only known systematic construction of points of infinite order are Heegner points and self-points [14,72], both using the modular parameterisation.We hope that our examples might inspire some new ideas for the construction of rational points.
In §2, we describe how to calculate local and global root numbers of elliptic curves, along with root numbers of quadratic twists of elliptic curves and twists of elliptic curves by self-dual Artin representations.We also give a first introduction as to how to use these to predict the existence of points of infinite order on elliptic curves.In §3, we illustrate how the parity conjecture predicts many unexplained arithmetic phenomena about rational points on elliptic curves.For a quick overview of these, see the table of contents.In §4, we propose an analogue of the minimalist conjecture for Artin twists of elliptic curves over Q, and discuss its consequences.The conjecture gives the expected Galois module structure of E(F ) ⊗ Z C, for a fixed Galois extension F/Q and a generic elliptic curve E/Q.
The parity conjecture is known assuming the finiteness of the Tate-Shafarevich group [25,47], and it is also known that these approaches to controlling the parity of the rank do not extend to controlling the Mordell-Weil rank modulo n for n > 2 [26].We will not be concerned with approaches to the proof of the parity conjecture; for a self-contained exposition on the proof of the parity conjecture assuming the finiteness of the Tate-Shafarevich group (and of the proof of p-parity conjecture for elliptic curves over Q) see [27].We will confine ourselves to the setting of elliptic curves, although the parity conjecture is formulated for abelian varieties.For some recent results on these conjectures see [8,9,20,23,24,25,29,33,41,48,49,50,51,52]; for results on root numbers of abelian varieties see [3,57].
1.4.Notation.We will use the following notation.

E
an elliptic curve; K a number field; where the product runs over all places v of K, including infinite ones.
Local root numbers of elliptic curves are defined using epsilon-factors of Weil-Deligne representations (see [15] for the original definitions and [27] Section 3.3 for a more down-to-earth introduction).For a finite place v, the local root number is computed by looking at the action of the absolute Galois group of K v on the ℓ-adic Tate module.For the purpose of this exposition, we will not concern ourselves with formal definitions, but will focus on how to calculate local root numbers in practice.
Remark 2.2.For an elliptic curve defined over Q, the local root number at a prime p agrees with the eigenvalue of the associated Atkin-Lehner involution for the associated modular form.This follows from the corresponding statement for modular forms (see [58] Theorem 3.2.2, for example) together with the local Langlands conjecture for GL 2 and the modularity of elliptic curves over Q.

2.2.
How to calculate local root numbers.Local root numbers of elliptic curves E/K have been classified for all places v of K, so, for most purposes, one can avoid the technical definitions.
Theorem 2.3.Let E be an elliptic curve over a local field K of characteristic zero.When K is non-Archimedean, let k be its residue field and let v : K × ։ Z denote the normalised valuation with respect to K. Let * k denote the quadratic residue symbol on k × and (a, b) K denote the Hilbert symbol in K.
In particular, if , where ∆ is the minimal discriminant of E. (viii) If E/K has additive, potentially good reduction and char(k) ≥ 3, let y 2 = x3 + ax 2 + bx + c be a Weierstrass equation for E and let ∆ E be the discriminant of this model.
• If the Kodaira type is II, IV, II * or IV * , then there exists a Weierstrass equation for which 3 ∤ v(c).For such an equation, we have , where δ ∈ {±1} and δ = 1 if and only if The semistable case follows from the definitions in [15], Rohrlich proved (ii)-(vii) in [54] §19 and [56] Theorem 2 (the case of K = Q 2 in (vi) can be found in [10] §3), and (viii) is a result of Kobayashi [44] Theorem 1.1.The statements of (i)-(v) and (vii), as they are written, are found in [25] Theorem 3.1.For elliptic curves over local fields of positive characteristic, there is a similar classification of root numbers in [11] Theorem 3.1.
Theorem 2.3 does not tell us how to calculate local root numbers when E has additive potentially good reduction over a local field of residue characteristic 2. For E/Q 2 , one can use a classification of Halberstadt [36], which was extended to non-minimal Weierstrass models and potentially multiplicative reduction by Rizzo (see [53] Table III).The tables that classify the local root number w(E/Q 2 ) can be found in Appendix A. Generalising Rizzo's and Halberstadt's tables to extensions of Q 2 seems unfeasible as there would be many cases to check for large extensions.However, in the case of potentially good reduction over an extension of Q 2 , the local root number can be described in terms of Gal(K(E [3])/K) ⊂ GL(2, F 3 ), see [21].There is also a formula ( [23] Theorem 1.12) for local root numbers in terms of the Tamagawa number of E and the curves that are 2-isogenous to E over subfields of K(E [2]); the Tamagawa numbers can be read off from Tate's algorithm [67] (or see [62] p. 365).
Corollary 2.4.Let E be an elliptic curve over a local field K of characteristic zero and residue characteristic p ≥ 5. Let k denote the residue field of K and let * k denote the quadratic residue symbol on k.Then where E is given by a Weierstrass equation Proof.For K = Q p this was proved in [55] Proposition 2. If E/K has Kodaira type I 0 , I * 0 , I * n , III or III * , these expressions follow immediately from the theorem.If E/K has type I n , the root number depends on whether the reduction type is split or non-split.We may assume the Weierstrass equation for E is minimal, since a change of model changes the value of b by a 6-th power.Write for the reduction of E modulo the maximal ideal of K. E/K has split multiplicative reduction if and only A similar argument shows that if E/K has type and similarly if E has type IV * , w(E/K) = −3 k , which completes the proof.
Corollary 2.5.Let E be a semistable elliptic curve defined over a number field K. Then where m is the number of primes of K where E has split multiplicative reduction and u is the number of infinite places of K.
Example 2.6.Let us take the curve This curve has split multiplicative reduction at 11 and good reduction everywhere else, so Assuming the parity conjecture, this tells us that the rank of E/Q is even.In fact, E has rank 0 over Q.
Example 2.7.Let us take the curve 11A3 as in the above example.We know that w(E/Q) = 1.Let us look at what happens to the global root number of E over K = Q( √ −2).Since −2 is a quadratic residue modulo 11, 11 splits in K into, say, v and ṽ.Then w(E/K v ) = w(E/K ṽ ) = −1 since the reduction over K v and K ṽ is split multiplicative.Hence, Assuming the parity conjecture, this tells us that E must acquire a point of infinite order over K.
Asking what happens to local root numbers when we look at the curve over an extension of the base field will inform many of our examples of curious parity phenomena in §3.The above example illustrates the following key feature of root numbers of elliptic curves.
Lemma 2.8.Let E/K be an elliptic curve and let F be a Galois extension of K. Let v be a prime of K and let v 1 and v 2 be primes above v in F .Then Proof.Since F/K is Galois, F v1 is isomorphic to F v2 .The isomorphism from F v1 to F v2 preserves the equation for E and the root number w(E/F vi ) only depends on the isomorphism class of E/F vi .
2.3.Root numbers of quadratic twists of elliptic curves.If E : y 2 = f (x) is an elliptic curve defined over a number field K and d ∈ K × , we call E d : dy 2 = f (x) the quadratic twist of E by d.Recall the following well-known relationship between the rank of E and the rank of E d .Lemma 2.9.Let E be an elliptic curve defined over a number field K and let Lemma 2.10.Let E/K and E d /K be as above.Then Remark 2.11.We caution the reader that the statement of Lemma 2.10 is not true on the level of local root numbers, see [8] Proposition 3.11 for the analogous statement in the local setting.The lemma follows from the local statement, along with the product formula for Hilbert symbols.
Lemma 2.10 tells us that the statement of the parity conjecture is consistent with what we know about the ranks of E and E d .One can exploit this relationship between global root numbers and use the parity conjecture to predict the existence of points of infinite order on quadratic twists of elliptic curves.
Example 2.12.Let us take E to be the curve 11A3 over Q and let K = Q( √ −2).We saw in Examples 2.6 and 2.7 that w(E/Q) = +1 and w(E/K) = −1.So, by Lemma 2.10, w(E −2 /Q) = −1 and the parity conjecture implies there is a rational point of infinite order on E −2 : y ) and since d 1 and d 2 are square-free, it follows that have the same number of infinite places.Hence (ii) First, we claim that we can find a d 0 < 0 such that (d 0 , 2N E ) = 1 and, for every d, we have Indeed, choose d 0 < 0 such that all the primes of bad reduction of has two infinite places and all finite places dividing the discriminant of E split into an even number of places in and we have proved the claim.Now, to finish proving the theorem, take d 0 as above and D as in the proof of (i).By part (i) of the theorem, the function d → w(E d /Q) is periodic on the set of positive (resp.negative) square-free integers with the period dividing D. The involution d ↔ d • d 0 changes the sign of the root number i.e. for any d such that w(E d /Q) = +1, we have w(E d•d0 /Q) = −1.Since d 0 is coprime to D, the density of square-free integers congruent to d modulo D is the same as the density of those congruent to dd 0 modulo D, see e.g.[39].The result follows.Theorem 2.13 gives a heuristic for Goldfeld's conjecture, which states that there is a 50/50 distribution of the rank being 0 and 1 in a quadratic twist family over Q, see [32] Conjecture B. Over general number fields, root numbers of quadratic twists are also periodic (by essentially the same proof).However, the distribution of the root numbers can be different.As an extreme case, it is possible for all quadratic twists to have the same root number; see §3.9 or [22] for the original discussion.In [42] Example 7.11 and [43], it is shown that the set of proportions that appear for elliptic curves is dense in [0, 1].In [42] and [43] this is phrased in terms of 2-Selmer ranks -this is equivalent to the corresponding statement for global root numbers because the 2-parity conjecture is known in this instance by [24] Theorem 1.3.
Remark 2.14.We expect that for elliptic curves over Q in a suitable ordering, there is a 50/50 distribution of the global root number being +1 and −1 (for the ordering used, see Notation 4.2).Despite the fact that this is a statement about root numbers rather than about ranks, this remains an open problem.
2.4.Root numbers of Artin twists of elliptic curves.Let E/K be an elliptic curve and let F be a Galois extension of K. Then E(F ) ⊗ Z C is naturally a Gal(F/K)-representation by letting elements of the Galois group act on the co-ordinates of the F -rational points of E. There is a parity conjecture for Artin twists of elliptic curves which describes the parity of the multiplicity of an irreducible, selfdual Gal this can tell us about the existence of rational points over F .The parity conjecture for twists is especially striking when the usual parity conjecture does not provide us with any information about the existence of rational points (see Example 2.19).We state the parity conjecture for Artin twists here and present a theorem that allows us to explicitly calculate global root number of twists of elliptic curves over Q by self-dual Artin representations when the conductor of ρ is coprime to the conductor of E.
Parity Conjecture for Twists.Let E/K be an elliptic curve over a number field and let ρ be a self-dual Artin representation of Gal( K/K) that factors through Gal(F/K), for a finite extension F/K.Then where w(E/K, ρ) is the global root number of the twist of E by ρ.
Theorem 2.15 ([28] Corollary 2).Let ρ be a self-dual Artin representation of Gal( Q/Q).Let E be an elliptic curve over Q whose conductor N E is coprime to the conductor of ρ.Then where * * is the Jacobi symbol and α ρ = 1 if det(ρ) = 1 and, otherwise, α ρ is such that the character det(ρ) We adopt the convention that since we assumed the conductor of ρ to be coprime to the conductor of E, and hence α ρ ≡ 1 or 5 (mod 8).For a statement that allows multiplicative reduction at primes dividing the conductor of ρ, see [28] Theorem 1. Local root numbers of Artin twists of elliptic curves have been classified by Rohrlich (with restrictions in residue characteristic 2 and 3), see [56] Theorem 2.
We present the following important properties of global root numbers of twists, which are also of use when predicting the existence of rational points of infinite order using the parity conjecture for twists (see Example 2.19).For reference, (i) and (ii) follow from the definition of local root numbers (see [15] for more details) and (iii) and (iv) can be found in [24] Proposition A.2.
Theorem 2. 16.Let E be an elliptic curve over a number field K. Let ρ and ρ ′ be Artin representations of G K = Gal( K/K).Let L/K be a finite extension and let τ be an Artin representation of , where ρ * is the dual representation.
We can relate the rank over subfields of F to the representation E(F ) ⊗ Z C using the following lemma.
Lemma 2.17.Let E be an elliptic curve over K and let F be a Galois extension with G = Gal(F/K).Then for every H ≤ G, ) follows from Frobenius reciprocity.
The following lemma tells us that in many cases (for instance symmetric groups) the parity conjecture for twists follows from the usual parity conjecture.Lemma 2.18.Let E/K be an elliptic curve and let F/K be a Galois extension of number fields.Let ρ be a representation of G = Gal(F/K) that can be written as a linear combination of permutation modules i.e.
for some H i , H ′ j ≤ G.If the parity conjecture holds for E over L i = F Hi for all i and L ′ j = F H ′ j for all j, then the twisted parity conjecture holds for E and ρ.
Proof.By Theorem 2.16 and assuming the usual parity conjecture for L i and L j , By Lemma 2.17, Thus the parity conjecture for twists holds for E and ρ.
When the condition of Lemma 2.18 is not satisfied, the parity conjecture for twists gives us more information about the arithmetic of elliptic curves than the usual parity conjecture, as we now illustrate.
Example 2.19.Let us take the elliptic curve of rank 1 over Q Let F be the splitting field of the polynomial which has discriminant −47 5 .Then G = Gal(F/Q) = D 10 , the dihedral group with 10 elements.We can predict the existence of points of infinite order on E/F using the parity conjecture for twists as follows.
The subgroups of D 10 , up to conjugacy, are the trivial group, C 2 , C 5 and D 10 .The irreducible representations of D 10 are the trivial representation 1, sign representation ǫ, and two 2-dimensional representations ρ 1 and ρ 2 .Now let V = E(F ) ⊗ Z C.This is a G-representation so we can decompose it as Since the character of this representation is rational, ρ 1 and ρ 2 must appear with the same multiplicity as their characters are G Q -conjugate.So we have By Lemma 2.17, for any subgroup E has non-split multiplicative reduction at 37 and good reduction everywhere else.Since 37 splits in , by Corollary 2.5 we find that So the parity conjecture and our calculations above imply that the rank of E is odd over all four intermediate fields.In particular, we could in principle have rk(E/F ) = rk(E/Q).We now use the parity conjecture for twists to show that c is odd, and hence rk(E/F ) ≥ rk(E/Q)+4 ≥ 5. Using the notation of Theorem 2.15, we have N E = 37 and α ρ1 = −47 since det(ρ 1 ) = ǫ.Hence, by Theorem 2.15, Thus, since ρ 1 , E(F ) ⊗ Z C = c, the parity conjecture for twists tells us that c must be odd and hence rk(E/F ) ≥ 5.In fact, assuming the parity conjecture for twists, the rank of E/F is larger than the rank of E over any of the subfields.

Parity phenomena
We now turn to examples of phenomena that are predicted by the parity conjecture as a result of root number calculations.Subsections 3.6 and 3.14-3.17deal with consequences of the parity conjecture for Artin twists.We stress that all 'parity phenomena' that we describe, including the statements in all of the subsection titles, are conditional on the parity conjecture and are effectively unsolved problems.Most of these have been observed before; only §3.8, §3.14 and §3.15 are new.The titles are designed to give specific examples of what will be discussed; the general statements will be contained within the subsections.The subsections are largely independent of one another and can be read in almost any order. ).
Example 3.1 (See also [27], Conjecture 8.7).The parity conjecture predicts that every elliptic curve E/Q has even rank over ).To see this, we note that every rational prime splits into an even number of primes in K. Indeed, the decomposition group of rational primes away from the discriminant of K is cyclic and so cannot be isomorphic to C 2 × C 2 .Hence these primes split into 2 or 4 primes in K.The primes that divide the discriminant of K are 3 and 13, and we note that 3 splits in Q( √ 13) and 13 splits in Q( √ −3).So, 3 and 13 each split into two primes in K. Thus, where v runs over primes of K and p runs over primes of Q.Here we have used the fact that if v and v ′ lie above the same prime p, then w(E/K v ) = w(E/K v ′ ) by Lemma 2.8.Assuming the parity conjecture, we deduce that E has even rank over K.
The same reasoning gives us the following lemma.
Lemma 3.2.Let E/K be an elliptic curve and let F be a finite Galois extension of K in which every place of K splits into an even number of places.Then, assuming the parity conjecture, E has even rank over F .
), namely (0, √ −3), which turns out to have infinite order.Example 3.1 tells us that rk(E/K) is even, so there should be another (independent) point of infinite order over K.In effect, the existence of a point of infinite order over Q( √ −3) forces the existence of another point of infinite order over Q( ).In this case, local root number calculations show that w(E/Q( √ 13)) = −1, so, assuming the parity conjecture, there should also be a K-rational point of infinite order coming from points on E over Q( √ 13).Indeed, the point 3, lies on E and generates the infinite part of the Mordell-Weil group of E/Q( √ 13).It is not at all clear how the points over Q( √ 13) relate to those over Q( √ −3), and why the existence in one of the fields guarantees the existence in the other.

Every
2 also gives us the following example.
Example 3.4.Let K be any Galois extension of Q with Galois group C d 2 , where d ≥ 4 and let E be an elliptic curve defined over Q.We claim that E should have even rank over K.As in Example 3.1, all unramified primes split into an even number of primes in K.We claim that every ramified prime splits into an even number of primes in K, too.If p > 2, note that the largest extension of Q p we can get when we localise at a prime v above p in K has Galois group Hence the decomposition group of v | p has size at most 4 and so there must be at least four primes above p in K. Similarly for p = 2, the largest extension of Q 2 we can get when we localise at a prime v above 2 has Galois group C 2 × C 2 × C 2 and so the decomposition group of v | 2 has size at most 8 and there are at least two primes above 2 in K. Since K has an even number of infinite places, Lemma 3.2 tells us that, assuming the parity conjecture, E has even rank over K.
Remark 3.5.The parity conjecture implies that if E/Q has a point of infinite order over a C 2 ×C 2 ×C 2 ×C 2extension of Q, it must automatically acquire a second one.This suggests that over any C 2 × C 2 × C 2 × C 2extension of Q, there might be some as yet unknown extra symmetry, for instance in the Mordell-Weil group, that would explain why the rank is even.As Example 3.3 illustrates, it is not clear what such a symmetry might be.Example 3.6.The parity conjecture predicts that every elliptic curve E/Q with split multiplicative reduction at 2 has infinitely many rational points over K = Q(ζ 8 ).Here, the only ramified prime is 2, which is totally ramified.So, if E has split multiplicative reduction at 2, for the unique prime v above 2 in K we have w(E/K v ) = −1 by Theorem 2.3.Note that Gal(K/Q) ∼ = C 2 × C 2 , so in the same vein as Example 3.1, all primes away from 2 split into an even number of primes in K, which means their product contributes a +1 to the root number.There are two infinite places and so these also contribute a +1 to the root number.Hence, w(E/K) = −1 and the parity conjecture predicts that E has odd rank over K.In particular, it says that E has infinitely many Q(ζ 8 )-rational points.We emphasise that it is not at all clear how to construct these points!
The above working tells us that every elliptic curve over Q of the form E : y 2 + xy = x 3 + Ax + B, where A ≡ B (mod 2), has infinitely many points over Q(ζ 8 ) since every such curve has split multiplicative reduction at 2. The example easily generalises to the following statement.Lemma 3.7.Suppose K/Q is a biquadratic extension and that exactly one prime p does not split in K/Q.If the parity conjecture holds, then every elliptic curve E/Q with split multiplicative reduction at p has a point of infinite order over K.
Example 3.8.The curves (i) E : each have split multiplicative reduction at 2, so by Example 3.6, the parity conjecture implies that these curves all have infinitely many points over Q(ζ 8 ).In fact, all of these curves have global root number +1 over Q, so the rank must grow in Q(ζ 8 ).One can perform descent calculations to show that the rank of each of these curves over Q(ζ 8 ) is 1.However, the point comes from a different quadratic subfield for each curve: E has a point of infinite order over Q( √ 2), E ′ has a point of infinite order over Q( √ −1) and E ′′ has a point of infinite order over Q( √ −2).
Remark 3.9 (See also [26] Remark 4).Consider the elliptic curve E : ) is formally a square, in the sense that each Euler factor appears an even number of times: If we define F (s) by the square root of this Euler product so that F (s) 2 = L(E/Q(ζ 8 ), s) for Re(s) > 3  2 , then F (s) shares many properties that one would expect from an L-function.However, it does not have analytic continuation to s = 1 since the order of vanishing of L(E/Q(ζ 8 ), s) at s = 1 is odd.
Example 3.10 (As considered in [28]).The elliptic curve has rank 0 over Q.We claim that, assuming the parity conjecture, it has infinitely many Q( √ m) has one real embedding and one complex embedding.This tells us that So, assuming the parity conjecture, E must have infinitely many Q( 3 √ m)-rational points.
We can numerically compute the points of infinite order on Without the parity conjecture, proving the existence of these points for all m seems completely out of reach.In [19], it is shown that for an elliptic curve E over a number field K, the rank of E goes up in infinitely many extensions of K obtained by adjoining a cube root of an element of K.However, these points account for very few m in Example 3.10.Remark 3.11.We caution the reader that in the above example there are no parametric solutions of the form ).Indeed, if f were analytic, it would give an analytic map P 1 → E(C) which contradicts the Riemann-Hurwitz formula.If f were only assumed to be continuous, it would give a map R → E(R).For m ∈ Q ×3 , this could only take values (0, 0), (0, 1) or O, which would force f to be constant.

E : y
Example 3.12.Example 3.10 tells us that the rank of E : Assuming the parity conjecture, it grows at every step of the tower Q( 3 n √ m) n≥1 .To see this, we claim that 19 splits into an odd number of places in ) is an odd degree extension of Q 19 and so e ṽ f ṽ is odd for every prime ṽ above 19 in , where e ṽ is the ramification degree and f ṽ is the residue degree of ṽ over 19.Let v be a prime above e vi f vi , the sum over all primes above 19 in Q( 3 n √ m), this tells us that k is odd and there are an odd number of primes above 19.Hence has an even number of infinite places when n is odd and an odd number when n is even.Since the sign of the global root number changes at each step, the parity conjecture implies that the rank of E must grow at every step of the tower Q( 3 n √ m) n≥0 .In particular, the rank of In [28], Example 3.12 is generalised to the setting of root numbers of elliptic curves defined over Q in the towers of extensions , where p is an odd prime.Here we present the results in the setting of semistable elliptic curves.Theorem 3.13 ([28] Theorem 6).Let E be a semistable elliptic curve over Q.Let p be an odd prime at which E has good reduction.Let m > 1 be an p-th power free integer 4 .Then the global root number for , where t is the number of primes of multiplicative reduction of E that do not divide m and that are nonsquares modulo p. Corollary 3.14.Let E be a semistable elliptic curve over Q and let p ≡ 3 (mod 4) be a prime at which E has good reduction.Suppose that every prime of multiplicative reduction is a square modulo p.If the parity conjecture holds then the rank of E over Q( p n √ m) is at least n for every p-th power free integer m.
√ m) for any positive 7-th power free integer m.Indeed, we have ∆ E = −11 and E has split multiplicative reduction at 11. Since 11 is a square modulo 7, the rank is at least n over Q( 7 n √ m).In particular, the parity conjecture implies that E has rank at least 1 over Q( 7 √ 3).Magma fails to find the point of infinite order on E/Q( 7 √ 3).
We claim that, assuming the parity conjecture for twists (see §2.4), E has rank at least Then, by Lemma 2.17, for any subgroup H ≤ Gal(K n /Q), we have We can use this, along with our knowledge from Example 3.12 about what happens to E over Q( 3 n √ m), to say something about the rank of E over K n .First, note that where ρ k is an irreducible representation of dimension p k−1 (p − 1), see [28] §5.2 for more details.By Theorem 2.16, we have Assuming the parity conjecture for twists, this tells us that ρ k appears in V with odd multiplicity for all k ∈ {1, . . ., n}.
and is orthogonal to the ρ k , along with our previous working this tells us that rk Remark 3.17.By Lemma 2.18, the result in the above example could have been deduced from the parity conjecture for E over intermediate fields rather than the parity conjecture for twists.
In [28] the methods demonstrated in the example above are generalised in order to prove the following result (which for simplicity we state just for semistable elliptic curves).Theorem 3.18 ([28] Corollary 13).Let p ≡ 3 (mod 4) be a prime number and let E be a semistable elliptic curve with good reduction at p and such that every prime of multiplicative reduction is a square modulo p.If the parity conjecture holds then, for every p-th power free integer m > 1, the rank of E over Remark 3.20.For a possible approach to explaining these points of infinite order using Heegner points, see [13].
It is not clear how one might prove this property of f (x) without the parity conjecture.
Example 3.21 (As in [17] Example 4).We claim that all positive quadratic twists of the curve have root number +1 and all negative quadratic twists have root number −1.The minimal discriminant of E is ∆ = 7 2 • 13 2 and E has type II reduction at both primes.By Theorem 2.3, 12 ⌋ = +1.By a similar calculation, if 7 is inert in K then w(E/K v ) = +1.If 7 splits, by Lemma 2.8 there are two distinct places with the same root number.In all cases v|7 w(E/K v ) = +1.We can do the same calculation for ṽ above 13 to find ṽ|13 w(E/K ṽ ) = +1.Thus and our claim follows from Lemma 2.10.In particular, the parity conjecture implies that all negative quadratic twists of E have infinitely many points.
is a cubic and the quadratic twist of the elliptic curve E : This is the quadratic twist by −1 of the curve in Example 3.21 so The parity conjecture implies that every positive Remark 3.24.Assuming Goldfeld's conjecture that there is a 50/50 distribution of the rank of quadratic twists of an elliptic curve E/Q being 0 and 1, from Example 3.23 we deduce5 that 0% of negative integers (up to squares) can be represented by x 3 − 91x − 182.This is a peculiar property for a cubic polynomial: its positive values hit all possible classes modulo squares whereas its negative values hit a very sparse set.
Proof.Take g(x) = d 0 f (x), where d 0 < 0 is such that all primes of bad reduction of E : is the quadratic twist of E by d 0 .By the proof of Theorem 2.13(ii), for any d ∈ Q × /Q ×2 the root number of the quadratic twists of E d0 and E by d satisfy w(E dd0 /Q) = −w(E d /Q).The parity conjecture implies that at least one of E d and E dd0 has a point of infinite order, and thus every We can also find examples of such curves that are not quadratic twists of each other.Theorem 2.13 tells us that the global root number in a quadratic twist family is periodic; we can use this to come up with curves that are not quadratic twists of each other, for which one of their quadratic twists always has root number −1 and where the root number does not only depend on the sign of d.
Example 3.27.Let E and E ′ be given by We claim that for every  3.9.All quadratic twists of 11) have infinitely many points.As mentioned at the end of §2.3, the 50/50 distribution of root numbers of quadratic twists of elliptic curves over Q can fail quite dramatically over number fields; there are elliptic curves over number fields for which all quadratic twists have the same root number.
Example 3.29 (As considered in [22]).Let 11).We claim that all quadratic twists of the curve over K have root number −1 and so should have positive rank.Indeed, the minimal discriminant of E/Q is 11 4 , and E acquires everywhere good reduction over K. Since K has three complex places and So, assuming the parity conjecture, E d has a point of infinite order for every d ∈ K × .See Theorem 3.33 below for a more general result concerning this behaviour.

It has complex multiplication defined over
, where i acts by [i] • (x, y) = (−x, iy).The rank of E over K is even, and similarly rk(E/F ) is even for any extension F of K, since E(F ) ⊗ Z Q is naturally a Q(i)-vector space, and so has even dimension over Q.Moreover, using the fact that rk( The parity conjecture forces some non-CM elliptic curves to exhibit the same behaviour. Example 3.31 (As considered in [22]).Let us take the curve Then ∆ E = 37 3 and j(E) = 2 12 , so E/Q has potentially good reduction at 37 and it has everywhere good reduction over K = Q( 4 √ −37) and over every extension F of K. Since F has an even number of infinite places, w(E/F ) = +1.In particular, assuming the parity conjecture, E has even rank over every extension of K. Remark 3.32.Such a field K exists for every elliptic curve E with integral j-invariant as this means that E has potentially good reduction everywhere, so we can find an extension of the base field over which E acquires everywhere good reduction.
As in §3.1 and §3.2, one would hope that there is some extra structure analogous to complex multiplication that would explain why E always has even rank over extensions of K in Example 3.31.However, it is not clear how to prove this behaviour of ranks for any non-CM curve without assuming the parity conjecture.Elliptic curves that display this phenomenon, and that described in §3.9, are classified by the following theorem.Theorem 3.33 ([22] Theorem 1).For an elliptic curve E over a number field K, the following are equivalent (i) All quadratic twists of E/K have the same root number; (ii) w(E/F ) = w(E/K) [F :K] for every finite extension F of K; (iii) K has no real places, and E aquires everywhere good reduction over an abelian extension of K.
Theorem 3.33 tells us that if E/K satisfies (iii) and w(E/K) = +1 then, assuming the parity conjecture, E will have even rank over every extension of K. Remark 3.34.It turns out that the criteria in Theorem 3.33 are equivalent to K having no real places, and for all primes ℓ and all places v ∤ ℓ of K, the action of the absolute Galois group of K v on the Tate module T ℓ (E) being abelian (see [22]).This is another way in which E resembles a CM curve, since (in view of the Tate conjecture) an elliptic curve has CM if and only if the action of the global absolute Galois group on ℓ-adic Tate module is abelian.
3.11.The rank of E : of even degree.Theorem 3.33 tells us that there exist elliptic curves over number fields for which all quadratic twists have the same root number.We can also use the theorem to come up with examples of elliptic curves E/K whose rank grows in every even degree extension of K.
Example 3.35 (As considered in [22]).Let 3 and E has additive, potentially good, type III reduction at 7. We claim that the rank of E must grow in every extension of K = Q( √ −1) of even degree.Indeed, 7 is inert in Q( √ −1) and by Theorem 2.3, w(E/K v ) = +1, where v is the unique prime above 7 in K, and so w(E/K) = −1.Note that K has no real places and E acquires everywhere good reduction over , which is an abelian extension of K.By Theorem 3.33, for any even degree extension F of K, w(E/F ) = w(E/K) [F :K] = (−1) even number = +1 = w(E/K), and the parity conjecture predicts that rk(E/F ) > rk(E/K).
3.12.Every positive integer n ≡ 5, 6, or 7 (mod 8) is a congruent number.The classical congruent number problem asks: for a natural number n, can it be realised as the area of a right-angled triangle with rational sides?We call such n congruent numbers.The congruent number problem dates back to Arab manuscripts from the 10th century and to this day it is not known in full generality, although it has been extensively studied.Heegner [37] proved that if p is a prime congruent to 5 modulo 8 then it is a congruent number, and other cases have been proved using the theory of Heegner points on modular curves, see [69].Smith [64] has announced a proof that at least 55.9% of positive square-free integers congruent to 5, 6, or 7 modulo 8 are congruent numbers.It has been long expected that every positive integer congruent to 5, 6, or 7 (mod 8) is a congruent number, and we will explain how this can be deduced from the parity conjecture.
Example 3.36.There is a right-angled triangle with sides of length 3, 4, 5 which has area 6, so that 6 is a congruent number.Similarly, the right-angled triangle with sides of length 35  12 , 24 5 and 337 60 has area 7, whence 7 is a congruent number.Definition 3.37.The elliptic curve E : y 2 = x 3 − x is called the congruent number curve because the quadratic twist of the curve by n ∈ N, E n : y 2 = x(x − n)(x + n), has a point of infinite order over Q if and only if n is a congruent number (see e.g.[70] for details on why this is the case).For these elliptic curves, a rational point (x, y) has infinite order if and only if y = 0. Example 3.38.We already know that 7 is a congruent number.Indeed, the elliptic curve has a rational point of infinite order, namely (25, 120).This tells us that 166 is a congruent number.The height of the generator E 166 is already very large.What if we wanted to know whether 800 006 was a congruent number?Magma [4] cannot find a point of infinite order on E 800006 and cannot tell us whether 800 006 is not a congruent number.Assuming the parity conjecture, we can easily show that it is.
Theorem 3.40.Let n be a positive, square-free integer.Then Proof.By Theorem 2.13(i), w(E n /Q) depends only on n (mod 16).So all one needs to do is calculate the root numbers for one square-free n in each of the congruence classes modulo 16.For E 1 : y 2 = x 3 − x, we have c 4 = 2 4 •3, c 6 = 0 and ∆ E1 = 2 6 .In the terminology of Notation A.1, we have c ′ 4 = 3 and c ′ 4 −4c 6,7 = 3 so that, by the table in Appendix A, w( which completes the proof. Corollary 3.41.Assuming the parity conjecture, every square-free positive integer n ≡ 5, 6, or 7 (mod 8) is the area of a right-angled triangle with rational sides.
We can deduce immediately from this corollary that, assuming the parity conjecture, 800 006 is a congruent number.Stephens was the first to apply parity-type conjectures to the congruent number problem [66] and proved that Selmer's second conjecture (see §1.2) implies that positive integers n ≡ 5, 6, 7 (mod 8) are congruent numbers.

E :
One incentive for studying root numbers is to study ranks in families of elliptic curves.Consider a one parameter family of elliptic curves E : y 2 = x 3 + A(t)x + B(t), for some fixed A(t), B(t) ∈ Q(t), where ∆ E = 0.A point of infinite order over Q(t) gives a parametric family of points over the fibres E t .A natural question to ask is 'if every fibre has positive rank, does it mean there is a point of infinite order over Q(t)?' Assuming the parity conjecture, the answer is 'no'.Theorem 3.42 (As considered in [7]).Let E : y 2 = x(x 2 − 49(1 + t 4 ) 2 ).Then (i) rk E /Q(t) = 0; (ii) Assuming the parity conjecture, for every t ∈ Q the fibre E t has rk(E t /Q) ≥ 1.
Proof.In [7], it is shown that Ẽ : y 2 = x(x 2 − (1 + t 4 ) 2 ) has rank 0 over C(t).We get (i) from the fact that this curve is isomorphic to E over C(t), and so rk E /Q(t) = 0.For (ii), let t = l/m where (l, m) = 1.Then E t is isomorphic to a curve of the form 4 .Clearly, n ≡ 6 or 7 (mod 8) so, by Theorem 3.40, w(E t /Q) = −1.
The family E in Theorem 3.42 has constant j-invariant.In [16] Theorem 1.2, Desjardins proved that, for a family with non-constant j-invariant, assuming Chowla's conjecture and the Squarefree conjecture (see [16] Conjectures 2.1 and 2.7), the function t → w(E t /Q) is never constant.She used this to prove that the sets are infinite and so, assuming the parity conjecture, the rational points E (Q) are Zariski dense in E .
Since the sets W + and W − are infinite when the j-invariant is non-constant, one might expect that the root numbers are equidistributed and that the average root number is 0. This is not always the case, as shown in [38], where Helfgott proved that if E has no places of multiplicative reduction, the average root number need not be zero.
For a family with non-constant j-invariant, if one restricts to t ∈ Z the function t → w(E t /Q) may be constant.For example, in [53] it is shown that the surface E : y 2 = x 3 + tx 2 − (t + 3)x + 1 has w(E t /Q) = −1 for every t ∈ Z.This example was first investigated by Washington [71] who proved via 2-descent that, for every t ∈ Z such that t 2 + 3t + 9 is square free and assuming the finiteness of the Tate-Shafarevich group, the rank of E t is odd.
3.14.y 2 + y = x 3 − x acquires new solutions over K( √ ∆ K ) whenever 37 ∤ ∆ K and √ ∆ K ∈ K.We saw in §3.5 and §3.6 that the parity conjecture can predict the growth of the rank of elliptic curves in towers of fields of the form Q( p n √ m) and Q(ζ p n , p n √ m).The parity conjecture also predicts that the rank of an elliptic curve over Q can grow in many generic number fields.Theorem 3.43.Let K be any number field that does not contain √ ∆ K and set By Frobenius reciprocity (see Lemma 2.17), ) so, assuming the parity conjecture for twists, it will suffice to show that one of w(E/Q, ρ), w(E/Q, ǫ) and w(E/Q, ǫ ⊗ ρ) is −1.In fact, by Lemma 2.18, it suffices to assume the parity conjecture for E over the subfields of K.
Let K = Q(α).The permutation action of G on the cosets of H agrees with the permutation action on the Galois conjugates of α, since both actions are transitive with H as a point-stabiliser.Even permutations are precisely those fixing √ ∆ K , and therefore det(Ind G H 1) = det ρ cuts out the extension Q( √ ∆ K )/Q so that det ρ = ǫ.It follows that det(ǫ⊗ρ) = ǫ ⊗ dim ρ ⊗det ρ is either 1 or ǫ if dim ρ is odd or even, respectively.By Theorem 2.15, if [K : Q] is even then dim ρ is odd and w Remark 3.44.Theorem 3.43 applies to generic extensions K/Q.If [K : Q] = n > 2 and the Galois closure of K over Q has Galois group S n then K does not contain √ ∆ K .We therefore expect the rank of every E/Q with w(E/Q) = −1 to increase in splitting fields of most polynomials.
Theorem 3.46.Let E/Q be an elliptic curve with w(E/Q) = −1.Assume the parity conjecture for twists.(i) Let ρ be an odd-dimensional irreducible self-dual Artin representation with det ρ = 1 and whose conductor is coprime to N E .Then ρ, E(F ) ⊗ Z C > 0, where ρ factors through F/Q.(ii) For a Galois extension where k is the sum of dimensions of the odd-dimensional irreducible self-dual representations of Gal(F/Q) and m is the number of one-dimensional representations of Gal(F/Q) of order 1 or 2. Remark 3.47.Theorem 3.46 implies that for a fixed elliptic curve E/Q with w(E/Q) = −1, every odddimensional irreducible self-dual Artin representation which has trivial determinant and whose conductor is coprime to N E must appear in the Mordell-Weil group of E over an extension of Q.
To generalise Example 3.45 to finding a lower bound on the rank of E over an S n -extension of Q, we need to understand the odd-dimensional irreducible representations of S n .For a given n, the dimensions of the irreducible representations of S n can be calculated using the Young tableaux, see [31] p. 50.We will instead use a known case of the McKay conjecture to find a crude lower bound on the expected rank of a rational elliptic curve E over an S n -extension of Q for general n.Proof.We have the wreath product with k copies of C 2 (see, for example, [18] Example 2.6.1),so we proceed by induction.For k = 1 we have P Theorem 3.51.Let E/Q be any elliptic curve with w(E/Q) = −1 and let K be any extension of where n = k a k 2 k is the binary expansion of n with a k = 0 or 1.
Proof.By Lemma 2.18, the parity conjecture for E over all subfields of K implies the parity conjecture for the twist of E by any irreducible self-dual representation ρ of Gal(F/Q) (the fact that all representations of S n can be written as linear combination of permutation modules follows from [40] §6.1).Thus, it suffices to prove that the parity conjecture for twists implies the result.The irreducible representations of S n are self-dual and there are two one-dimensional irreducible representations.Thus, by Theorem 3.46, to find a lower bound on the rank, it suffices to find a lower bound on the number of pairs of odd-dimensional irreducible representations of S n .Theorem 3.48 says that the number of odd-dimensional irreducible representations of S n is equal to the number of one-dimensional representations of a 2-Sylow subgroup of S n , since the normaliser of a 2-Sylow subgroup of S n is itself and the dimension of an irreducible representation divides the order of the group.Since n = k a k 2 k , the 2-Sylow subgroup of S n is the 2-Sylow subgroup of k:a k =0 S 2 k .By Lemma 3.50, the number of one-dimensional representations of the 2-Sylow subgroup of S 2 k is 2 k .Hence, the number of one-dimensional representations of the 2-Sylow subgroup of k:a k =0 S 2 k is k 2 ka k .Theorem 3.48 then tells us that there are 1 2 k 2 ka k pairs of odd-dimensional representations of S n .By Lemma 3.49, for n > 6 other than the trivial and the sign representation, all of the odd-dimensional representations have dimension at least n, unless n is even in which case there are also two (n − 1)dimensional representations.Hence, assuming the parity conjecture for twists, for n > 6 we have rk Using Theorem 3.46(ii), a simple check shows that the result is also true for n ≤ 6.
3.16.Heegner hypothesis.When studying Heegner points, one often imposes the Heegner hypothesis, that the quadratic field K is imaginary and that all primes of bad reduction of E/Q split in K (e.g. in Kolyvagin's work on the Birch-Swinnerton-Dyer conjecture, see [35]).From root number calculations, this ensures the odd order of vanishing of L-functions of E over K and of certain twists of E, as we now explain.Proposition 3.54.Let E/Q be an elliptic curve and let K be a quadratic extension of Q that satisfies the Heegner hypothesis.Let F be a C n -extension of K so that Gal(F/Q) = D 2n , the dihedral group with 2n elements.Assume that (N E , ∆ F ) = 1.Then (i) For χ any of the 1-dimensional representations of Gal(F/K), ord s=1 L(E/K, χ, s) is odd.
(ii) The order of vanishing of the L-function of E/F satisfies ord s=1 L(E/F, s) ≥ n.
Proof.Let H = Gal(F/K) and G = Gal(F/Q) = D 2n .The representations of D 2n are all self-dual.The 1-dimensional representations are 1, the representation ǫ that factors through Gal(K/Q), and, if n is even, two more representations ǫ 1 and ǫ 2 .There are n−1 2 irreducible 2-dimensional representations if n is odd and n−2 2 irreducible 2-dimensional representations if n is even.Call the irreducible 2-dimensional representations ρ i .
(i) By the inductivity of L-functions, for χ any of the 1-dimensional representations of Gal(F/K), all of which have determinant character ǫ.Hence, by the Heegner hypothesis on K and in the notation of Theorem 2.15, α ρ is negative and satisfies By the inductivity of root numbers (Theorem 2.16(iii)), w(E/K, χ) = −1 and, since we know the analytic continuation of L(E/K, χ, s) and that it satisfies a functional equation whose sign is w(E/K, χ) (see e.g [60] Proposition 4•13), this implies that ord s=1 L(E/K, χ, s) is odd.
(ii) By the inductivity of L-functions, where χ runs over the 1-dimensional representations of Gal(F/K).The result follows from part (i).
3.17.Artin representations that always appear with even multiplicity in E(K) ⊗ Z C. The parity conjecture for twists tells us the parity of the multiplicity of a representation ρ in E(K) ⊗ Z C. Conversely, if ρ is self-dual, irreducible and has Schur index 2 (e.g.ρ is symplectic) then it must appear with even multiplicity in E(K) ⊗ Z C.This has been checked to be compatible with root numbers in most cases, see [56] Proposition E and [57].On the other hand, root numbers suggest that certain representations must always appear with even multiplicity despite having Schur index 1.
Proposition 3.55 (Rohrlich,[56] Proposition D).Let K be a Galois extension of Q where Gal(K/Q) = D 2q × D 2r × D 2s × D 2t for distinct primes q, r, s, t ≥ 5 and let τ be an irreducible 16-dimensional representation of Gal(K/Q).Then w(E/Q, τ ) = +1 for every E over Q. Proposition 3.55 tells us that, assuming the parity conjecture for twists, the multiplicity of τ in E(K)⊗ Z C is always even, which is as mysterious as the fact that every E/Q should have even rank over any number field which is Galois over Q and in which all places split into an even number of places (see Lemma 3.2).

Minimalist conjecture for twists
There is a folklore 'minimalist conjecture', which says that, generically, the rank of an elliptic curve E/Q is 0 or 1 depending on whether w(E/Q) = +1 or −1 (see e.g.[27] Conjecture 8.2).We propose a minimalist conjecture for Artin twists, which gives the expected Galois module structure of E(F ) ⊗ Z C for most elliptic curves E/Q, where F is a finite Galois extension of Q (see Conjecture 4.3 and Theorems 4.7 and 4.10 below).It says that, generically, if E/Q is an elliptic curve and ρ is an irreducible self-dual Artin representation that factors through F/Q, then ρ appears in E(F ) ⊗ Z C with multiplicity 0 or 1, depending on whether the global root number of the twist of E by ρ is +1 or −1.
Example 4.1.Let E/Q be any elliptic curve with w(E/Q) = −1, and let F be any extension of Q with Gal(F/Q) = A 5 and (N E , ∆ F ) = 1.By the conductor-discriminant formula, this means that N E is coprime to the conductors of the representations of Gal(F/Q).We claim that rk(E/F ) ≥ 12.
We have det(σ) = det(τ i ) = 1 since A 5 has no subgroup of index 2. Hence, by Theorem 3.46, the parity conjecture for twists implies that a, c, d and e are odd.In particular, rk(E/F ) ≥ 12.
We typically expect to have exactly , and rk(E/F ) = 12.This is based on a 'minimalist conjecture for twists'.The idea is that for any elliptic curve E/Q and any self-dual irreducible Artin representation τ that factors through an extension F/Q, the contribution to the rank of E/F by τ should be as small as the parity conjecture for twists will allow.Since the global root number of the twist is the conjectured sign in the functional equation relating L(E/Q, ρ, s) and L(E/Q, ρ * , 2 − s), if ρ is self-dual the parity of the order of vanishing of L(E/Q, ρ, s) at s = 1 is governed by w(E/Q, ρ).For most elliptic curves, one expects the order of vanishing of the twisted L-function to be as small as is allowed by the functional equation.If ρ is not self-dual, the functional equation does not tell us anything about the order of vanishing of L(E/Q, ρ, s) at s = 1.In this case, for most elliptic curves we expect the order of vanishing to be zero and that ρ does not appear in E(F ) ⊗ Z C.
Remark 4.4.If ρ is a complex irreducible representation with Schur index n that factors through F/Q, the multiplicity of ρ in E(F ) ⊗ Z C is divisible by n.The Schur index of an irreducible self-dual character can only be 1 or 2 by the Brauer-Speiser theorem [30,5,65].So in the context of root numbers and the minimalist conjecture, Schur indices never force the multiplicity of an irreducible self-dual representation ρ in E(F ) ⊗ Z C to be greater than 1.
Lemma 4.5.Let F be a Galois extension of Q, let ρ be an irreducible Artin representation that factors through F/Q and let D be a non-zero integer.The minimalist conjecture for twists implies that for 100% of elliptic curves E/Q with (N E , D) = 1, Proof.It is known that, for any integer D, when ordered by the height of the coefficients there is a positive proportion of elliptic curves over Q whose conductor is coprime to D (see e.g.Theorem 4.2(2) of [12]; their ordering is slightly different, which may result in a different density, but we only need to know that this density is non-zero).In other words, lim X→∞ #N #T = k > 0 where T is the set of curves E A,B such that H(E A,B ) < X and N ⊂ T is the set of elliptic curves E A,B such that (N E , D) = 1.Let M ⊂ T be the set of curves E A,B that do not satisfy the conclusion of Conjecture 4.3, so that where we have used Conjecture 4.3 for #M/#T → 0. This proves the lemma.
Lemma 4.6.Let E/Q be an elliptic curve.Let F be a Galois extension of Q that does not contain any quadratic number field.For any self-dual Gal(F/Q)-representation ρ whose conductor is coprime to N E , w(E/Q, ρ) = w(E/Q) dimρ .
Theorem 4.7.Let F be a Galois extension of Q that does not contain any quadratic number field.Write W G for the direct sum of the odd-dimensional irreducible self-dual representations of G = Gal(F/Q) and k = dim W G .Assuming the minimalist conjecture for twists, for 100% of elliptic curves E/Q with (N E , ∆ F ) = 1 In particular, for 100% of elliptic curves E/Q with (N E , ∆ F ) = 1

3. 8 .
Every d ∈ Q × /Q ×2 can be represented by 4x 3 − 32x − 35 or 9x 3 + 16x + 16.Example 3.23 and Example 3.21 show that, assuming the parity conjecture, for every d ∈ Q at least one of the equations d • y 2 = x 3 − 91x + 182 and d • y 2 = x 3 − 91x − 182 has a rational solution.In particular, every d ∈ Q × /Q ×2 can be represented by x 3 − 91x + 182 or x 3 − 91x − 182.It turns out that every cubic f (x) has an auxiliary cubic g(x) with this property: Theorem 3.25.Let f (x) ∈ Q[x] be a separable cubic polynomial.Assuming the parity conjecture, there exists a separable cubic polynomial g

Theorem 3 . 48 (
McKay conjecture for p = 2, see[45]).Let G be a finite group and let P ≤ G be a 2-Sylow subgroup.For H ≤ G, let Irr(H) denote the set of isomorphism classes of complex irreducible representations ofH.Let Irr ′ (H) = {ρ ∈ Irr(H) : 2 ∤ dim ρ}.Then | Irr ′ (G) |=| Irr ′ (N G (P )) |where N G (P ) is the normaliser of P in G. Lemma 3.49 (See Exercise 4.14 in[31]).For n > 6, there are precisely four representations of S n of dimension less than n.Two of these have dimension 1 and two have dimension n − 1. Lemma 3.50.Let P k be the 2-Sylow subgroup of S 2 k .Then P k /[P k , P k ] ∼ = C k 2 , where [P k , P k ] denotes the commutator subgroup.

Notation 4 . 2 .
For E/Q an elliptic curve, it is isomorphic to a unique curve E A,B : y 2 = x 3 + Ax + B where A, B ∈ Z and for all primes p, either p 4 ∤ A or p 6 ∤ B. The naive height of E A,B isH(E A,B ) = max(4|A 3 |, 27B 2 ).We say that 100% of elliptic curves over Q satisfy a property T iflim X→∞ #{E A,B |H(E A,B ) < X and E A,B satisfies property T } #{E A,B |H(E A,B ) < X} = 1,where for all primes p, either p 4 ∤ A or p 6 ∤ B.

fact, E −2 has rank 1 over Q. Theorem 2.13. Let E be an elliptic curve over
Q. (i) The function d → w(E d /Q) isperiodic on the set of positive (resp.negative) square-free integers.The period divides p|NE p 2 if N E is odd and 4 p|NE p 2 if N E is even; in particular, it divides 4N 2 Let D = p|NE p 2 if N E is odd and D = 4 p|NE p 2 if N E is even.Suppose d 1 ≡ d 2 (mod D) and d 1 and d 2 have the same sign, and let p | N E .First suppose p > 2. Since d 1 ≡ d 2 (mod p 2 ) and d 1 and d 2 are square-free, it follows that d 1 The parity conjecture predicts that E has odd rank over Q(ζ 8 ) in a similar way to Example 3.6.The fact that E/Q(ζ 8 ) has odd analytic rank has the following consequence for L-functions.Since all primes away from 2 split in Q(ζ 8 ) and E has additive reduction at the prime above 2 in Q(ζ 8 ), the Euler product of the L-function of E/Q(ζ 8 which has Kodaira type IV over Q 2 .By computing root numbers of the quadratic twists of E by −1, 2 and −2 and using Lemma 2.10, we find that w(E/Q(ζ 8 )) = −1.
3√ m)-rational solutions for every cube-free m > 1.First, note that ∆ E = 19, and E has split multiplicative reduction at 19.Hence, w(E/K v ) = −1 at primes v above 19.So, to calculate the global root number, all we need to know is how many primes there are above 19 in Q( 3 √ m).If 19 ∤ m, the Kummer-Dedekind theorem tells us that there is either one or three primes above 19, corresponding to whether X 3 − m is irreducible or splits completely over F 19 .If 19 | m, it is totally ramified.The product of local root numbers at the infinite places is +1 since Q( 3