On elliptic curves of prime power conductor over imaginary quadratic fields with class number 1

Abstract

The main result of this paper is to extend from $\mathbb{Q}$ to each of the nine imaginary quadratic fields of class number 1 a result of [Serre, Duke Math. J. 54 (1987) 179–230] and [Mestre–Oesterlé, J. reine. angew. Math. 400 (1989) 173–184], namely that if $E$ is an elliptic curve of prime conductor, then either $E$ or a 2‐, 3‐ or 5‐isogenous curve has prime discriminant. For four of the nine fields, the theorem holds with no change, while for the remaining five fields the discriminant of a curve with prime conductor is (up to isogeny) either prime or the square of a prime. The proof is conditional in two ways: first that the curves are modular, so are associated to suitable Bianchi newforms; and second that a certain level‐lowering conjecture holds for Bianchi newforms. We also classify all elliptic curves of prime power conductor and non‐trivial torsion over each of the nine fields: in the case of 2‐torsion, we find that such curves either have CM or with a small finite number of exceptions arise from a family analogous to the Setzer–Neumann family over $\mathbb{Q}$.

Introduction

The theory of elliptic curves plays a crucial role in modern number theory. An important advance came with the systematic construction of tables, as done by the first author for elliptic curves defined over $\mathbb{Q}$ 7. The original purpose of the present article was to prove analogues over imaginary quadratic fields of class number 1 to well‐known bounds relating the conductor and discriminant of an elliptic curve over $\mathbb{Q}$ with prime power conductor. A useful application of such a result is an effective algorithm to construct tables of elliptic curves of prime power conductor, via solving Thue equations. While succeeding in carrying out our original aim, we also found several examples of phenomena for curves over imaginary quadratic fields that do not occur over $\mathbb{Q}$, and were able to classify all elliptic curves with prime power conductor and nontrivial torsion over the fields in question, extending the results of Miyawaki 23 for elliptic curves over $\mathbb{Q}$ and Shumbusho 32 over imaginary quadratic fields.

Over $\mathbb{Q}$, there are only finitely many elliptic curves with complex multiplication (CM curves) of prime power conductor $p^r$. The reason is, first, that there are finitely many imaginary quadratic orders $\mathcal{O}$ of class number 1, and second, that if $p$ divides the discriminant of $\mathcal{O}$, then all curves $E$ with endomorphism ring isomorphic to $\mathcal{O}$ have additive reduction at $p$, and the field where $E$ attains good reduction is not abelian. In particular, any other rational curve isomorphic to $E$ (that is, a twist of $E$) will also have bad reduction at $p$, so the prime power discriminant condition gives at most one curve per order.

Over an imaginary quadratic field $K$, this finiteness statement no longer holds true. If $E/K$ has CM by ${\mathcal{O}}_K$, the ring of integers of $K$, and $p$ ramifies in $K$, the curve attains good reduction over an abelian extension of the completion $K_p$. In particular, some twist of $E$ (quadratic, quartic or sextic, depending on the number of roots of unity in $K$) attains good reduction at $p$. However the required local extension does not come from a global one, so we do not get a curve over $K$ of conductor 1 (which do not exist), but an infinite family, all twists of each other, of CM curves of prime square conductor; moreover in each case the density of primes of $K$ whose squares arise in this way is in each case $1/\#{\mathcal{O}}_K^{\ast }$. In Section 3, we explain this phenomenon, and describe precisely all elliptic curves of prime power conductor and complex multiplication.

Another notable new phenomenon that occurs over imaginary quadratic fields is the existence of elliptic curves of prime conductor whose residual Galois representations modulo $p$ are irreducible but not absolutely irreducible. When $p$ is odd, this phenomenon does not occur for elliptic curves over $\mathbb{Q}$: the reason is the existence of complex conjugation in the absolute Galois group, whose image under the mod‐ $p$ representation is similar to $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$. Over an imaginary quadratic field, or any number field which is totally complex (that is, has no real embedding), the absolute Galois group has no such complex conjugation elements and this argument does not apply.

At the prime 2 the situation is more exciting! Over $\mathbb{Q}$, if $E$ is a semistable elliptic curve whose discriminant is a square, then its residual image is reducible. There are two different ways to prove such an assertion: one comes from the study of finite flat group schemes over $Spec(\mathbb{Z})$. The hypotheses ( $E$ being semistable and the discriminant being a square) imply by a theorem of Mazur (see Theorem 2.2) that $E[2]$ is a finite flat group scheme of type (2,2). As explained in 22, there are only four such group schemes, and only two of them satisfy that the determinant of the group scheme is isomorphic to the group scheme ${\mu }_2$; they both have an invariant subspace. A different approach comes from the study of the mod‐2 Galois representation itself: if it is irreducible, then by Ribet's level‐lowering result, there would exist a modular form of level 1 and weight 2, while if the image were irreducible but not absolutely irreducible (that is, it is the cyclic subgroup of order 3 in ${GL}_2({\mathbb{F}}_2)$), then there would exist a cubic extension of $\mathbb{Q}$ unramified outside 2; but such an extension does not exist.

Over an imaginary quadratic field $K$ of class number 1, in four cases such an extension does not exist either, so the same result holds (for modular elliptic curves, assuming an analogous level‐lowering conjecture). However for $K=\mathbb{Q}(\sqrt{-d})$ for $d\in \{11,19,43,67,163\}$ where 2 is inert in $K$, there is no reason for such Galois representation not to exist. In particular, they supply new examples of irreducible finite flat group schemes of type (2,2) over $Spec({\mathcal{O}}_K)$ (the irreducibility comes from the fact that irreducible 2‐groups are only the additive and the multiplicative ones as proved in [37, Corollary, p. 21], which give rational 2‐torsion). One simple example of this phenomenon comes from the elliptic curves over $\mathbb{Q}$ of conductor 11, which have minimal discriminant either $-11$ or $-{11}^5$; after base change to $K=\mathbb{Q}(\sqrt{-11})$ the conductor is still prime but the minimal discriminants are all now square. For an example which is not a base change, the curve 2.0.11.1‐47.1‐a1 (using its LMFDB label, see 20):

$$E:y^2+y=x^3+\alpha x^2-x,$$

where $\alpha =(1+\sqrt{-11})/2$, has prime conductor $\mathfrak{p}=(\pi )$ of norm 47, with $\pi =7-2\alpha$, and discriminant ${\pi }^2$, and is alone in its isogeny class. This phenomenon is studied in detail in Section 6. We suspect that there are infinitely many examples like this over each of these five fields.

Another interesting problem that arose is that of classifying elliptic curves (up to 2‐isogenies and twists) with a rational 2‐torsion point. The prime 2 is always hard to handle, and we develop tools (including a result characterizing such curves, see Theorem 5.4) that allow us to give a complete description of them. The ideas developed here could be adapted to more general situations, but the fact that $K$ has finitely many units appears to play an important role.

The main results of the article are Theorem 2.3 and Corollary 2.4, which establish Szpiro's conjecture for ‘modular’ prime power conductor elliptic curves over $K$, assuming a form of level‐lowering result (Conjecture 2). The general strategy of the proof is as follows: let $E/K$ be a modular elliptic curve (see Conjecture 1) of prime power conductor. If $E$ has potentially good reduction, there is a well‐known bound for its minimal discriminant $\mathcal{D}(E)$, so we can focus on the semistable case. If $\ell$ is a rational prime dividing the valuation of $\mathcal{D}(E)$, by a theorem of Mazur, $E[\ell ]$ is a finite flat group scheme. If the residual Galois representation is absolutely irreducible, then the level‐lowering conjecture implies the existence of a Bianchi modular form modulo $\ell$ of level 1 and weight 2 whose Galois representation matches that of $E$. Since for $\ell$ odd, there are no such forms over the fields in question, we get a contradiction. We want to emphasize that the level‐lowering result might be hard to prove for small primes (which divide the order of elliptic points). For this reason, we include an unconditional proof (not relying on modularity) for $\ell =2$ and 3 in Theorem 2.3 for the nine fields considered.

Assume otherwise that the residual image is absolutely reducible, that is, it is either reducible over ${\mathbb{F}}_{\ell }$, or irreducible but reducible over ${\mathbb{F}}_{{\ell }^2}$. For the first possibility, we prove that either such curves do not exist, or otherwise there is an isogenous curve whose discriminant valuation is prime to $\ell$ (the case $\ell =2$ being hardest). When $\ell$ is odd, the second case is eliminated by a detailed study of the residual Galois representation. Finally, for $\ell =2$ we prove that the discriminant valuation is at most 2, where (for certain of the fields $K$ only) the exceptional curves with square discriminant arise as described above.

The article is organized as follows. The first section contains a brief description of Bianchi modular forms and modularity of elliptic curves. It also contains the conjectural level‐lowering statement in the spirit of Ribet's result, which we expect to hold in the setting of Bianchi modular forms. An important difference in our statement comes from the fact that we do not expect forms of minimal level to always lift to characteristic zero: this is why the statement of Conjecture 2 is in terms of group cohomology over ${\mathbb{F}}_{\ell }$.

The second section contains the main theorem, and its proof when the image is absolutely irreducible. The third section studies elliptic curves over $K$ with complex multiplication. The main results includes a complete classification of all CM curves of odd prime power discriminant. The case of small image at an odd prime is treated in the fourth section, where elliptic curves over $K$ of odd prime power conductor with a rational point of odd order $\ell$ are considered. In Theorem 4.3 we prove that if $\ell =3$ and $K=\mathbb{Q}(\sqrt{-3})$, then any such curve either has discriminant valuation not divisible by 3, or its 3‐isogenous curve satisfies this property. For all other choices of $\ell$ and $K$, we give a finite list of possible curves: see Table 4.1.

Table 4.1. Prime power conductor curves with an $\ell$‐torsion point ( $\ell$ odd)

Label	Weierstrass Coefficients	$\ell$	$\mathcal{D}(E)$	Field
19.a2	$[0,1,1,-9,-15]$	3	$-{19}^3$	$1,7,11,19,43,163$
19.a3	$[0,1,1,1,0]$	3	$-19$	$1,7,11,19,43,163$
27.a2	$[0,0,1,-30,63]$	3	$-3^5$	$1,7,19,43,67,163$
27.a3	$[0,0,1,0,-7]$	3	$-3^9$	$1,7,19,43,67,163$
27.a4	$[0,0,1,0,0]$	3	$-3^3$	$1,7,19,43,67,163$
37.b2	$[0,1,1,-23,-50]$	3	${37}^3$	$2,19,43,163$
37.b3	$[0,1,1,-3,1]$	3	37	$2,19,43,163$
243.a2	$[0,0,1,0,20]$	3	$-3^{11}$	$1,7,19,43,67,163$
243.b2	$[0,0,1,0,2]$	3	$-3^7$	$1,7,19,43,67,163$
2.0.4.1‐757.1‐a1	$[1+i,-1+i,1,-15+5i,-17+6i]$	3	${(-26+9i)}^3$	1
2.0.4.1‐757.1‐a2	$[1+i,1+i,1,2i,i]$	3	$-26+9i$	1
2.0.8.1‐9.1‐CMa1	$[\sqrt{-2},1-\sqrt{-2},1,-1,0]$	3	${(-1-\sqrt{-2})}^6$	2
2.0.11.1‐9.3‐CMa1	$[0,\frac{1-\sqrt{-11}}{2},1,\frac{-5-\sqrt{-11}}{2},-2]$	3	${(\frac{-1+\sqrt{-11}}{2})}^6$	11
11.a2	$[0,-1,1,-10,-20]$	5	$-{11}^5$	$1,3,11,67,163$
11.a3	$[0,-1,1,0,0]$	5	$-11$	$1,3,11,67,163$
2.0.4.1‐25.3‐CMa1	$[1+i,i,i,0,0]$	5	${(2i+1)}^3$	1
2.0.3.1‐49.3‐CMa1	$[0,\frac{-3+\sqrt{-3}}{2},\frac{1+\sqrt{-3}}{2},\frac{1-\sqrt{-3}}{2},0]$	7	${(\frac{-1-3\sqrt{-3}}{2})}^2$	3

The fifth section studies curves of odd conductor over $K$ with a rational 2‐torsion point. An important result is Theorem 5.4, where a description of all such curves (up to twist) is given. This result is very general, and can be applied to different situations. Using this characterization, we describe all curves of prime power conductor ${\mathfrak{p}}^r$ with a rational 2‐torsion point, according to the following possible cases: either a twist of the curve has good reduction at $\mathfrak{p}$, or all twists have additive reduction at $\mathfrak{p}$, or the curve has a twist of multiplicative reduction. We prove that in the first case all curves have CM and come in families, described explicitly in Theorem 5.7, using the results from Section 3. For the additive ones, we prove in Theorem 5.9 that there are only some sporadic cases (also all CM), and for multiplicative ones, we prove that there are potentially (and probably) infinitely many, almost all belonging to a family analogous to the so‐called Setzer–Neumann family (as in 31) for curves defined over $\mathbb{Q}$. As in the classical case, all such curves have rank 0.

The important consequence of this detailed case by case study for our main result is that if $E$ is an elliptic curve defined over $K$ of odd prime conductor with a rational 2‐torsion point, then either $E$ or a 2‐isogenous curve has odd discriminant valuation.

Finally, in the last section, we study the case of curves whose Galois image modulo 2 is cyclic of order 3, giving curves of prime conductor but prime square discriminant, as explained above.

Notation and terminology

$K$ will denote an imaginary quadratic field $\mathbb{Q}(\sqrt{-d})$ of class number 1, with ring of integers ${\mathcal{O}}_K$. As is well known, there are nine such fields, with $d\in \{1,2,3,7,11,19,43,67,163\}$. Many of the results of Section 5 also apply to the case $K=\mathbb{Q}$.

We say that an ideal or element of ${\mathcal{O}}_K$ is odd if it is coprime to 2. Let $e_2$ be the ramification degree of 2 in $K/\mathbb{Q}$, so that $e_2=1$ except for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-2})$. Note also that 2 splits only in $K=\mathbb{Q}(\sqrt{-7})$. The primes dividing 2 play a crucial role in Section 5, where we will denote by $\mathfrak{q}$ a prime dividing 2, and by $\mathfrak{p}$ any prime ideal (which might divide 2 or not). The valuation at $\mathfrak{p}$ is denoted $v_{\mathfrak{p}}()$. We denote by $\epsilon$ a generator of the finite unit group ${\mathcal{O}}_K^{\ast }$ so that $\epsilon =-1$ except for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-3})$ when $\epsilon =\sqrt{-1}$, or $\epsilon$ is a sixth root of unity, respectively.

All curves explicitly mentioned will be labeled with their LMFDB label, see 20.

Remark.It is well known that given an elliptic curve $E/K$, if the class number of $K$ equals $1$, then $E$ has a global minimal model. Such model might not be unique, hence there is in general no notion of a minimal discriminant. However, over an imaginary quadratic field, all units are annihilated by 12, and hence the value of the minimal discriminant $\mathcal{D}(E)$ is well defined.

1 Bianchi modular forms and modularity of elliptic curves

For background on Bianchi modular forms and modularity of elliptic curves defined over imaginary quadratic fields, we refer to the survey article of Şengün 29 and the work of the first author 5, 6, 9. For our purposes we may restrict our attention to the space $S_2(\mathfrak{n})$ of Bianchi modular forms which are cuspidal and of weight 2 for the congruence subgroup ${\mathrm{\Gamma }}_0(\mathfrak{n})⩽{GL}_2({\mathcal{O}}_K)$, where the level $\mathfrak{n}\subseteq {\mathcal{O}}_K$ is an integral ideal of $K$. This space is a finite‐dimensional vector space equipped with a Hecke action. Its newforms subspace is spanned by eigenforms which are simultaneous eigenvectors for the algebra of Hecke operators. Bianchi modular forms can also be seen within the context of cohomological automorphic forms, since we have the isomorphism $S_2(\mathfrak{n})\cong H^1(Y_0(\mathfrak{n}),\mathbb{C})$, where $Y_0(\mathfrak{n})$ is the quotient of hyperbolic 3‐space ${\mathcal{H}}_3$ by ${\mathrm{\Gamma }}_0(\mathfrak{n})$. A more concrete description of these Bianchi modular forms is as real analytic functions ${\mathcal{H}}_3\to {\mathbb{C}}^3$ satisfying certain conditions. In the work of the second author and his students 5, 9 explicit methods were developed to compute the spaces $S_2(\mathfrak{n})$ over $K$ for each of the nine imaginary quadratic fields $K$ of class number 1. Recall that the orders of the isotropy groups of points of $Y_0(1)$ are only divisible by the primes 2 and 3, hence for $\ell ⩾5$ the cohomology $H^1(Y_0(1),{\mathbb{F}}_{\ell })$ matches the group cohomology $H^1({PGL}_2({\mathcal{O}}_K),{\mathbb{F}}_{\ell })=H^1({PGL}_2({\mathcal{O}}_K),\mathbb{Z})\otimes {\mathbb{F}}_{\ell }$ (see for example [1, Section 1.4]). Then the previous computations, together with results of 28, establish the following result:

Theorem 1.1.For each of the nine imaginary quadratic fields of class number 1, the space $S_2(1)$ of weight 2 cuspidal Bianchi modular forms of level 1 is trivial. Moreover for all primes $\ell ⩾5$ the space $H^1(Y_0(1),{\mathbb{F}}_{\ell })$ is also trivial.

It is known that these Bianchi modular forms have associated $\ell$‐adic Galois representations ${\rho }_{F,\ell }$. These were first constructed by Taylor et al. in 12, 38 with subsequent results by Berger and Harcos in 2. Below we only need to refer to the residual mod‐ $\ell$ representations $\overline{{\rho }_{F,\ell }}$.

For an elliptic curve $E$ defined over an imaginary quadratic field $K$, we say that $E$ is modular if $L(E,s)=L(F,s)$ for some $F\in S_2(\mathfrak{n})$ over $K$, where $\mathfrak{n}$ is the conductor of $E$. The following conjecture, a version of which was first made by Mennicke, is part of [29, Conjecture 9.1], following 6:

Conjecture 1.Let $E$ be an elliptic curve defined over an imaginary quadratic field $K$ of class number 1 that does not have complex multiplication by an order in $K$. Then $E$ is modular.

Some cases of the conjecture have been proved (see 10), and under some mild hypothesis it can be shown that any such curve is potentially modular (see [3, Theorem 1.1]). Curves with complex multiplication correspond to Hecke characters and belong to non‐cuspidal modular forms. These will be considered in Section 3.

Following the classical result of Ribet 25, we make the following conjecture.

Conjecture 2.Let $F$ be a weight 2 Bianchi newform of level ${\mathrm{\Gamma }}_0(\mathfrak{n}\mathfrak{p})$, with $\mathfrak{p}$ a prime number. Suppose that $\ell ⩾5$ is a prime for which the representation ${\rho }_{F,\ell }$ satisfies.

• (i) The residual representation is absolutely irreducible.
• (ii) The residual representation is finite at $\mathfrak{p}$.

Then there is an automorphic form $G\in H^1(Y_0(\mathfrak{n}),{\mathbb{F}}_{\ell })$ whose Galois representation is isomorphic to $\overline{{\rho }_{F,\ell }}$.

Some results in the direction of the conjecture are proven in 4. Note that the conjecture refers to an element in the mod $\ell$ cohomology. In contrast to modular forms for ${SL}_2(\mathbb{Z})$, for Bianchi modular forms the homology (and the cohomology) of $Y_0(\mathfrak{n})$ contains a big torsion part. By results of Scholze 27 the torsion forms do have Galois representations attached, but these are residual representations that in general do not admit a lift of level $\mathfrak{n}$ to characteristic zero, so the form in Conjecture 2 is not expected to be global.

2 Modular elliptic curves of prime power conductor

Recall the following result due to Serre and Mestre–Oesterlé (see 22).

Theorem 2.1.Let $E/\mathbb{Q}$ be a modular elliptic curve of prime conductor $p$. If $p>37$, then $\mathcal{D}(E)=\pm p$ up to 2‐isogenies (that is, either $\mathcal{D}(E)=\pm p$, or there exists a curve 2‐isogenous to $E$ over $\mathbb{Q}$ with discriminant $\pm p$).

A key ingredient in the proof is the following result of Mazur.

Theorem 2.2. (Mazur)Let $K$ be a number field, let $E/K$ be a semistable elliptic curve with discriminant $\mathcal{D}(E)$, and let $\ell$ be a prime number. Then $E[\ell ]$ is a finite flat group scheme if and only if $\ell \mid v_{\mathfrak{q}}(\mathcal{D}(E))$ for all primes $\mathfrak{q}$ in ${\mathcal{O}}_K$, that is, if the ideal $(\mathcal{D}(E))$ is an $\ell$th power.

Proof.See [21, Proposition 9.1]. $\square$

If $E/K$ is an elliptic curve of prime conductor $\mathfrak{p}$, and $\ell$ is a prime dividing the valuation $v_{\mathfrak{p}}(\mathcal{D}(E))$ of $\mathcal{D}(E)$, then $E[\ell ]$ is a finite flat group scheme over $Spec({\mathcal{O}}_K)$ by Mazur's theorem. To apply a level‐lowering result such as Conjecture 2, one also needs a big image hypothesis, that is, that the residual representation at $\ell$ is absolutely irreducible. While working over the rationals, either the residual image is reducible, in which case the curve has a $\mathbb{Q}$‐point (which does not happen for $\ell >7$) or it is absolutely irreducible. The reducible case for small primes can be discarded by a result of Fontaine [11, Theorem B].

One of the main results of this article is a generalization of Theorem 2.1 to $K$.

Theorem 2.3.Let $K$ be an imaginary quadratic field of class number 1. Let $E/K$ be a modular elliptic curve of prime conductor. Assume that Conjecture 2 holds. Then there exists a curve isogenous to $E$ over $K$ with prime or prime square discriminant.

Proof.Let $\mathfrak{p}$ be the conductor of $E$. For the case $\mathfrak{p}\mid 2$, Koutsianas used the methods of 16 to compute for us all elliptic curves with conductor a power of $\mathfrak{p}$, and found that over each of the nine fields there are none with conductor $\mathfrak{p}$. This is consistent with the tables of automorphic forms of 5, 9.

From now on we may assume that $\mathfrak{p}\nmid 2$. Suppose that $\mathcal{D}(E)$ is not prime, and let $\ell$ be a prime number dividing $v_{\mathfrak{p}}(\mathcal{D}(E))$. Recall that by semistability and Mazur's result, the Galois module $E[\ell ]$ is unramified away from $\ell$, and also that its determinant is the $\ell$th cyclotomic character. We claim that the hypotheses of the theorem imply that for every prime number $\ell$ dividing $v_{\mathfrak{p}}(\mathcal{D}(E))$, the module $E[\ell ]$ is not absolutely irreducible.

Assume the claim. For odd $\ell$, Theorem 2.5 below implies that either $E$, or an $\ell$‐isogenous curve, has a rational torsion point of order $\ell$. In Section 4 all curves of odd prime conductor with an $\ell$‐torsion point are computed (see Table 4.1) with the exception of curves with a rational 3‐torsion point over $\mathbb{Q}(\sqrt{-3})$. It is easy to verify from Table 4.1 that all curves of prime conductor listed there have (up to an $\ell$‐isogeny) discriminant with valuation prime to $\ell$. Finally, Theorem 4.3 proves that any curve with a rational 3‐torsion point over $\mathbb{Q}(\sqrt{-3})$ either has discriminant valuation prime to 3, or a 3‐isogenous curve does. Hence, up to isogeny, the claim (for all odd $\ell$) implies that $v_{\mathfrak{p}}(\mathcal{D}(E))$ has no odd prime factors.

Suppose that $v_{\mathfrak{p}}(\mathcal{D}(E))$ is even. The hypothesis implies that $\mathcal{D}(E)=\epsilon {\pi }^2$ for some $\epsilon \in {\mathcal{O}}_K^{\ast }$ and $\pi \in {\mathcal{O}}_K$. If $E[2]$ is absolutely irreducible, then $\mathcal{D}(E)$ is not a square, and the extension $K(\sqrt{\epsilon })/K$ has degree 2 and is unramified away from 2. However, using explicit class field theory (as implemented in 24), we may check that none of the nine fields $K$ has an extension $L/K$ with Galois group ${GL}_2({\mathbb{F}}_2)\cong S_3$ which is unramified away from 2 and has quadratic subfield $K(\sqrt{\epsilon })$. Hence $E[2]$ is not absolutely irreducible, establishing the claim for $\ell =2$. Now either $E[2]$ is reducible, in which case Corollary 5.13 shows that up to 2‐isogeny $v_{\mathfrak{p}}(\mathcal{D}(E))=1$; or $E[2]$ is irreducible (but not absolutely irreducible) in which case Theorem 6.2 implies that $v_{\mathfrak{p}}(\mathcal{D}(E))=2$.

It remains to prove the claim for odd primes $\ell$ which divide $v_{\mathfrak{p}}(\mathcal{D}(E))$. We consider first the case $\ell =3$, where as with $\ell =2$ an elementary argument establishes the claim unconditionally. If $K\ne \mathbb{Q}(\sqrt{-3})$, then every unit in $K$ is a cube, so the hypothesis that $3\mid v_{\mathfrak{p}}(\mathcal{D}(E))$ implies that $\mathcal{D}(E)$ is a cube in $K$. In case $K=\mathbb{Q}(\sqrt{-3})$ we can only say that $\mathrm{\Delta }(E)$ is a unit times a cube, and we may have $K(\sqrt[3]{\mathcal{D}(E)})=\mathbb{Q}({\zeta }_9)$. In general, $K(\sqrt[3]{\mathcal{D}(E)})$ is the subfield of $K(E[3])$ cut out by the Sylow 2‐subgroup, which has index 3 and order 16, in ${GL}_2({\mathbb{F}}_3)$; this subgroup is the normalier of the non‐split Cartan subgroup. Hence the field $K(E[3])$ is obtained by a succession of quadratic extensions of $K(\sqrt[3]{\mathcal{D}(E)})$ each unramified away from 3. Neither $\mathbb{Q}(\sqrt{-3})$ nor $\mathbb{Q}({\zeta }_9)$ possesses any such quadratic extension, so for $K=\mathbb{Q}(\sqrt{-3})$ the image of the mod‐3 representation is either trivial (when $\mathrm{\Delta }$ is a cube) or of order 3, in which case it is conjugate to $\left(\begin{array}{cc}1& \ast \\ 0& 1\end{array}\right)$.

For the other fields, the residual image is contained in the normalizer of the non‐split Cartan subgroup, of order 16. If $E[3]$ were absolutely irreducible, then the residual image could not be contained in the non‐split Cartan subgroup itself, since that is absolutely reducible; hence there would be a (non‐trivial) quadratic character unramified outside 3 whose kernel has residual image equal to the non‐split Cartan subgroup. However, for each of the eight fields $K$ in question, the only quadratic extension ramified only above 3 is $K({\zeta }_3)$, which cuts out a different index 2 subgroup of the non‐split Cartan normalizer, since the non‐split Cartan contains matrices of determinant $-1$. (Here we are using the fact that the determinant of the representation is the cyclotomic character.) This contradiction shows that $E[3]$ is not absolutely irreducible.

Finally, we prove the claim when $v_{\mathfrak{p}}(\mathcal{D}(E))$ is divisible by a prime $\ell ⩾5$. By Theorem 2.2, $E[\ell ]$ is a finite flat group scheme. If the residual representation of $E$ at the prime $\ell$ is absolutely irreducible, the modularity assumption on $E$ gives the hypothesis of Conjecture 2, so there must exist a mod $\ell$ Bianchi modular form of level 1 whose Galois representation matches the residual representation of $E$ modulo $\ell \mathit{\text{;}}$ this contradicts Theorem 1.1. $\square$

As a corollary, we have the following version of Szpiro's conjecture.

Corollary 2.4.Let $K$ be an imaginary quadratic field of class number 1. Assume that Conjecture 2 holds. Let $E/K$ be a modular elliptic curve of prime power conductor. Then $\mathcal{N}({\mathcal{D}}_{min}(E))⩽\mathcal{N}{(cond(E))}^6$, except for the curve 11.a2 over $\mathbb{Q}(\sqrt{-11})$, whose conductor has valuation 1 while its discriminant has valuation 10.

Proof.If $E$ has potentially good reduction, then the result is well known, see for example [33, p. 365, Table 4.1]. Otherwise, either $E$ is semistable or is a quadratic twist of a semistable curve. Recall that there are no curves of prime conductor dividing 2, so we can restrict to odd conductor. It is enough to consider the semistable case, since if $E$ has prime conductor $\mathfrak{p}$ and satisfies $\mathcal{N}({\mathcal{D}}_{min}(E))⩽{\mathcal{N}\mathfrak{p}}^6$, the twisted curve $E^{\prime }$ satisfies $\mathcal{N}({\mathcal{D}}_{min}(E^{\prime }))=\mathcal{N}({\mathcal{D}}_{min}(E))·{\mathcal{N}\mathfrak{p}}^6$ and $cond(E^{\prime })={\mathfrak{p}}^2$.

By Theorem 2.3 the result holds for some curve in the isogeny class. Furthermore, when the isogeny class of a semistable curve has at most 2 elements, Theorem 6.2 and Corollary 5.13 prove that the discriminant valuation is at most 2.

When there is a 3‐isogeny, the isogenous curve has discriminant valuation 3 or 6 (depending on whether or not the conductor is ramified in $K$), and there is a unique semistable case with a 5‐isogeny (see Theorem 4.7), which is precisely the curve 11.a2. $\square$

We end this section with a result used in the proof of Theorem 2.3.

Theorem 2.5.Let $K$ be an imaginary quadratic field of class number 1 and $E/K$ be a semistable elliptic curve. Let $\ell ⩾3$ be a prime number such that the residual representation of $E$ at $\ell$ is not absolutely irreducible. Then either $E$, or a curve $\ell$‐isogenous to $E$ over $K$, has a point of order $\ell$ defined over $K$.

Proof.The residual representation associated to $E$ takes values in $GL(2,{\mathbb{F}}_{\ell })$, and may be either reducible (over ${\mathbb{F}}_{\ell }$), irreducible but absolutely reducible (that is, reducible over ${\mathbb{F}}_{{\ell }^2}$) or absolutely irreducible. The hypothesis in the theorem excludes only the absolutely irreducible case. Over $\mathbb{Q}$, or any number field with at least one real place, the second case cannot occur, due to the action of complex conjugation: any invariant line over ${\mathbb{F}}_{{\ell }^2}$ must be defined over ${\mathbb{F}}_{\ell }$. In our situation, we need to consider both the reducible and the absolutely reducible cases separately.

• Reducible case: Without loss of generality, the residual representation is of the form

  $$\overline{{\rho }_{E,\ell }}\simeq \left(\begin{array}{cc}{\theta }_1& \ast \\ 0& {\theta }_2\end{array}\right),$$(2.1)
  for ${\theta }_i$ characters of $Gal(\overline{K}/K)$ such that ${\theta }_1{\theta }_2={\chi }_{\ell }$ (the cyclotomic character). Since $E$ is semistable, the conductors of the characters ${\theta }_i$ are supported in $\ell$ (see [18, Lemme 1]). Also since $K$ has class number 1, the only unramified character is the trivial character. Hence we must show that at least one of the characters is unramified, since ${\theta }_1$ trivial implies that $E$ has a point of order $\ell$ while if ${\theta }_2$ is trivial then the isogenous curve has such a point. If $\ell$ is unramified in $K$, then the result follows from the proof of [18, Corollaire 1, p. 249, 250], as we now explain.

  If $\ell$ is inert in $K$, then only one of the ${\theta }_i$ can be ramified at $\ell$ (see [18, Lemme 1]), hence the statement. Then we can restrict to the case when both characters are ramified at a prime dividing $\ell$.

  If $\ell ={\mathfrak{l}}_1{\mathfrak{l}}_{\mathfrak{2}}$ splits, we can assume that ${\theta }_i$ has conductor ${\mathfrak{l}}_i$ since by [18, Lemme 1] they cannot both be ramified at the same prime. Then on one hand the restriction of ${\theta }_i$ to the inertia group at ${\mathfrak{l}}_i$ is the cyclotomic character, so ${\theta }_i(-1)=-1$, and on the other hand it is a character of ${({\mathcal{O}}_K/{\mathfrak{l}}_i)}^{\times }$ which is trivial in ${\mathcal{O}}_K^{\times }$ (as it factors through the Artin map), so ${\theta }_i(-1)=1$. This is impossible since $\ell \ne 2$.

  Lastly, suppose that $\ell$ ramifies in $K$, so $\ell \equiv 3(mod4)$ and $K=\mathbb{Q}(\sqrt{-\ell })$. In particular, the restriction of the cyclotomic character to $Gal(\overline{K}/K)$ has (odd) order $(\ell -1)/2$. Let $\mathfrak{l}$ denote the prime of $K$ dividing $\ell$ and $I_{\mathfrak{l}}$ the associated inertia subgroup. Then the ${\theta }_i$ are both characters of level 1 at $\mathfrak{l}$ (since characters of level 2 have irreducible image), ${\theta }_1{\theta }_2={\chi }_{\ell }$ and $E$ has good supersingular reduction at $\ell$. In the notation of 30, let $a_{\ell }$ denote the $\ell$th coefficient in the series for multiplication by $\ell$ in the formal group of $\stackrel{\sim }{E}$, the reduced curve over ${\mathcal{O}}_K/\mathfrak{l}\cong {\mathbb{F}}_{\ell }$. By [30, Proposition 10, p. 272], if $v_{\ell }(a_{\ell })>1$, then ${\theta }_i{|}_{I_{\mathfrak{l}}}$ for $i=1,2$ would both be squares of fundamental characters of level 2, whose image is not in ${\mathbb{F}}_{\ell }$; hence the valuation is 1, and the same proposition implies that ${\theta }_i{|}_{I_{\mathfrak{l}}}$ equals the fundamental character of level 1 for $i=1,2$, and $\ast$ is unramified at $\ell$. So ${\theta }_1/{\theta }_2$ is unramified, hence trivial, and so ${\theta }_1={\theta }_2$ and ${\theta }_1^2={\chi }_{\ell }$.

  The order of ${\theta }_1$ equals $\ell -1$ and it factors through a cyclic degree $(\ell -1)$ extension of $K$ unramified outside $\ell$, containing ${\zeta }_{\ell }$ (the $\ell$th roots of unity). This implies in particular the existence of a quadratic extension of $K$ unramified outside $\ell$. It can be easily verified (using 24 for example) that there are no such quadratic extensions, hence this case cannot occur.

• Irreducible but absolutely reducible case. In this case, there is a character of $K$ of order ${\ell }^2-1$ or $\frac{{\ell }^2-1}{2}$ unramified outside $\ell$ (by the same argument as before). This implies the existence of a quadratic extension of $K$ unramified outside $\ell$ which, as we saw in the previous case, does not exist. $\square$

3 Curves with complex multiplication

Let $K$ denote one of the nine imaginary quadratic fields with class number 1. Let $E/K$ be an elliptic curve with complex multiplication by an order $\mathcal{O}$ in $K$. Since $K$ has class number 1, the $j$‐invariant $j(E)$ is rational, in fact in $\mathbb{Z}$, so $E$ is a twist of the base extension of an elliptic curve defined over $\mathbb{Q}$.

For each field $K$, we fix one such curve $E$, choosing it to have bad reduction only at the unique ramified prime $\mathfrak{p}$ of $K$, and to have endomorphism ring isomorphic to the maximal order ${\mathcal{O}}_K$. Every other elliptic curve with CM by an order in $K$ is then isogenous to a twist of this base curve $E$. For $d=1,2,3,7,11,19,43,67,163$, respectively, we take the base curve to be the base change to $K$ of the elliptic curve over $\mathbb{Q}$ with LMFDB label 64.a4, 256.d1, 27.a4, 49.a4, 121.b2, 361.a2, 1849.b2, 4489.b2, 26569.a2, respectively. Our goal in this section is to determine which twists of these base curves $E$ have odd prime power conductor, recalling that for $d=1$ and $d=3$, respectively, we must consider quartic and sextic twists, not only quadratic twists.

Remark 3.1.For $d>3$ the base curve listed is uniquely determined (up to isomorphism over $K$) by the condition that it has CM by ${\mathcal{O}}_K$ and bad reduction only at the ramified prime $\mathfrak{p}=(\sqrt{-d})$. However for $d=1,2,3$ there are several choices. In the results of this section, we consider elliptic curves with prime power conductor as explicit twists of the base curve, so it is important to fix this choice.

The automorphic form attached to $E/K$ consists of the sum of two conjugate Hecke Grossencharacters $\{\chi ,\overline{\chi }\}$ of infinity types (1,0) and (0,1), and conductor $\mathfrak{n}$ which is a power of $\mathfrak{p}$. In particular, $cond(E)={\mathfrak{n}}^2$. Note that the Grossencharacters take values in $K^{\times }$. The character $\chi$ is unramified outside $\mathfrak{p}$, so has conductor a power of $\mathfrak{p}$, and the local character ${\chi }_{\mathfrak{p}}$ restricted to ${\mathcal{O}}_{\mathfrak{p}}^{\times }$ is a finite character taking values in the roots of unity of $K$. In particular, it is quadratic except for $d=-1,-3$.

3.1 $K=\mathbb{Q}(\sqrt{-d})$ with $d\ne 1,3$

The character $\chi$ is quadratic, and unramified outside the prime $\mathfrak{p}=(\sqrt{-d})$, the unique ramified prime of $K$. If we twist $E$ by a quadratic character whose ramification at $\mathfrak{p}$ matches that of $\chi$, we get a curve with good reduction at $\mathfrak{p}$. Note that there is no global quadratic Grossencharacter unramified outside $\mathfrak{p}$ which locally matches ${\chi }_{\mathfrak{p}}$, as the Archimedean part of all such characters is trivial. In particular, although we can move the ramification by twisting, there is no twist with everywhere good reduction.

Remark 3.2.Rational 2‐torsion is preserved under twisting, so the quadratic twists of $E$ have rational 2‐torsion if and only if $E$ does. The fields $K$ for which the curves $E$ have a $K$‐rational two torsion point are $K=\mathbb{Q}(\sqrt{-d})$ for $d=2,7$.

Theorem 3.3.Let $K=\mathbb{Q}(\sqrt{-d})$, with $d\ne 1,3$, and let $E/K$ be the base elliptic curve with CM by ${\mathcal{O}}_K$ defined above. Then all elliptic curves with complex multiplication over $K$ of odd prime power conductor are isogenous to $E$ or to the quadratic twist of $E$ by $\pi \sqrt{-d}$, where $\pi$ is a prime such that $\pi \equiv u^2\sqrt{-d}(mod4)$ for $d\ne 2$, respectively $\pi \equiv u^2(1+\sqrt{-d})(mod4)$ for $d=2$, with $u$ odd. In particular, for $d=7$, the condition reads $\pi \equiv \sqrt{-7}(mod4)$, for $d=2$ the condition reads $\pi \equiv \pm 1+\sqrt{-2}(mod4)$ and for $d⩾11$, the condition reads $\pi \equiv w^{2k}\sqrt{-d}(mod4)$, for $0⩽k⩽2$, where $w=\frac{1+\sqrt{-d}}{2}$.

Before giving the proof, we need an auxiliary result.

Lemma 3.4.Let $K$ be a 2‐adic field, and $\alpha \in {\mathcal{O}}_K$ be a 2‐adic integer which is a unit and is not a square. Then the extension $K(\sqrt{\alpha })$ is unramified if and only if there exist a unit $u\in {\mathcal{O}}_K$ such that $u^2\equiv \alpha (mod4)$.

Proof.Let $L=K(\sqrt{\alpha })$ be the quadratic extension. The ring ${\mathcal{O}}_K[\sqrt{\alpha }]\subset {\mathcal{O}}_L$ has discriminant $4\alpha$ over ${\mathcal{O}}_K$. Then the extension ${\mathcal{O}}_L$ is unramified if and only if $[{\mathcal{O}}_L:{\mathcal{O}}_K[\sqrt{\alpha }]]=2$, if and only if $\frac{u+v\sqrt{\alpha }}{2}\in {\mathcal{O}}_L$ for some $u,v\in {\mathcal{O}}_K^{\ast }$. The minimal polynomial of any such element is $x^2-ux+\frac{u^2-\alpha v^2}{4}$, hence the index is 2 if and only if there exist units $u,v$ in ${\mathcal{O}}_K$ such that $u^2\equiv \alpha v^2(mod4)$. Multiplying by the inverse of $v$ we get the result. $\square$

Proof of Theorem 3.3.Since all curves with complex multiplication over $K$ are isogenous to a quadratic twist of $E$, we are led to determine which quadratic twists have bad reduction at exactly one odd prime. Any global character corresponds to a quadratic extension $K(\sqrt{\alpha })$. If the twist has good reduction at $\mathfrak{p}$, then $\mathfrak{p}\mid \alpha$ (and the twist attains good reduction at $\mathfrak{p}$), and the curve will have bad reduction at all other primes dividing $\alpha$. Thus $(\alpha )=\mathfrak{p}\mathfrak{q}$, for $\mathfrak{q}$ an odd prime. Finally, we need to check whether the character is unramified at 2, which follows from Lemma 3.4, and a description in each case of the squares modulo 4.

Explicitly, for odd $d$ we require $\alpha =\pi \sqrt{-d}$, where $\mathfrak{q}=(\pi )$, such that $\alpha \equiv u^2(mod4)$. For $d=7$ the only odd square modulo 4 is 1. For $d⩾11$, since 2 is inert in $K$ the odd squares modulo 4 are the squares of the odd residues modulo 2, which are $1,w,w^2$. For $d=2$, the twist of the base curve by $(1+\sqrt{-2})\sqrt{-2}$ has odd conductor ${(1+\sqrt{-2})}^2$, so we must twist by $\pi \sqrt{-2}$ where $\pi \equiv (1+\sqrt{-2})u^2(mod4)$; since the odd squares modulo 4 are 1 and $-1+2\sqrt{-2}$ we obtain the condition stated. $\square$

Remark 3.5.In each case in Theorem 3.3, the condition on $\pi$ is satisfied by one quarter of the odd residue classes modulo 4. Since $\mathfrak{q}=(\pi )$ has two generators $\pm \pi$, our construction gives curves of conductor ${\mathfrak{q}}^2$ for half the odd primes of $K$.

3.2 $K=\mathbb{Q}(\sqrt{-1})$

The elliptic curve with complex multiplication by $\mathbb{Z}[\sqrt{-1}]$ is

$$E:y^2=x^3+x\text{with}\text{label}\text{64a4}.$$

Its conductor over $K$ equals ${\mathfrak{p}}^8$, where $\mathfrak{p}=(1+\sqrt{-1})$. In particular, ${\chi }_{\mathfrak{p}}$ has conductor ${\mathfrak{p}}^4$ and order 4. Note that since the automorphism group of $E$ is cyclic of order 4 we must consider quartic twists. For $\alpha \in K^{\ast }$, the quartic twist of $E$ by $\alpha$ equals

$$E_{\alpha }:y^2=x^3+\alpha x.$$(3.1)

Note that this operation does not coincide with the twist of the L‐series by a quartic character (as such L‐series do not satisfy a functional equation). Indeed, if $E$ corresponds to the automorphic form $\chi \oplus \overline{\chi }$ (where $\chi$ is a Grossencharacter), then $E_{\alpha }$ corresponds to the automorphic form $\chi \psi \oplus \overline{\chi }\overline{\psi }$, where $\psi ={(\frac{}{\alpha })}_4$ (the quartic Legendre symbol). It is still true that the curve $E_{\alpha }$ is isomorphic to $E$ over the extension $K(\sqrt[4]{\alpha })$.

Remark 3.6.All the quartic twists of $E$ have a non‐trivial $K$‐rational 2‐torsion point.

We first need a local result about when a pure quartic extension is unramified above 2.

Lemma 3.7.Let $K={\mathbb{Q}}_2(\sqrt{-1})$, and let $\alpha \in {\mathcal{O}}_K$ be a unit. Then the extension $K(\sqrt[4]{\alpha })$ is unramified over $K$ if and only if $\alpha \equiv 1,1+4\sqrt{-1}(mod8)$.

Proof.The extension $K(\sqrt[4]{\alpha })$ depends on $\alpha$ up to fourth powers, that is, two elements give the same extension if and only if they differ by a fourth power. By Hensel's lemma, an odd element of ${\mathbb{Q}}_2(\sqrt{-1})$ is a fourth power if and only if it is congruent to 1 modulo ${(1+\sqrt{-1})}^7$, hence the extension is characterized by $\alpha$ modulo ${(1+\sqrt{-1})}^7$. Also, $K(\sqrt[4]{\alpha })$ is unramified if and only if it is contained in the unique unramified extension of $K$ of degree 4, which is $K({\zeta }_5)=K(\sqrt[4]{1+4\sqrt{-1}})$, as can be easily checked. Thus, for $K(\sqrt[4]{\alpha })$ to be unramified, $\alpha$ must be congruent to a power of $1+4\sqrt{-1}$, that is, $\alpha \equiv 1,1+4\sqrt{-1},9,\text{or}9+4\sqrt{-1}(mod{(1+\sqrt{-1})}^7)$; this simplifies to $\alpha \equiv 1,1+4\sqrt{-1}(mod8)$ as stated. $\square$

Theorem 3.8.Let $K=\mathbb{Q}(\sqrt{-1})$ and let $E/K$ be the elliptic curve 64.a4. Then all elliptic curves with complex multiplication over $K$ of odd prime power conductor are isogenous to the quartic twist of $E$ by $\pi$, where $\pi$ is a prime power such that $\pi \equiv -1\pm 2\sqrt{-1}(mod8)$.

Proof of Theorem 3.8.Let $\pi =-1+2\sqrt{-1}$. One may check that the quartic twist of $E$ by $\pi$ has good reduction at 2 and bad additive reduction at $(-1+2\sqrt{-1})$. Then any quartic twist of $E$ of odd conductor is a twist of $E_{-1+2\sqrt{-1}}$ by a quartic character of odd conductor, which by Lemma 3.7 correspond to elements which are congruent to 1 or $1+4\sqrt{-1}(mod8)$. Multiplying by $-1+2\sqrt{-1}$ gives the classes $-1\pm 2\sqrt{-1}(mod8)$ as stated. $\square$

Remark 3.9.Of all odd primes $\mathfrak{q}$ of $K$, one quarter have a generator $\pi$ satisfying the condition in Theorem 3.8. Hence our construction gives elliptic curves of conductor ${\mathfrak{q}}^2$ for one quarter of all primes $\mathfrak{q}$ of $K$.

3.3 $K=\mathbb{Q}(\sqrt{-3})$

The elliptic curve with complex multiplication by $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ is the curve 27.a4, with (non‐minimal) equation

$$E:y^2=x^3+16.$$

Its conductor over $K$ is ${\mathfrak{p}}^4$, where $\mathfrak{p}=(\sqrt{-3})$. In particular, ${\chi }_{\mathfrak{p}}$ has conductor ${\mathfrak{p}}^2=(3)$ and order 6 (note that ${({\mathcal{O}}_K/3)}^{\times }\simeq \mathbb{Z}/6\mathbb{Z}$). Since $E$ has automorphism group of order 6, we must consider sextic twists, where the sextic twist by $\alpha \in K^{\ast }$ of the previous model is

$$E_{\alpha }:y^2=x^3+16\alpha .$$(3.2)

Similar considerations as for the quartic twists apply. In particular, the curve $E_{\alpha }$ is isomorphic to $E$ over the extension $K(\sqrt[6]{\alpha })$; such twists are needed to cancel the CM character ${\chi }_{\mathfrak{p}}$.

As before, we need local results, now at both 2 and 3:

Lemma 3.10.Let $K=\mathbb{Q}(\sqrt{-3})$, let $w=(1+\sqrt{-3})/2\in K$ be a sixth root of unity, and let $\alpha \in {\mathcal{O}}_K$ be a 2‐adic and 3‐adic unit. Then the extension $K(\sqrt[6]{\alpha })/K$ is unramified over both 2 and 3 if and only if

• (i) $\alpha \equiv 1,w^2$, or $w^4(mod4)$ (equivalently, $\alpha$ is congruent to a square modulo 4); and
• (ii) $\alpha \equiv \pm 1(mod{\sqrt{-3}}^3)$ (equivalently, $\alpha$ is congruent to a cube modulo ${\sqrt{-3}}^3$).

Proof.Since $K(\sqrt[6]{\alpha })=K(\sqrt{\alpha },\sqrt[3]{\alpha })$ we require both $K(\sqrt{\alpha })/K$ and $K(\sqrt[3]{\alpha })/K$ to be unramified. The first is certainly unramified over 3, and over 2 we may apply Lemma 3.4 to obtain the first condition stated.

Similarly, $K(\sqrt[3]{\alpha })/K$ is always unramified over 2, so we need the condition for it to be unramified also over 3. By Hensel's lemma, a unit of ${\mathbb{Q}}_3(\sqrt{-3})$ is a cube if and only if it is congruent to $\pm 1$ modulo 9, hence the extension is characterized by $\alpha$ modulo 9. We may check that $K(\sqrt[3]{{\alpha }_1})$ is unramified, for ${\alpha }_1=2+3w$. Hence $K(\sqrt[3]{\alpha })/K$ is unramified at 3 if and only if $\alpha \equiv \pm 1,\pm {\alpha }_1,\pm {\alpha }_1^2(mod9)$, which is if and only if $\alpha \equiv \pm 1(mod{\sqrt{-3}}^3)$. $\square$

Theorem 3.11.Let $K=\mathbb{Q}(\sqrt{-3})$, and let $E/K$ be the elliptic curve 27.a4. Then all elliptic curves with complex multiplication over $K$ of odd prime power conductor are isogenous to $E$ or to the sextic twist $E_{\alpha }$ of $E$ by $\alpha ={\sqrt{-3}}^3\pi$, where $\pi$ is a prime power such that:

• (i) $\pi \equiv \sqrt{-3},\sqrt{-3}{w}^2$, or $\sqrt{-3}{w}^4(mod4)$, and
• (ii) $\pi \equiv \pm 4(mod{\sqrt{-3}}^3)$,

where $w=(1+\sqrt{-3})/2$ is a sixth root of unity.

Proof. $E$ itself has good reduction except at $\sqrt{-3}$; by Lemma 3.10, the sextic twist $E_{\alpha }$ will also have good reduction at 2 provided that $\alpha$ is an odd square modulo 4, equivalently $\alpha \equiv 1,w^2,w^4(mod4)$. Hence the first condition on $\pi$ ensures that $E_{\alpha }$ has good reduction except at $\pi$ and (possibly) at $\sqrt{-3}$.

The twist $E_{{\alpha }_1}$ with ${\alpha }_1={\sqrt{-3}}^3·4$ has good reduction at $\sqrt{-3}$. Hence by Lemma 3.10, $E_{\alpha }$ has good reduction at $\sqrt{-3}$ if $\alpha /{\alpha }_1$ is a cube modulo ${\sqrt{-3}}^3$, or equivalently $\alpha /{\alpha }_1\equiv \pm 1(mod{\sqrt{-3}}^3)$. This is ensured by the second condition, since $\pi /4=\alpha /{\alpha }_1$. $\square$

Remark 3.12.The curves in the previous family never have a rational 2‐torsion point, since $2\alpha$ is not a cube.

Remark 3.13.Of all primes of $K$ other than (2) and $(\sqrt{-3})$, half have a generator $\pi$ satisfying the 2‐adic condition in Theorem 3.11, and one third have a generator satisfying the 3‐adic condition. Hence our construction gives elliptic curves of conductor ${\mathfrak{p}}^2$ for one‐sixth of all primes $\mathfrak{p}$ of $K$.

3.3.1 Curves with complex multiplication by $K$ over an imaginary quadratic field $L$

A natural question is what happens if we consider a curve $E$ with complex multiplication by an order in $K$, over a possibly different imaginary quadratic field $L\mathit{\text{:}}$ Are there twists of $E$ with good reduction at primes dividing $cond(E)$?

The proofs of the previous results are of a local nature, hence if $L$ has the same completion at a prime dividing $cond(E)$ as $K$ we are in exactly the same situation.

Theorem 3.14.Let $E/\mathbb{Q}$ be an elliptic curve of conductor $p^r$ with complex multiplication by an order in $K$. Let $L=\mathbb{Q}(\sqrt{-t})$ be an imaginary quadratic field different from $K$ and $\mathfrak{p}$ a prime ideal of $L$ dividing $p$. If the completion of $L$ at $\mathfrak{p}$ is isomorphic to the completion of $K$ at the prime dividing $p$, then there exists $\alpha \in L$ such that:

• (i) if $K=\mathbb{Q}(\sqrt{-d})$, $d\ne 1,3$, then the quadratic twist of $E/L$ by $\sqrt{-t}\alpha$ has good reduction at $\mathfrak{p}$;
• (ii) if $K=\mathbb{Q}(\sqrt{-1})$, let $\sqrt{-1}$ denote an element of $L$ whose square is congruent to $-1$ modulo 8. Then the quartic twist of the curve 64.a4 by $\alpha$ has good reduction at 2 for $\alpha \equiv -1\pm 2\sqrt{-1}(mod8)$;
• (iii) if $K=\mathbb{Q}(\sqrt{-3})$, let $\sqrt{-3}$ denote an element of $L$ whose square is congruent to $-3$ modulo 9. Then the sextic twist of the curve 27.a4 by $\alpha$ has good reduction at 3 for $\alpha \equiv \pm 4{\sqrt{-3}}^3(mod{\sqrt{-3}}^3)$.

On the other hand, if the completions are not isomorphic, no such twist exists.

Proof.The proof of the first facts mimics that of Theorems 3.3, 3.8 and 3.11, as the completions being isomorphic implies that the local reduction types are the same. Suppose that the local completions are not isomorphic. Consider the Weil representation at $p$ attached to our elliptic curve $E/\mathbb{Q}$ (recall that CM elliptic curves have no monodromy, hence we do not need to consider the whole Weil–Deligne representation). Then it is easy to check that the image of inertia at $p$ equals:

• (i) a cyclic group of order 4 if $d\ne 1,3$,
• (ii) the dihedral group of order 8 if $d=1$,
• (iii) the dihedral group of order 12 if $d=3$.

Recall that the curve $E/L$ will have good reduction at $\mathfrak{p}$ if the restriction of the Weil representation to the inertia subgroup of $L$ at $\mathfrak{p}$ is trivial.

In the first case, since the completion of $L$ at $\mathfrak{p}$ is not isomorphic to the completion of $K$ at $\mathfrak{p}$, the restriction to the inertia subgroup of $L$ at $\mathfrak{p}$ still has order 4. But since $d\ne 1$, we cannot take quartic twists, hence we cannot cancel the ramification, and all such curves $E/L$ will have bad reduction at $\mathfrak{p}$. In the other two cases, the image of the inertia subgroup of $\mathbb{Q}$ at $p$ is not abelian, and the unique quadratic extension whose restriction becomes abelian is $K_{\mathfrak{p}}$, hence by twisting we cannot kill the ramification for any other quadratic extension of ${\mathbb{Q}}_p$. $\square$

Corollary 3.15.If $E/\mathbb{Q}$ is an elliptic curve with complex multiplication by an order in $K$ and $L$ is an imaginary quadratic field of class number 1 different from $K$, then $E$ is the unique curve in its family of appropriate twists (quadratic, quartic or sextic) which has prime power conductor.

Proof.Let $D=\{1,2,3,7,11,19,43,67,163\}$. The field $\mathbb{Q}(\sqrt{-d})$ for each odd $d\in D$ has a different ramification set, hence we cannot get good reduction at $d$ by Theorem 3.14. For $d$ equal to 1 or 2 the completions are also different, hence we cannot get odd conductor from the curve with CM by an order of $\mathbb{Q}(\sqrt{-1})$ over $\mathbb{Q}(\sqrt{-2})$, or vice versa. $\square$

4 Prime power conductor curves with rational odd torsion

Recall the following result of 14, 15:

Theorem 4.1.Let $K$ be a quadratic field, and $E/K$ be an elliptic curve. Then $E{(K)}_{tors}$ is isomorphic to one of the following groups:

• (i) $\mathbb{Z}/N$, with $1⩽N⩽18$ but $N\ne 17$;
• (ii) $\mathbb{Z}/2N\times \mathbb{Z}/2$ with $1⩽N⩽6$;
• (iii) $\mathbb{Z}/4\times \mathbb{Z}/4$;
• (iv) $\mathbb{Z}/3\times \mathbb{Z}/3N$ with $N=1,2$.

In particular the primes dividing the order of the torsion subgroup are $2,3,5,7,11$, and 13. Let $\mathfrak{q}\mid 2$ be a prime and $E$ be an elliptic curve of odd conductor. By Hasse's bound

$$\#E({\mathbb{F}}_{\mathfrak{q}})=|\mathcal{N}\mathfrak{q}+1-a_E(\mathfrak{q})|⩽\mathcal{N}\mathfrak{q}+1+2\sqrt{\mathcal{N}\mathfrak{q}}<11.$$

In particular a curve of odd prime power conductor over $K$ can only have a torsion point of odd prime order $\ell$ for $\ell \in \{3,5,7\}$.

While studying the possible torsion of an elliptic curve over $K$, the case $\ell =3$ and $K=\mathbb{Q}(\sqrt{-3})$ is quite different from the others. The reason is that since $K$ contains the sixth roots of unity, the determinant of the Galois representation acting on 3‐torsion points is trivial. The main results of the present section are Theorem 4.3, which implies that curves over $\mathbb{Q}(\sqrt{-3})$ of odd prime conductor and with a rational 3‐torsion point, have (up to 3‐isogeny) discriminant valuation not divisible by 3; and a list of all elliptic curves of prime power conductor and a point of order $\ell \in \{3,5,7\}$ is given for all the other fields. The complete list (omitting Galois conjugates) is given in Table 4.1, whose completeness will be proved in this section, in Theorems 4.5, 4.7, and 4.8.

Remark 4.2.Besides curves over $\mathbb{Q}(\sqrt{-3})$ with a rational 3‐torsion point, the only curves of prime conductor over imaginary quadratic fields with a rational $\ell$‐torsion point are those over $\mathbb{Q}(\sqrt{-1})$ with a 3‐torsion point, and base changes of elliptic curves defined over $\mathbb{Q}$.

Let $p$ be an odd prime, $K/\mathbb{Q}$ be a quadratic extension, and $E/K$ be an elliptic curve with a global minimal model. If $P\in E(K)$ has order $p$, then by [34, VII, Theorem 3.4] $P$ has algebraic integer coordinates in the minimal model, except when $p=3$ and $K/\mathbb{Q}$ is ramified at 3 where, if ${\mathfrak{p}}_3$ denotes the prime dividing 3, the case $v_{{\mathfrak{p}}_3}(x(P),y(P))=(-2,-3)$ might occur.

Theorem 4.3.Let $K=\mathbb{Q}(\sqrt{-3})$, and let $E/K$ be a curve with a point $P$ of order 3 and prime power conductor ${\mathfrak{p}}^r$ and let $\stackrel{\sim }{E}$ be the 3‐isogenous curve obtained by taking the quotient of $E$ by the group generated by $P$. Then the valuations at $\mathfrak{p}$ of $\mathcal{D}(E)$ and $\mathcal{D}(\stackrel{\sim }{E})$ are not both divisible by 3, unless $E$ is one of the CM curves 2.0.3.1‐729.1‐CMa1 or 2.0.3.1‐729.1‐CMb1.

Proof.Suppose $(\mathcal{D}(E))={\mathfrak{p}}^{3r}$. Let $P$ denote the point of order 3 in $E(K)$. If $P$ has integral coordinates, we use the parametrization of elliptic curves with a rational point of order 3 given by Kubert in [19, Table 1]: such curves have a minimal model of the form

$$E:y^2+a_{1}xy+a_{3}y=x^3,$$(4.1)

where $a_i$ are algebraic integers, $P=(0,0)$ has order 3, with discriminant

$$\mathcal{D}(E)=a_3^3(a_1^3-27a_3).$$(4.2)

If $P$ does not have integral coordinates, then we take the minimal equation to be of the form 4.1, with $v_{\mathfrak{p}}(a_1)⩾0$ and $v_{\mathfrak{p}}(a_3)=-3$.

Note that over a field containing the 3rd roots of unity the cyclotomic character modulo 3 is trivial so the representation of the Galois group acting on $E[3]$ has image in $SL(2,3)$. Then if it is reducible, with upper triangular matrices, the diagonal entries are both $+1$ or both $-1$. So if the curve $E$ has a rational point of order 3, so does the isogenous curve $\stackrel{\sim }{E}$. We find that $\stackrel{\sim }{E}$ has equation $y^2+a_{1}xy+a_{3}y=x^3-5a_1{a}_{3}x-a_1^3{a}_3-7a_3^2$, and discriminant

$$\mathcal{D}(\stackrel{\sim }{E})=a_3{(a_1^3-27a_3)}^3.$$(4.3)

Suppose that both $\mathcal{D}(E)$ and $\mathcal{D}(\stackrel{\sim }{E})$ generate ideals which are cubes. Then

$$\begin{array}{cc}\hfill a_1^3-27a_3& =u{\alpha }^3\hfill \\ \hfill a_3& =v{\beta }^3,\hfill \end{array}$$

for $u,v$ units and $\alpha ,\beta \in K^{\times }$; in fact, $\alpha ,\sqrt{-3}\beta \in {\mathcal{O}}_K$. In particular, $(a_1:-\alpha :-3\beta )$ is a $K$‐rational point on the cubic curve

$$x^3+u{y}^3+v{z}^3=0,$$(4.4)

a twist of the Fermat cubic. By Lemma 4.4 below, all $K$‐rational points $(x:y:z)$ on all curves of the form 4.4 either satisfy $xyz=0$, or (after scaling so that $x,y,z\in {\mathcal{O}}_K$ are coprime) that $x,y,z$ are all units.

Since $\alpha \beta \ne 0$ the first case is possible only when $a_1=0$. Then $a_3$ is a unit times a cube, so by minimality is a unit: this leads to the three isomorphism classes of curves with conductor (9) or (27). The first of these is 2.0.3.1‐81.1‐CMa1, whose discriminant valuation is 6 but has a 3‐isogenous curve with discriminant valuation 10; the other two, which are given in the statement of the theorem, are each isomorphic to their 3‐isogenous curves (the isogeny being an endomorphism) and have discriminant valuation 6.

In case none of the coordinates is zero, we consider separately the finitely many cases where $\beta$ is integral or has valuation $-1$, and find that there are no more solutions. $\square$

Lemma 4.4.Let $K=\mathbb{Q}(\sqrt{-3})$ and $u,v\in {\mathcal{O}}_K^{\ast }$. The cubic curve 4.4 has either 3 or 9 rational points, which either lie on one of the lines $x=0$, $y=0$ or $z=0$, or have projective coordinates which are all units.

Proof.After permuting the coordinates, scaling by units and absorbing cubes, there are only three essentially different equations, those with $(u,v)=(1,1)$, $(1,\zeta )$, and $(\zeta ,{\zeta }^2)$ where $\zeta \in K$ is a sixth root of unity. When $u=1$, it is well known that all points have one zero coordinate (see [13, Proposition 17.8.1]). There are nine such points (all the flexes) when $v=1$, and three when $u=\zeta$. The curve with $(u,v)=(\zeta ,{\zeta }^2)$ is isomorphic to the one with $(u,v)=(1,1)$, since the isomorphism class depends only on $uv$ modulo cubes, and hence also has nine points; these are $({\zeta }^{2k}:{\zeta }^{2l+1}:1)$ for $k,l\in \{0,1,2\}$. $\square$

Theorem 4.5.Let $K=\mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field of class number 1, $K\ne \mathbb{Q}(\sqrt{-3})$ and $E/K$ an elliptic curve of odd prime power conductor with a point of order 3. Then $E$ is isomorphic to one of the following:

• (i) the curve 19.a2 or 19.a3 over $\mathbb{Q}(\sqrt{-d})$ for $d=1,7,11,19,43,163\text{;}$
• (ii) the curve 27.a2 or 27.a3 or 27.a4 over $\mathbb{Q}(\sqrt{-d})$ for $d=1,7,19,43,67,163\text{;}$
• (iii) the curve 37.b2 or 37.b3 over $\mathbb{Q}(\sqrt{-d})$ for $d=2,19,43,163\text{;}$
• (iv) the curve 243.a2 or 243.b2 over $\mathbb{Q}(\sqrt{-d})$ for $d=1,7,19,43,67,163\text{;}$
• (v) $K=\mathbb{Q}(\sqrt{-1})$ and $E$ is one of the 3‐isogenous curves
  $$\begin{array}{ccc}\hfill y^2+(1+i)xy+iy& =& x^3+(-1+i)x^2+(-14-8i)x+(-10-20i),\hfill \\ & & \text{with}\text{label}\text{2.0.4.1757.1a1},\hfill \\ \hfill y^2+(1+i)xy+iy& =& x^3+(1+i)x^2+(1-i)x+2,\hfill \\ & & \text{with}\text{label}\text{2.0.4.1757.1a2},\hfill \end{array}$$
  or their Galois conjugates with labels 2.0.4.1‐757.2‐a1 and 2.0.4.1‐757.2‐a2.
• (vi) $K=\mathbb{Q}(\sqrt{-2})$ and $E$ is the curve
  $$\begin{array}{c}\hfill y^2+\sqrt{-2}xy+y=x^3+(1-\sqrt{-2})x^2-x,\text{with}\text{label}\text{2.0.8.19.1CMa1},\end{array}$$
  or its Galois conjugate with label 2.0.8.1‐9.3‐CMa1;
• (vii) $K=\mathbb{Q}(\sqrt{-11})$ and $E$ is the curve
  $$\begin{array}{c}\hfill y^2+y=x^3+\frac{1-\sqrt{-11}}{2}{x}^2+\frac{-5-\sqrt{-11}}{2}x-2,\text{with}\text{label}\text{2.0.11.19.3CMa1},\end{array}$$
  or its Galois conjugate with label 2.0.11.1‐9.1‐CMa1.

Proof.As in Theorem 4.3, we use the parametrization of elliptic curves with a rational point of order 3 given by Kubert in [19, Table 1]: such curves have a model of the form

$$E:y^2+a_{1}xy+a_{3}y=x^3,$$(4.5)

where $P=(0,0)$ has order 3, with discriminant

$$\mathcal{D}(E)=a_3^3(a_1^3-27a_3).$$(4.6)

By scaling, we can choose a model of this form such that for all primes $\mathfrak{q}$ either $\mathfrak{q}\nmid a_1$ or ${\mathfrak{q}}^3\nmid a_3$. Then the model is minimal at all primes, as we now show. To be non‐minimal at a prime $\mathfrak{q}$ implies that ${\mathfrak{q}}^6\mid c_6$ and ${\mathfrak{q}}^{12}\mid \mathcal{D}(E)$, where $c_6$ is the usual invariant of the model. The ideal generated by $c_6$ and $\mathcal{D}(E)$ in the ring $\mathbb{Z}[a_1,a_3]$ contains both $a_1^{15}$ and $3^3{a}_3^5$, so $\mathfrak{q}\mid a_1$ and $\mathfrak{q}\mid a_3$: in case $\mathfrak{q}\mid 3$, we need the fact that 3 is not ramified in $K$. By minimality, $v_{\mathfrak{q}}(a_3)\in \{1,2\}$; this implies $v_{\mathfrak{q}}(\mathcal{D}(E))⩽11$, contradiction.

Let $\mathfrak{p}=(\pi )$ be the unique prime dividing $\mathcal{D}(E)$. As above, we can assume that either $\mathfrak{p}\nmid a_1$ or ${\mathfrak{p}}^3\nmid a_3$. We can also scale by units, replacing $(a_1,a_3)$ by $(u{a}_1,u^3{a}_3)$, and we note that for the fields under consideration every unit is a cube. Our strategy is to prove that $(a_1,a_3)$ lies in a small finite set, and then systematically search through all possible values.

• (i) If $a_3$ is a unit, we may assume by scaling that $a_3=1$. Then we can factor 4.6 as
  $$\mathcal{D}(E)=a_1^3-27=(a_1-3)(a_1^2+3a_1+9).$$
  We consider the following cases.
  • (a) If $\mathfrak{p}\nmid 3$, then the factors are coprime, so one is a unit. If $u=a_1-3\in {\mathcal{O}}_K^{\ast }$ we get four solutions $(a_1,a_3)=(4,1)$, (2,1), $(3\pm \sqrt{-1},1)$ which give the curves 37.b3, 19.a3, 2.0.4.1‐757.1‐a2, and 2.0.4.1‐757.2‐a2. If $a_1^2+3a_1+9=u\in {\mathcal{O}}_K^{\ast }$, then ${(2a_1+3)}^2=4u-27\in \{-23,-31,-27\pm 4\sqrt{-1}\}$ which has no solution in $K$.
  • (b)

    If $\mathfrak{p}\mid 3$, then $\mathfrak{p}\mid a_1$. If $v_{\mathfrak{p}}(a_1)=1$, we write $a_1=\pi b$. In the inert case, $\pi =3$ and $\mathcal{D}(E)=27(b^3-1)=27(b-1)(b^2+b+1)$. If either factor is a unit one finds no solutions, otherwise both are divisible by 3, so write $b=1+3c$; now the second factor is $3(1+3c+3c^2)$ so $1+3c+3c^2$ is a unit. Only $u=1$ gives a solution: $c=-1$, $b=-2$, and $a_1=-6$. The pair $(a_1,a_3)=(-6,1)$ yields the curve 27.a4. In the split case we get $\mathcal{D}(E)={\pi }^3(b^3-{\overline{\pi }}^3)={\pi }^3(b-\overline{\pi })(b^2+b\overline{\pi }+{\overline{\pi }}^2)$. Elementary computations reveal two solutions: $b=\sqrt{-2}$ giving $(a_1,a_3)=(\sqrt{-2}-2,1)$ and the curve 2.0.8.1‐9.1‐CMa1; and $b=-2$ giving $(a_1,a_3)=(-1-\sqrt{-11},1)$ and the curve 2.0.11.1‐9.1‐CMa1; together with their Galois conjugates.

    If $a_1=0$ or $v_{\mathfrak{p}}(a_1)⩾2$, then $v_{\mathfrak{p}}(\mathcal{D}(E))=3$. This gives the equation $a_1^3=27+{\pi }^{3}u$ with $u$ a unit. When 3 is inert we have $\pi =3$ and ${(a_1/3)}^3=1+u\in \{2,0,1\pm \sqrt{-1}\}$, giving just one solution $(a_1,a_3)=(0,1)$ and the curve 27.a4. When $3=\pi \overline{\pi }$ we have ${(a_1/\pi )}^3={\overline{\pi }}^3\pm 1$ but this is not a cube. (Here, $\pi =1\pm \sqrt{-2}$ or $\pi =(1\pm \sqrt{-11})/2$.)

• (ii)

  Now suppose that $a_3$ is not a unit. If $a_1=0$, then 3 is inert in $K$ and $a_3=3^j$, with $j\in \{1,2\}$ by the minimality of the model, corresponding to the curves 243.b2 and 243.a2 respectively. Suppose $a_1\ne 0$.

  • (a)

    If $\mathfrak{p}\mid a_1$, the minimality of the model implies that $v_{\mathfrak{p}}(a_3)<3$; scaling by units we can assume that $a_3={\pi }^j$ with $j\in \{1,2\}$.

    If $\mathfrak{p}\nmid 3$, then $v_{\mathfrak{p}}(a_1^3-27a_3)=v_{\mathfrak{p}}(a_3)=j$ so $a_1^3={\pi }^j(27+v)$ for some unit $v$, but none of these expressions is a cube. Hence $\mathfrak{p}\mid 3$.

    If $v_{\mathfrak{p}}(a_1)=1$, we have $v_{\mathfrak{p}}(a_1^3-27a_3)=3$ so $a_1^3=27a_3+{\pi }^{3}u$ with $u$ a unit. In the inert cases ${(a_1/3)}^3=a_3+u$, whose only solution is $(a_1,a_3)=(6,9)$ which yields the curve 27.a3. In the split cases we find no solutions.

    If $v_{\mathfrak{p}}(a_1)⩾2$, we have $v_{\mathfrak{p}}(a_1^3-27a_3)=v_{\mathfrak{p}}(27a_3)=3+j$ so $0\ne a_1^3=27a_3+{\pi }^{3+j}u={\pi }^j(27+{\pi }^{3}u)$ with $u$ a unit. This has no solutions.

  • (b)

    If $\mathfrak{p}\nmid a_1$, then $a_1^3-27a_3$ is a unit and we can scale so the unit is 1, so $a_1^3=1+27a_3$.

    If $\mathfrak{p}\mid 3$, then from $27a_3=a_1^3-1\equiv {(a_1-1)}^3(mod3)$ we have $\mathfrak{p}\mid a_1=1$ and either $a_1-1$ or $a_1^2+a_1+1$ has valuation 1, but no solutions arise.

    Hence $\mathfrak{p}\nmid 3$. Now $27a_3=a_1^3-1=(a_1-1)(a_1^2+a_1+1)$, where the gcd of the factors divides 3. One of the factors is coprime to $\mathfrak{p}$, so divides 27, and both factors are divisible by the prime or primes dividing 3, so $3\mid (a_1-1)$. Suppose that we are in the case that $a_1-1\mid 27$. In case 3 is inert, write $a_1-1=3^{k}u$ with $u$ a unit and $k\in \{1,2,3\}$. The only cases where $a_1^3\equiv 1(mod27)$ are $(a_1,a_3)=(10,37)$ and $(-8,-19)$, giving the curves 37.b2 and 19.a2, respectively. In case 3 splits as $3=\omega \overline{\omega }$ we have $a_1=1\pm {\omega }^k{\overline{\omega }}^l$ with $k,l\in \{1,2,3\}$; the only solutions are those which have $k=l=2$ which have already been seen. Secondly, if $a_1^2+a_1+1=d\mid 27$, then the quadratic in $a_1$ has discriminant $4d-3$ which must be a square; enumeration of cases shows no solutions.

In summary we find that the only solutions $(a_1,a_3)$, up to scaling by units and Galois conjugates, are $(a_1,1)$ for $a_1\in \{0,2,4,-6,3+\sqrt{-1},-2+\sqrt{-2},-1-\sqrt{-11}\}$, and (0,3), (0,9), (6,9),(10,37), and $(-8,-19)$. $\square$

Remark 4.6.It is an interesting question to determine whether there are infinitely many curves over $\mathbb{Q}(\sqrt{-3})$ with a point of order 3 and prime conductor. Based on numerical evidence, it seems quite plausible that this is indeed the case, but we did not focus on this particular problem.

Theorem 4.7.Let $K=\mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field of class number 1 and $E/K$ an elliptic curve of odd prime power conductor with a point of order 5. Then either $E$ is isomorphic to the curve 11.a2 or 11.a3 for $d=1,3,11,67,163$ or $K=\mathbb{Q}(\sqrt{-1})$ and $E$ is the curve:

$$\begin{array}{c}\hfill y^2+(i+1)xy+iy=x^3+i{x}^2,\text{with}\text{label}\text{2.0.4.125.3CMa1},\end{array}$$(4.7)

or its Galois conjugate with label 2.0.4.1‐25.1‐CMa1.

Note that the curve 2.0.4.1‐25.3‐CMa1 has CM by $\mathbb{Z}[i]$; in particular, the conductor valuation is 2.

Proof.Again we use Kubert's parametrization from [19, Table 1]: such curves have a model of the form

$$y^2+(1-d)xy-dy=x^3-d{x}^2,$$

with $d\in K$. An integral model is then given by

$$E_{a,b}:y^2+(b-a)xy-a{b}^{2}y=x^3-ab{x}^2,\text{with}gcd(a,b)=1.$$

This model has discriminant $\mathcal{D}(E_{a,b})=a^5{b}^5(a^2-11ab-b^2)$, and $c_4$‐invariant $a^4-12a^{3}b+14a^2{b}^2+12a{b}^3+b^4$. In the polynomial ring $\mathbb{Z}[a,b]$ the ideal these generate contains $5a^{15}$ and $5b^{15}$, so they are coprime away from 5. Hence at all primes except possibly those dividing 5 the model $E_{a,b}$ is minimal, and has multiplicative reduction.

Let $\mathfrak{p}$ be a prime above 5 and suppose that $\mathfrak{p}$ divides both $\mathcal{D}(E_{a,b})$ and $c_4(E_{a,b})$. Then $\mathcal{D}(E_{a,b})\equiv a^5{b}^5{(a+2b)}^2(mod\mathfrak{p})$ and $c_4(E_{a,b})\equiv {(a+2b)}^4(mod\mathfrak{p})$, so $\mathfrak{p}\mid (a+2b)$. Writing $a=-2b+c$ where $\mathfrak{p}\mid c$ and using the fact that 5 is not ramified in $K$, we find that $c_4\equiv -5b^4(mod{\mathfrak{p}}^2)$, so ${\mathfrak{p}}^2\nmid c_4$. Hence $E_{a,b}$ is also minimal at $\mathfrak{p}$. Examples show that the reduction at such a prime may be either good or additive.

In the factorization $\mathcal{D}(E_{a,b})=a^5{b}^5(a^2-11ab-b^2)$, the three factors are pairwise coprime. Then for $\mathcal{D}(E_{a,b})$ to be a prime power, two of $a$, $b$, and $a^2-11ab-b^2$ are units. Since we may scale $a$ and $b$ simultaneously by a unit we may assume that either $a=1$ or $b=1$. When $a=1$, $b=\pm 1$ leads to discriminant $-11$ while $a^2-11ab-b^2=\pm 1$ leads to discriminant $-{11}^5$, giving the curves 11.a2 or 11.a3 over any field in which 11 is not split. Over $\mathbb{Q}(\sqrt{-1})$, additionally, $(a,b)=(1,\pm \sqrt{-1})$ gives the curves with conductor ${(2\pm \sqrt{-1})}^2$ as stated, while over $\mathbb{Q}(\sqrt{-3})$ none of the additional units gives a solution. Lastly if $a$ is not a unit we may assume that $b=1$ and require $u=a^2-11a-1\in {\mathcal{O}}_K^{\ast }$ so that $125+4u$ is a square in $K\mathit{\text{;}}$ the only possibility is $u=-1$ and $a=11$ giving discriminant $-{11}^5$ again. $\square$

Theorem 4.8.Let $K$ be an imaginary quadratic field of class number 1 and $E/K$ be an elliptic curve of odd prime power conductor with a point of order 7. Then $K=\mathbb{Q}(\sqrt{-3})$ and $E$ is isomorphic to

$$\begin{array}{c}\hfill y^2+ay=x^3+(a-2)x^2+(1-a)x\text{with}\text{label}\text{2.0.3.149.3CMa1},\end{array}$$(4.8)

where $a=\frac{1+\sqrt{-3}}{2}$.

Proof.In this case, a general elliptic curve with a 7‐torsion point is given by

$$E_{a,b}:y^2+(b^2+ab-a^2)xy-(a^3{b}^3-a^2{b}^4)y=x^3-(a^{3}b-a^2{b}^2)x^2,$$

where $a,b\in {\mathcal{O}}_K$ and $gcd(a,b)=1$ (see 19). Its discriminant is given by $\mathcal{D}(E_{a,b})=a^7{b}^7{(a-b)}^7(a^3-8a^{2}b+5a{b}^2+b^3)$. As in the previous theorem, we find that $gcd(\mathcal{D}(E_{a,b}),c_4(E_{a,b}))$ is not divisible by any prime except possibly those dividing 7, and that the model $E_{a,b}$ is minimal even at such primes.

Since $(a,b)=1$, two of $a$, $b$, and $c={(a-b)}^7(a^3-8a^{2}b+5a{b}^2+b^3)$ are units. If $a$ is a unit but not $b$, then we can scale so $a=1$ and now $1-b$ is a unit. None of the possibilities gives an odd prime power discriminant. Similarly if $b=1$ and $a$ is not a unit. Lastly if $a=1$ and $b$ is a unit, the only possibility which works is when $b$ is a sixth root of unity, giving the curve as stated in the theorem (which is isomorphic to its Galois conjugate, while not being a base change from $\mathbb{Q}$). $\square$

Remark 4.9.Note that in all the exceptional cases, the discriminant valuation is at most 5, except for the curve 11.a2 over $\mathbb{Q}(\sqrt{-11})$, where it is 10 (as 11 ramifies in the extension).

5 Prime power conductor curves with rational 2‐torsion

Our goal in this section is to classify all elliptic curves $E$ defined over $K$ with odd prime power conductor which have a $K$‐rational point of order 2. As a result we will see (Corollary 5.13 below) that each isogeny class of such curves contains a curve whose discriminant has odd valuation. Our results here extend those of Shumbusho, who in his 2004 thesis 32 considered elliptic curves over the same fields as we do, with prime conductor and rational 2‐torsion.

In this section we will make essential use of the local criterion of Kraus from [17, Théorème 2], which we state here for the reader's convenience.

Proposition 5.1. (Kraus)Let $K$ be a finite extension of ${\mathbb{Q}}_2$ with valuation ring ${\mathcal{O}}_K$, normalized valuation $v$, and ramification degree $e=v(2)$. Let $c_4$, $c_6$, $\mathrm{\Delta }$ in ${\mathcal{O}}_K$ satisfy $c_4^3-c_6^2=1728\mathrm{\Delta }\ne 0$. Then there exists an integral Weierstrass model of an elliptic curve over $K$ with invariants $c_4$ and $c_6$ if and only if one of the following holds:

• (i) $v(c_4)=0$, and there exists $a_1\in {\mathcal{O}}_K$ such that $a_1^2\equiv -c_6(mod4);$
• (ii) $0<v(c_4)<4e$, and there exist $a_1,a_3\in {\mathcal{O}}_K$ such that
  $$\begin{array}{ccc}\hfill d=-a_1^6+3a_1^2{c}_4+2c_6& \equiv & 0(mod16);\hfill \\ \hfill a_3^2& \equiv & d/16(mod4);\hfill \\ \hfill 4a_1^{2}d& \equiv & {(a_1^4-c_4)}^2(mod256).\hfill \end{array}$$
• (iii) $v(c_4)⩾4e$, and there exists $a_3\in {\mathcal{O}}_K$ such that $a_3^2\equiv c_6/8(mod4)$.

5.1 Preliminaries on curves with odd conductor and rational 2‐torsion

In this subsection and the next we assume that $E$ is an elliptic curve defined over $K$, with odd conductor, and with a rational 2‐torsion point; later we will specialize to the case where the conductor is an odd prime power. Such an elliptic curve $E$ has an equation of the form

$$E_{a,b}:y^2=x(x^2+ax+b),$$(5.1)

where $a,b\in {\mathcal{O}}_K$ and the given 2‐torsion point is (0,0). However, while it is easy to see that this model can be taken to be minimal at all odd primes, we need to be more precise concerning the primes dividing 2 where such a model cannot have good reduction. To this end, we need to consider carefully the transformation from a global minimal model for $E$ (which exists since $K$ has class number 1) to this form. Let

$$y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6$$(5.2)

be a global minimal model for $E$; its discriminant ${\mathcal{D}}_{min}(E)$ is odd since $E$ has odd conductor. To transform this model into $E_{a,b}$ we first complete the square, then scale to make the equation integral, and finally translate the $x$‐coordinate so that the 2‐torsion point has $x=0$.

After completing the square, the right‐hand side of the equation is

$$x^3+(a_2+{(a_1/2)}^2)x^2+(a_4+(a_1/2)a_3)x+(a_6+{(a_3/2)}^2)$$(5.3)

which is still integral, and minimal, at all odd primes. Let $\mathfrak{q}=(\tau )$ be a prime of $K$ dividing 2. Then $a_1$ and $a_3$ are not both divisible by $\mathfrak{q}$, as otherwise $\mathfrak{q}$ would divide the discriminant, so 5.3 is not integral at $\mathfrak{q}$. To make the equation integral we scale $x$ by ${\tau }^{2r}$ for some $r⩾1$ chosen to be minimal. The minimal $r$ depends on whether $E$ is ordinary or supersingular at $\mathfrak{q}$, or equivalently whether $v_{\mathfrak{q}}(a_1)=0$ or $v_{\mathfrak{q}}(a_1)>0$.

Proposition 5.2.Let $E$ be an elliptic curve of odd conductor over $K$ with a $K$‐rational point of order 2 with minimal equation 5.2. Let $\mathfrak{q}=(\tau )$ be a prime of $K$ dividing 2.

• (i) If $E$ has ordinary reduction at $\mathfrak{q}$ (that is, if $v_{\mathfrak{q}}(a_1)=0$), then the minimal scaling to make 5.3 integral is $x\mapsto 2^{2}x$ (with scaling valuation $r=2e_2$).
• (ii) If $E$ has supersingular reduction at $\mathfrak{q}$ (that is, if $v_{\mathfrak{q}}(a_1)>0$), then $K$ is ramified at 2 and $v_{\mathfrak{q}}(a_1)=1$. The minimal scaling is $x\mapsto {\tau }^{2}x$ (with $r=2$).

In both cases, after scaling, 5.3 reduces to $x^2(x+u)$ modulo $\mathfrak{q}$ with $u$ odd. In the supersingular case, there is only one $K$‐rational point of order 2, whose $x$‐coordinate (after scaling) is odd.

Proof.After scaling $x$ by ${\tau }^{2r}$, the coefficient of $x^{3-j}$ is multiplied by ${\tau }^{2rj}$.

First suppose that $v_{\mathfrak{q}}(a_1)=0$ (the ordinary case). From the coefficient of $x^2$ in 5.3 it is immediate that $x\mapsto 4x$ is the minimal scaling which gives integral coefficients. After scaling, the coefficient of $x^2$ is $u=a_1^2+4a_2$, with $\mathfrak{q}$‐valuation 0, and the others are divisible by 8 and 16, respectively.

Now suppose $v_{\mathfrak{q}}(a_1)>0$ (the supersingular case), which implies $v_{\mathfrak{q}}(a_3)=0$. If $v_{\mathfrak{q}}(a_1)⩾e_2$, then all coefficients in 5.3 are $\mathfrak{q}$‐integral except the last one, which has $\mathfrak{q}$‐valuation $-2e_2$. But then all roots have valuation $-2e_2/3$, which is not an integer, contradicting the fact that the polynomial has a root in $K$. It follows that this supersingular case can only occur if $e_2=2$ and $v_{\mathfrak{q}}(a_1)=1$. The coefficients in 5.3 now have valuations $-2$, $-1$, $-4$, from which it follows that the roots (whether in $K$ or an extension) have valuations $-2$, $-1$, $-1$; since the $x$‐coordinates of non‐integral $K$‐rational points must have even valuation, there can be only one $K$‐rational point of order 2, with $x$‐coordinate of valuation $-2$. To achieve integrality we must scale $x$ by ${\tau }^2$, after which the cubic reduces to $x^2(x+1)$ modulo $\mathfrak{q}$ and the $x$‐coordinate of the $K$‐rational point of order 2 is odd. $\square$

We will refer to the two cases of this proposition as ‘the ordinary case’ and ‘the supersingular case’, respectively.

The model $E_{a,b}$ has invariants

$$\mathrm{\Delta }=2^4{b}^2(a^2-4b),c_4=2^4(a^2-3b),c_6=2^{5}a(9b-2a^2).$$(5.4)

We set $\mathcal{D}(E)=b^2(a^2-4b)$, and compare this with ${\mathcal{D}}_{min}(E)$. In the ordinary case, $\mathrm{\Delta }=2^{12}{\mathcal{D}}_{min}(E)$ so $\mathcal{D}(E)=2^8{\mathcal{D}}_{min}(E)$; in the supersingular case, $\mathrm{\Delta }={\tau }^{12}{\mathcal{D}}_{min}(E)$ and $\mathcal{D}(E)=\pm 2^2{\mathcal{D}}_{min}(E)$, with sign $-1$ for $\mathbb{Q}(\sqrt{-1})$ when $\tau =1+\sqrt{-1}$ and $+1$ for $\mathbb{Q}(\sqrt{-2})$ when $\tau =\sqrt{-2}$.

Corollary 5.3.Let $(x_0,0)$ be the coordinates of the given 2‐torsion point on the scaled model, so $x_0\in {\mathcal{O}}_K$. We obtain a model of the form $E_{a,b}$ by shifting the $x$‐coordinate by $x_0$.

• (i) In the ordinary case, if $v_{\mathfrak{q}}(x_0)>0$, then we obtain $(a,b)$ with $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(0,4e_2)$, while if $v_{\mathfrak{q}}(x_0)=0$, then $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(e_2,0)$.
• (ii) In the supersingular case we always have $v_{\mathfrak{q}}(x_0)=0$, and $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(k,0)$ with $k⩾3$. (We include the possibility that $a=0$ here.)

Proof.After the shift by $x_0$ we have $\mathfrak{q}\mid b$ if $x_0$ reduces to the double root modulo $\mathfrak{q}$ and $\mathfrak{q}\mid a$ otherwise; $\mathfrak{q}$ does not divide both since there is no triple root modulo $\mathfrak{q}$. In the supersingular case, $\mathfrak{q}$ must divide $a$.

Suppose $\mathfrak{q}\mid b$. Then we are in the ordinary case, and $8e_2=v_{\mathfrak{q}}(\mathcal{D}(E))=2v_{\mathfrak{q}}(b)$ so $v_{\mathfrak{q}}(b)=4e_2$.

Alternatively, suppose $\mathfrak{q}\mid a$. Then in the ordinary case, $v_{\mathfrak{q}}(a^2-4b)=v_{\mathfrak{q}}(\mathcal{D}(E))=8e_2>2e_2=v_{\mathfrak{q}}(4b)$, so $v_{\mathfrak{q}}(a^2)=2e_2$ and $v_{\mathfrak{q}}(a)=e_2$. In the supersingular case, $v_{\mathfrak{q}}(a^2-4b)=4=v_{\mathfrak{q}}(4b)$ so $v_{\mathfrak{q}}(a)⩾2$; however it is easy to see that $v_{\mathfrak{q}}(a)=2$ leads to a contradiction since the residue field at $\mathfrak{q}$ has only size 2. $\square$

Note that $a=0$ can only happen in the supersingular case. Such curves have CM by $\mathbb{Z}[\sqrt{-1}]$ and were considered in Section 3.

In what follows, it would be enough to determine curves up to quadratic twist, since given one elliptic curve it is straightforward (see 8) to find all of its twists with good reduction outside a fixed set of primes. The quadratic twists of $E_{a,b}$ have the form $E_{\lambda a,{\lambda }^{2}b}$ for $\lambda \in K^{\ast }$. Taking $\lambda ={\mu }^{-1}$ with $\mu \in {\mathcal{O}}_K$, where $\mu \mid a$ and ${\mu }^2\mid b$, we obtain a curve with a smaller discriminant, by a factor ${\mu }^6$. In our situation of curves with odd conductor, such a factor $\mu$ must be odd, supported on primes of bad reduction, and also square‐free (by minimality of the equation at all odd primes). However, if $\mathfrak{p}=(\pi )$ is an odd prime factor of the conductor such that $\pi \mid a$ and ${\pi }^2\mid b$, it can happen11 This can be avoided over $\mathbb{Q}$, since for every odd prime ideal $(p)$, either $\pm p\equiv 1(mod4)$, and $\mathbb{Q}(\sqrt{\pm p})$ is unramified at 2.
that the twist $E^{\pi }$ acquires bad reduction at a prime $\mathfrak{q}$ dividing 2: this is the case if $\mathfrak{q}$ ramifies in $K(\sqrt{\pi })$.

For a prime $\mathfrak{p}=(\pi )$ we say that the pair $(a,b)$ is minimal at $\mathfrak{p}$ (or $\pi$) if either $\pi \nmid a$ or ${\pi }^2\nmid b$. When $(a,b)$ are the parameters obtained from a curve of odd conductor, we have already seen that $(a,b)$ is minimal at $\mathfrak{q}$ for all $\mathfrak{q}\mid 2$ since either $a$ or $b$ is not divisible by $\mathfrak{q}$, while for the primes $\mathfrak{p}$ dividing the conductor, $(a,b)$ may not be $\mathfrak{p}$‐minimal. However, as already observed, we can always assume that either ${\pi }^2\nmid a$ or ${\pi }^4\nmid b$.

Since we cannot assume that a minimal twist of a curve with odd conductor still has odd conductor, we will need to consider curves with non‐minimal $(a,b)$. In the prime power conductor case, this means that we will consider curves in sets of four twists, a base curve $E=E_{a,b}$ which is $\mathfrak{p}$‐minimal, may have bad reduction at primes dividing 2, and may even have good reduction at $\mathfrak{p}$; and the twists $E^s$ of $E$ by $s\in \{\epsilon ,\pi ,\epsilon \pi \}$, where $\mathfrak{p}=(\pi )$.

For example, over $K=\mathbb{Q}(\sqrt{-2})$ we have seen in a previous section that the elliptic curve 256.d2, which has conductor ${(\sqrt{-2})}^{10}$, has infinitely many quadratic twists of odd prime square conductor.

5.2 Classification of curves with odd conductor and 2‐torsion

We continue with the notation of the previous subsection: $E$ is an elliptic curve with odd conductor and a $K$‐rational point of order 2, and $(a,b)$ are parameters for the model $E_{a,b}$ for $E$ constructed above. Write $(2)={\mathfrak{q}}_a{\mathfrak{q}}_b$ where ${\mathfrak{q}}_a$, with generator ${\tau }_a$, (respectively ${\mathfrak{q}}_b$, with generator ${\tau }_b$), is divisible only by primes dividing $a$ (respectively $b$), and $2={\tau }_a{\tau }_b$. For $K\ne \mathbb{Q}(\sqrt{-7})$ we have $({\tau }_a,{\tau }_b)=(1,2)$ or (2,1) according to whether $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(0,4e_2)$ or $(e_2,0)$ for the unique prime $\mathfrak{q}$ dividing 2 in the ordinary case, and also $({\tau }_a,{\tau }_b)=(2,1)$ in the supersingular case.

The following result completely classifies curves with odd conductor and a point of order 2 in terms of the solutions to a certain equation 5.5.

Theorem 5.4.Let $E$ be an elliptic curve defined over $K$, with a $K$‐rational 2‐torsion point and odd conductor. Let $D={\mathcal{D}}_{min}(E)$. Set

• (i) $P=gcd(D,b,a^2-4b)=A{s}^2$ with $A$ square‐free;
• (ii) $B=(a^2-4b)/({\tau }_a^{2}P)\text{;}$
• (iii) $C=4b/({\tau }_a^{2}P)\text{;}$
• (iv) $\stackrel{\sim }{a}=as/({\tau }_{a}P)$.

Then

$${\stackrel{\sim }{a}}^{2}A=B+C$$(5.5)

and $\stackrel{\sim }{a},A,B,C\in {\mathcal{O}}_K$ satisfy the following conditions:

• (i) $gcd(B,C)=1$ and $A$ is square‐free;
• (ii) $A,B,C$ are only divisible by primes dividing $2D$.

Furthermore,

• (i) For each prime $\mathfrak{p}\mid D$ with $k=v_{\mathfrak{p}}(D)$ and $k^{\prime }=k-6\text{:}$
  $$(v_{\mathfrak{p}}(A),v_{\mathfrak{p}}(B),v_{\mathfrak{p}}(C))=(0,k,0),(0,k^{\prime },0),(0,0,\frac{1}{2}k),(0,0,\frac{1}{2}{k}^{\prime })\text{or}(1,0,0);$$
• (ii) In the ordinary case, for each prime $\mathfrak{q}\mid 2$,
  $$(v_{\mathfrak{q}}(A),v_{\mathfrak{q}}(B),v_{\mathfrak{q}}(C))=(0,6e_2,0)\text{or}(0,0,6e_2).$$
• (iii) In the supersingular case, for $\mathfrak{q}\mid 2$, $(v_{\mathfrak{q}}(A),v_{\mathfrak{q}}(B),v_{\mathfrak{q}}(C))=(0,0,0)$.

Conversely, given integral $A,B,C,\stackrel{\sim }{a}$ satisfying ${\stackrel{\sim }{a}}^{2}A=B+C$ and the above conditions, if we set $a=2A\stackrel{\sim }{a}/gcd(2,C)$ and $b=AC/gcd{(2,C)}^2$, then $E_{a,b}$ and its twists $E_{sa,s^{2}b}$ for all square‐free $s\in {\mathcal{O}}_K$ dividing $D$ all have good reduction outside $2D$.

Proof.We have $a^2=(a^2-4b)+4b=B_0+C_0$ where $B_0=a^2-4b=\mathcal{D}(E)/b^2$ and $C_0=4b$, and consider $gcd(B_0,C_0)$ one prime at a time, these primes all being divisors of $2D$.

First consider odd primes $\mathfrak{p}=(\pi )$. These contribute to $P$ if $\mathfrak{p}\mid B_0$ and $\mathfrak{p}\mid C_0$, so that $\mathfrak{p}\mid b$ and $\mathfrak{p}\mid a$. In general, the contribution to $P$ from $\mathfrak{p}$ is ${\pi }^j$ where $j=min\{v_{\mathfrak{p}}(b),v_{\mathfrak{p}}(a^2-4b)\}$, with $0⩽j⩽3$ as in Table 5.1, where $k=v_{\mathfrak{p}}(D)$. The entries above the line correspond to $(a,b)$ being $\mathfrak{p}$‐minimal while those below are non‐minimal. We include $a=0$ as a possibility in each row with ‘ $⩾$’ in the first column. Let $P$ be the product of ${\pi }^j$ over all these odd prime ideals. We can write $P=A{s}^2$ with $A$ and $s$ square‐free (since $j⩽3$ in all cases); $P$ is the odd part of $gcd(a^2-4b,4b)$, which is $gcd(D,b,a^2-4b)$.

Now consider the prime (or primes) $\mathfrak{q}\mid 2$, which divide $a$ or $b$ but not both by Corollary 5.3. The possible valuations are given in Table 5.2. Since $e_2=2$ in the supersingular case, this prime(s) contributes ${\tau }_a^2$ to $gcd(B_0,C_0)$.

Hence $gcd(B_0,C_0)={\tau }_a^{2}P\mathit{\text{;}}$ dividing through gives ${\stackrel{\sim }{a}}^{2}A=B+C$ with $B,C,\stackrel{\sim }{a}$ as given. In the factorization $P=A{s}^2$, $A$ is the product of those odd $\pi$ for which $j$ is odd while $s$ is the product of those for which $j⩾2$; both are square‐free divisors of $D$.

For all primes $\mathfrak{p}$ dividing $2D$, the values of $v_{\mathfrak{p}}(A)$, $v_{\mathfrak{p}}(B)$, $v_{\mathfrak{p}}(C)$ in the tables may easily be deduced from the previous columns.

For the converse, it suffices to observe that with $a,b$ as defined we have $b^2(a^2-4b)=4A^{3}B{C}^2/gcd{(2,C)}^6$, whose support lies in $2D$, and hence determines a base curve $E_{a,b}$, which has good reduction away from $2D$. The same is true of twists by square‐free $s$ dividing $D$. $\square$

Table 5.1. Possible parameter valuations at an odd prime

$v_{\mathfrak{p}}(a)$	$v_{\mathfrak{p}}(b)=v_{\mathfrak{p}}(C_0)$	$v_{\mathfrak{p}}(a^2-4b)=v_{\mathfrak{p}}(B_0)$	$j$	$k$	$v_{\mathfrak{p}}(A)$	$v_{\mathfrak{p}}(B)$	$v_{\mathfrak{p}}(C)$
0	0	$⩾0$	0	$⩾0$	0	$k$	0
$⩾1$	0	0	0	0	0	0	0
0	$⩾1$	0	0	$⩾2$, even	0	0	$\frac{1}{2}k$
$⩾1$	1	1	1	3	1	0	0
1	2	$⩾2$	2	$⩾6$	0	$k-6$	0
$⩾2$	2	2	2	6	0	0	0
1	$⩾3$	2	2	$⩾8$, even	0	0	$\frac{1}{2}(k-6)$
$⩾2$	3	3	3	9	1	0	0

Table 5.2. Possible parameter valuations at a prime dividing 2

Case	$v_{\mathfrak{p}}(a)$	$v_{\mathfrak{p}}(b)$	$v_{\mathfrak{p}}(B_0)=v_{\mathfrak{p}}(a^2-4b)$	$v_{\mathfrak{p}}(C_0)=v_{\mathfrak{p}}(4b)$	$v_{\mathfrak{p}}(A)$	$v_{\mathfrak{p}}(B)$	$v_{\mathfrak{p}}(C)$
Ordinary	0	$4e_2$	0	$6e_2$	0	0	$6e_2$
	$e_2$	0	$8e_2$	$2e_2$	0	$6e_2$	0
Supersingular	$⩾3$	0	4	4	0	0	0

Remark 5.5.We can scale solutions $(A,B,C)$ to 5.5 by units without affecting the conditions, and scaling by squares of units gives isomorphic curves. Since $K$ has finitely many units, for each odd $D\in {\mathcal{O}}_K$, we can use Theorem 5.4 to compute all curves of discriminant $D$ with a $K$‐rational two‐torsion point. This will be our strategy in the following subsections, where we restrict to the case where there is only one odd prime factor.

Recall that each curve $E_{a,b}$ has a 2‐isogenous curve $E_{-2a,a^2-4b}$, via the 2‐isogeny with kernel (0,0), which has the same conductor as $E$. At odd primes $\mathfrak{p}$ it is immediate that $(a,b)$ is minimal at $\mathfrak{p}$ if and only if $(a^{\prime },b^{\prime })=(-2a,a^2-4b)$ is minimal. When $E_{a,b}$ has odd conductor with $(a,b)$ given by our construction, at primes $\mathfrak{q}$ dividing 2 we see that in the ordinary case, $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(0,4e_2)$ implies $(v_{\mathfrak{q}}(-2a),v_{\mathfrak{q}}(a^2-4b))=(e_2,0)$, while conversely if $(v_{\mathfrak{q}}(a),v_{\mathfrak{q}}(b))=(e_2,0)$, then $(v_{\mathfrak{q}}(a^{\prime }),v_{\mathfrak{q}}(b^{\prime }))=(2e_2,8e_2)$ and the associated minimal pair is $(a^{\prime }/{\tau }^{2e_2},b^{\prime }/{\tau }^{4e_2})$ with valuations $(0,4e_2)$. In the supersingular case with valuations $(⩾3,0)$, a minimal pair for the isogenous curve is $(a^{\prime }/(-2),b^{\prime }/4)=(a,{(a/2)}^2-b)$ with the same valuations as $(a,b)$.

Proposition 5.6.Following the notation of Theorem 5.4, if $E_{a,b}$ is a curve with odd conductor associated to a triple $(A,B,C)$ satisfying 5.5, then the 2‐isogenous curve $E_{-2a,a^2-4b}$ has associated triple $(A,C,B)$.

Proof.It is clear that the parameter $A$ is the same for both curves, and then a straightforward computation gives the result, using the remarks about minimality stated above. $\square$

Given an odd prime ideal $\mathfrak{p}=(\pi )$ of bad reduction, we have two different situations depending on whether $v_{\mathfrak{p}}(A)=0$ or $v_{\mathfrak{p}}(A)=1$. In the first case, one curve in each isogenous pair has double the discriminant valuation of the other, and one of the two isogenous pairs, which are $\pi$‐twists of each other, has good or multiplicative reduction at $\mathfrak{p}$ (when the parameter $s$ of Theorem 5.4 is not divisible by $\pi$), while the other has additive reduction (when  $s$ is divisible by $\pi$). In the second case, all curves have additive reduction at $\mathfrak{p}$, one isogenous pair (with $\pi \nmid s$) has discriminant valuation 3 and the other (with $\pi \mid s$) has valuation 9.

Each solution to 5.5 falls into one of these cases, according to whether $B+C$ has even or odd valuation at $\mathfrak{p}$, unless $B+C=0$, corresponding to $a=0$, which we have treated separately. When seeking curves with conductor a power of the odd prime $\mathfrak{p}$ in subsequent subsections, we will also treat separately solutions to 5.5 where all of $A$, $B$, and $C$ are units at $\mathfrak{p}$, since in such cases we cannot recover $\mathfrak{p}$ from the solution. Such cases occur when an elliptic curve $E$ has conductor ${\mathfrak{p}}^2$, so has additive reduction at $\mathfrak{p}$, but is a quadratic twist of a curve $E_0$ with good reduction at $\mathfrak{p}$. Here we must consider twists of $E_0$ by all odd primes to see which have good reduction above 2.

In the next three subsections we will determine all elliptic curves with odd prime power conductor ${\mathfrak{p}}^r$, treating first this ‘good twist’ case where $E$ has a twist with good reduction at $\mathfrak{p}$ (this includes the case $a=0$), then the ‘additive twist’ cases where all twists have additive reduction at $\mathfrak{p}$ (here $A=\pi$) and lastly the ‘multiplicative twist’ case where $E$ has a twist with multiplicative reduction at $\mathfrak{p}$ (here $A=1$).

In each case we determine which $(B,C)$ pairs give an appropriate solution to 5.5, thus obtaining a ‘base curve’ $E_{a,b}$, and then determine which of its twists have good reduction at primes dividing 2.

5.3 Curves with odd prime power conductor: the good twist case

In this subsection we determine all elliptic curves defined over one of the fields $K$ with $K$‐rational 2‐torsion, and conductor a power of an odd prime $\mathfrak{p}$, such that a quadratic twist of $E$ by a generator $\pi$ of $\mathfrak{p}$ has good reduction at $\mathfrak{p}$. In fact all such curves have CM by an order in $K$ and have been fully described previously.

Theorem 5.7.Let $K$ be an imaginary quadratic field with class number 1, and let $\mathfrak{p}$ be an odd prime of $K$. Let $E$ be an elliptic curve defined over $K$, with a $K$‐rational point of order 2 and conductor a power of $\mathfrak{p}$. If $E$ has a quadratic twist with good reduction at $\mathfrak{p}$, then $E$ belongs to one of the following complex multiplication families as studied in Section 3.

• (i) $K=\mathbb{Q}(\sqrt{-7})$ and $E$ is a twist of the base change to $K$ of one of the curves in the isogeny class 49.a over $\mathbb{Q}$, with conductor ${\mathfrak{p}}^2={(\pi )}^2$ where $\pi \equiv 1(mod4)\text{;}$ or
• (ii) $K=\mathbb{Q}(\sqrt{-1})$ and $E$ is a twist of the base change to $K$ of one of the curves in the isogeny class 64.a over $\mathbb{Q}$. $E$ has equation $y^2=x(x^2+b)$ and conductor ${\mathfrak{p}}^2={(\pi )}^2$, where $b\equiv -1\pm 2i(mod8)$ with $b=\pi$ or $b={\pi }^3\text{;}$ or
• (iii) $K=\mathbb{Q}(\sqrt{-2})$ and $E$ is a twist of the base change to $K$ of one of the curves in the isogeny class 256.a over $\mathbb{Q}$, with conductor ${\mathfrak{p}}^2={(\pi )}^2$ where $\pi \equiv \pm 1+\sqrt{-2}(mod4)$, for $(a,b)=(2\pi \sqrt{-2},-{\pi }^2)$.

In each case, the elliptic curves have CM by an order in the field of definition $K$: their $j$‐invariants are either $-{15}^3$ or ${255}^3$ in the first case, either ${12}^3$ or ${66}^3$ in the second, and in the third they have $j={20}^3$ and CM by $\mathbb{Z}[\sqrt{-2}]$.

Remark 5.8.There are four curves in the isogeny class 49.a over $\mathbb{Q}$, linked by 2‐ and 7‐isogenies and in two pairs of $-7$‐twists, so that over $\mathbb{Q}(\sqrt{-7})$ they become isomorphic in pairs. The first two are 49.a1, which has parameters $(a,b)=(21,112)$, and 49.a2, with $(a,b)=(-42,-7)$; the other two are their $-7$‐twists and are 7‐isogenous to these.

Proof.By Theorems 3.3 and 3.8 we know that the three families of curves do have good reduction away from $\mathfrak{p}$, hence we are led to prove that these are the unique ones.

In the notation of Table 5.1 such curves have discriminant valuation $k=6$ so come from solutions with $\mathfrak{p}$‐valuations given in the first two lines of the second half of the table. Hence $B$ and $C$ are not divisible by $\pi$, and from Table 5.2 they have valuation $6e_2$ or 0 at the prime(s) above 2 in the ordinary case, while in the supersingular case they are units. Up to scaling by units, and interchanging $B$ and $C$ (corresponding to applying a 2‐isogeny), we reduce to considering the following finite number of possibilities for $(B,C)$:

• (1) over all fields, $(B,C)=(64\eta ,1)$ with $\eta \in {\mathcal{O}}_K^{\ast }$;
• (2) over $\mathbb{Q}(\sqrt{-7})$ where $2=\tau \overline{\tau }$ with $\tau =\frac{1+\sqrt{-7}}{2}$, $(B,C)=(\pm {\tau }^6,{\overline{\tau }}^6)$;
• (3) over $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-2})$, $(B,C)=(\pm 1,1)$ (the supersingular case).

For a solution we require $B+C$ to be either 0 or a non‐zero square times a unit.

Case (1) yields no solutions with $\eta =1$ since $B+C=65=5·13$ is not a square since neither 5 nor 13 ramifies in any of the fields. Taking $\eta =-1$ in (1) gives $B+C=-63=3^2·7$, which is valid when 7 is ramified, and leads to the base curves $E=E_{a,b}$ with $(a,b)=(6\sqrt{-7},1)$ (and its Galois conjugate). Such curves lie in the first family.

A simple check shows that none of the additional units in $\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-3})$ gives a value of $B+C$ of the required form.

In case (2) we have $B+C=\pm {\tau }^6\pm {\overline{\tau }}^6\in \{\pm 9,\pm 5\sqrt{-7}\}$, giving a potential solution with $A=\pm 6$ and $b={\overline{\tau }}^6$ (or its Galois conjugate). Taking $(a,b)=(6,{\tau }^6)$ we find a twist of 49.a1 and hence no curves not already encountered.

In case (3) with $B+C=0$ we obtain curves with $a=0$. All such curves have CM by $\mathbb{Z}[\sqrt{-1}]$, hence we get curves in the second family from Theorem 3.8.

In case (3) with $B+C=2$ we obtain a solution when 2 is ramified, with base curve $E=E_{a,b}$ where $(a,b)=(2(1+i),i)$ or $(2\sqrt{-2},-1)$. Both cases are isomorphic to the curve 256.a1 with CM by $\mathbb{Z}[\sqrt{-2}]$. Then we get the third family from Theorem 3.11 and Corollary 3.15. $\square$

5.4 Curves with odd prime power conductor: the additive twist case

We continue to consider elliptic curves $E$ whose conductor is a power of the odd prime $\mathfrak{p}=(\pi )$, using Theorem 5.4 to find all such curves by considering solutions to the parametrizing equation 5.5.

In this subsection, we consider the ‘additive twist’ case in which the parameter $A$ is divisible by $\pi$ so that the curves and their twists by $\pi$ and by units all have additive reduction at $\mathfrak{p}$. The discriminant valuations are 3 or 9. We find that the only such curves are again the base changes of CM elliptic curves over $\mathbb{Q}$ with conductor 49, but unlike the previous subsection, $K$ must be one of the six fields in which 7 is inert. This corresponds to looking at elliptic curves with CM by an order in $K$ over a field $L\ne K$, which furthermore have a 2‐torsion point and odd prime power conductor. By the results of Section 3 (specifically Corollary 3.15), we have to restrict to odd values of $d$, and the unique curve with a 2‐torsion point corresponds to $d=7$.

Theorem 5.9.Let $K$ be an imaginary quadratic field with class number 1, and let $\mathfrak{p}$ be an odd prime of $K$. Let $E$ be an elliptic curve defined over $K$, with a $K$‐rational point of order 2 and conductor a power of $\mathfrak{p}$, such that no quadratic twist of $E$ has good or multiplicative reduction at $\mathfrak{p}$. Then

• (i) $K=\mathbb{Q}(\sqrt{-d})$ for $d=1,2,11,43,67,163$, $E$ has conductor ${\mathfrak{p}}^2$ where $\mathfrak{p}=(7)$, and $E$ is a base change to $K$ of one of the curves in the isogeny class 49.a over $\mathbb{Q}$.

Proof.Inspecting Table 5.1, we are led to the same set of pairs $(B,C)$ as considered in the proof of Theorem 5.7, except that now we require $B+C$ to be non‐zero, with odd valuation at exactly one odd prime $\mathfrak{p}$. We use the same numbering of cases as before and recall that these are all possibilities, up to scaling by units and switching $B$ and $C$.

Case (1), where $B+C\in \{\pm 63,\pm 65\}$ again yields no solutions with $B+C=65=5·13$ since neither 5 nor 13 ramifies in any of the fields. However, $B+C=-63=3^2·7$ is valid when 7 is inert in $K$. This gives the base curve with $(a,b)=(-42,-7)$, which is the elliptic curve defined over $\mathbb{Q}$ with label 49.a2. Note that this curve also appeared in the good twist case over $\mathbb{Q}(\sqrt{-7})$, but here we require 7 to be inert. The quadratic twist by $-7$ (with label 49.a4) also has good reduction away from 7, so all four curves in the isogeny class 49.a have conductor ${(7)}^2$ over the fields listed.

A simple check shows that none of the additional units in $\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-3})$ gives a value of $B+C$ of the required form.

In case (2) we have $B+C=\pm {\tau }^6\pm {\overline{\tau }}^6\in \{\pm 9,\pm 5\sqrt{-7}\}$. Since 5 is inert this gives no solutions.

Case (3), with $B+C\in \{0,\pm 2\}$, also provides no solutions. $\square$

5.5 Curves with odd prime power conductor: the multiplicative twist case

We now consider curves of odd prime power conductor ${\mathfrak{p}}^r$ which in our parametrization have $A=1$, such that the base curve $E_{a,b}$ (with $\mathfrak{p}$‐minimal $(a,b)$) has multiplicative reduction at $\mathfrak{p}$.

The main result of this subsection is that these elliptic curves are of two types, up to quadratic twist by a generator of $\mathfrak{p}$:

• (1) one of a finite number of ‘sporadic’ curves, with conductor either a prime dividing 17 (over all fields where 17 does not split), or a prime of norm 257 over $\mathbb{Q}(\sqrt{-1})$ only, or a prime of norm 241 over $\mathbb{Q}(\sqrt{-3})$ only;
• (2) one of a family analogous to the Setzer–Neumann family over $\mathbb{Q}$.

The sporadic curves are all given by a more general construction which we discuss first. Over any number field $K$ let $u\in K\setminus \{0,-16\}$ and define $E_u=E_{-(u+32)/4,u+16}$, an elliptic curve with invariants $c_4=u^2+16u+256$, $c_6=(u-16)(u+8)(u+32)$ and discriminant ${\mathrm{\Delta }}_u=u^2{(u+16)}^2$.

Lemma 5.10. $E_u$ has full 2‐torsion over $K$; the three curves 2‐isogenous to $E$ are isomorphic to $E_{a,b}$ for $(a,b)=(2(u-16),u^2+32u+256)$, $(2(u+32),u^2)$, and $(u+8,16)$, with discriminants $-u{(u+16)}^4$, $u^4(u+16)$, and $u(u+16)$, respectively.

Proof.Elementary: note that ${\mathrm{\Delta }}_u$ is a square. $\square$

In fact the family of curves $E_u$ is the universal family of elliptic curves with full 2‐torsion over $K$, as it is easy to check that $E_u$ has Legendre parameter $\lambda =(u+16)/u$. Our reason for writing the family this way is that if we specialize the parameter $u$ to a unit with certain properties, then we obtain elliptic curves with square‐free odd conductor.

Proposition 5.11.Let $K$ be any number field and $u\in {\mathcal{O}}_K^{\ast }$. The quadratic twist $E_u^{(-u)}$ of $E_u$ by $-u$, together with its three 2‐isogenous curves, is semistable with bad reduction only at primes dividing $u+16$. The same is true of $E_u$ itself if $-u$ is congruent to a square modulo 4.

Proof.From the invariants given above we see that ${\mathrm{\Delta }}_u$ is only divisible by primes dividing $u+16$ which is odd, and that ${\mathrm{\Delta }}_u$ is coprime to $c_4$ so the reduction is multiplicative at all bad primes. Also since $c_4$ and $c_6$ are odd the condition that $c_4$ and $c_6$ are the invariants of an integral model, which then has good reduction at primes dividing 2, is that $-c_6$ is a square modulo 4, which is the case when $-u$ is a square modulo 4 since $c_6\equiv u^3(mod4)$. Twisting by $-u$ gives a curve whose $c_6\equiv -u^6(mod4)$ which satisfies Kraus's condition unconditionally. $\square$

For example, over $\mathbb{Q}$ we take $u=1$ and find that $E_1^{(-1)}$ is the elliptic curve 17.a2 of conductor 17, with 2‐isogenous curves 17.a1, 17.a3, and 17.a4. Since $17\equiv 1(mod4)$, its quadratic twists by 17 also have good reduction at 2. Taking $u=-1$ gives curves of conductor 15, which are not relevant for us.

More generally we consider the curves given by this proposition over imaginary quadratic fields, for units $u$ such that $u+16$ is a prime power, so that we obtain curves of prime conductor. When $\pm 1$ are the only units, the only case is the one just considered with $u=1$, leading to curves whose conductors are divisible only by the primes above 17, which are primes except when 17 splits in $K$. Since $17\equiv 1(mod4)$, the quadratic twists of such curves also have good reduction at 2.

Over $K=\mathbb{Q}(\sqrt{-1})$ we can also take $u=\pm \sqrt{-1}$ since $16\pm \sqrt{-1}$ have prime norm 257. This gives eight elliptic curves, four in one isogeny class 2.0.4.1‐257.1‐a with conductor $\mathfrak{p}=(16+\sqrt{-1})$, linked by 2‐isogenies, and their Galois conjugates in isogeny class 2.0.4.1‐257.2‐a. The quadratic twists by $1\pm 16\sqrt{-1}$ have good reduction at 2 and give curves of conductor ${\mathfrak{p}}^2$ in isogeny classes 2.0.4.1‐66049.1‐a and 2.0.4.1‐66049.3‐a.

Over $K=\mathbb{Q}(\sqrt{-3})$ let $\epsilon$ be a sixth root of unity generating the unit group. Taking $u={\epsilon }^2$ or its Galois conjugate, we obtain elliptic curves with prime conductors $\mathfrak{p}$ of norm 241. Again there are two Galois conjugate isogeny classes 2.0.3.1‐241.1‐a and 2.0.3.1‐241.3‐a, each containing four elliptic curves linked by 2‐isogenies. The quadratic twists by $16\pm u$ have good reduction at 2, conductor ${\mathfrak{p}}^2$, in isogeny classes 2.0.3.1‐58081.1‐a and 2.0.3.1‐58081.3‐a.

The next result shows that, apart from these sporadic cases, all elliptic curves with odd prime conductor and rational 2‐torsion come from an analogue of the Setzer–Neumann family over $\mathbb{Q}$.

Theorem 5.12.Let $K$ be an imaginary quadratic field with class number 1, and $\epsilon$ a generator of its unit group. Let $E$ be an elliptic curve defined over $K$ with conductor an odd prime power ${\mathfrak{p}}^r$ and a $K$‐rational 2‐torsion point. Assume that $E$ has a quadratic twist with multiplicative reduction at $\mathfrak{p}$. Then $E$ is either

• (i) one of the sporadic curves listed above, where $\mathfrak{p}$ has norm 17 (over all fields), or 257 (over $K=\mathbb{Q}(\sqrt{-1})$ only) or 241 (over $K=\mathbb{Q}(\sqrt{-3})$ only); or
• (ii) isomorphic or 2‐isogenous to $E_{a,b}$ where $b=16\epsilon$ and $a$ satisfies an equation of the form
  $$a^2=u{\pi }^r+64\epsilon ,$$
  with $r$ odd, $u$ a unit, and $u{\pi }^r\equiv 1(mod\frac{8}{e_2})\text{;}$ or
• (iii) the quadratic twist by $u\pi$ of the previous case, without any congruence condition.

Proof.We start with the observation that in each case $\epsilon$ is not congruent to a square modulo 4, which may be checked easily and which will be used repeatedly.

As before we use Theorem 5.4 to first find the curves with minimal parameters, arising from solutions to 5.5 with $A=s=1$. Up to 2‐isogeny, we may assume that $a=\stackrel{\sim }{a}$ is odd, that $B$ is odd, and $C$ divisible by 64 with $C/64$ odd, except in the case $K=\mathbb{Q}(\sqrt{-7})$ where $B$ and $C$ are each divisible by the sixth power of one of the two primes dividing 2, or the supersingular case over $\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-2})$. We will leave these last cases to the end.

Scaling by squares of units, we must solve each of the following equations:

$$a^2=P+64$$(5.6)

$$a^2=1+64P$$(5.7)

$$a^2=P+64\epsilon$$(5.8)

$$a^2=\epsilon +64P$$(5.9)

where $P$ is an odd prime power, that is, an element of ${\mathcal{O}}_K$ with precisely one prime factor. We immediately see that 5.9 has no solution modulo 4.

5.6 factors as $(a-8)(a+8)=P$. Without loss of generality (changing $a$ for $-a$ if necessary) we have $P\mid (a-8)$; writing $a=8+Pt$ leads to $t(16+Pt)=1$, so $t$ is a unit, and $16-t^{-1}=-Pt$. Setting $u=-t^{-1}$ leads to one of the sporadic cases (we have one of the curves 2‐isogenous to $E_{-t^{-1}}$) and its quadratic twists.

5.7 factors as $(a-1)(a+1)=64P$. Now $P$ divides one factor, and also one factor is divisible exactly by 2, the other by 32. By symmetry this gives two cases to consider: if $a=1+32Pt$ with $t$ odd, then $t(1+16Pt)=1$ so $t$ is a unit and $16Pt=t^{-1}-1$ which is impossible. Otherwise $a=1+2Pt$ with $t$ odd, and $t(1+Pt)=16$ so again $t$ is a unit and we have a sporadic case (a twist of $E_{-t}$).

In 5.8 we divide according to whether the valuation $r$ of $P$ is even or odd. If even then we must have $P=Q^2$ with $Q$ a prime power, since $P=\epsilon Q^2$ gives a contradiction modulo 4. Now $(a-Q)(a+Q)=64\epsilon$; by symmetry $a=Q+32t$ with $t$ odd, so $t(Q+16t)=\epsilon$, leading to the third sporadic case (a twist of $E_{-\epsilon t^2}$).

Otherwise in 5.8 we have $P=u{\pi }^r$ with $u$ a unit and $r$ odd, leading to the Setzer–Neumann family. Recall that $c_4$ is odd and $2c_6=a(9b-2a^2)$, hence Proposition 5.1 implies that $a\equiv \square (mod4)$ so $a^2\equiv 1(mod8)$ if 2 is unramified in $K$ and $a^2\equiv 1(mod4)$ otherwise. In any case, the same criterion implies that the quadratic twist by $u\pi$ has good reduction at 2.

Over $K=\mathbb{Q}(\sqrt{-7})$ we must also consider the equation

$$a^2=\pm T+UP$$(5.10)

(up to Galois conjugation and 2‐isogeny) where $T={\alpha }^6$ with $\alpha =(1+\sqrt{-7})/2$ and $U=\overline{T}$ so that $TU=64$; here $P$ again denotes a prime power. The minus sign is impossible modulo ${\overline{\alpha }}^2$, and with the plus sign we can factor as $(a-{\alpha }^3)(a+{\alpha }^3)=UP$. Arguing as in earlier cases one finds that this equation has no solutions.

Lastly we consider curves which are supersingular at $\mathfrak{q}\mid 2$, which by Theorem 5.4 and Corollary 5.3 arise from solutions to the following equations:

$${\stackrel{\sim }{a}}^2=P+1$$(5.11)

$${\stackrel{\sim }{a}}^2=P+\epsilon$$(5.12)

with $a=2\stackrel{\sim }{a}$.

5.11 factors as $(\stackrel{\sim }{a}-1)(\stackrel{\sim }{a}+1)$, and one of the factors is a unit. This gives solutions $P=3$ and $P=-1\pm 2i$ over $\mathbb{Q}(i)$ but the associated curves with $(a,b)=(4,1)$ and $(2\pm 2i,1)$ have bad reduction at $1+i$ as do all their quadratic twists.

In 5.12 the base curve has $(a,b)=(2\stackrel{\sim }{a},\epsilon )$ with $(c_4,c_6)=(2^4(4P+\epsilon ),2^6\stackrel{\sim }{a}(-8P+\epsilon ))$. We scale by $\tau =1+i$ (respectively $\sqrt{-2}$) to get $(c_4,c_6)=({\tau }^4(4P+i),{\tau }^6\stackrel{\sim }{a}(8P-i))$ over $\mathbb{Q}(i)$ or $(c_4,c_6)=({\tau }^4(4P-1),{\tau }^6\stackrel{\sim }{a}(-8P-1))$ over $\mathbb{Q}(\sqrt{-2})$, respectively. We must test whether these, or their twists by $s\in \{1,\epsilon ,\pi ,\epsilon \pi \}$ have good reduction at $\tau$. Note that $v_{\mathfrak{q}}(c_4)=4$ and $v_{\mathfrak{q}}(c_6)⩾7$, and that we are in the second case of Proposition 5.1, with $a_1=\tau$ (since we are in the supersingular case).

Over $\mathbb{Q}(i)$ the first congruence in Proposition 5.1 reduces to $1+i{s}^2\equiv 0(mod2)$ which is impossible.

Over $\mathbb{Q}(\sqrt{-2})$, in the notation of Proposition 5.1 the first condition on $d$ is always satisfied (since $s$ is odd), while the second is that either $(1-s^2)/2$ or $(1-s^2)/2+2\tau$ is a square modulo 4, depending on whether $v_{\mathfrak{q}}(\stackrel{\sim }{a})⩾2$ or $v_{\mathfrak{q}}(\stackrel{\sim }{a})=1$. At least one of these is satisfied provided that $s\equiv \pm 1(mod{\tau }^3)$, and in either case $d/16\equiv 0(mod4)$. But now the final condition implies $s^2\equiv -1(mod4)$, contradiction. $\square$

The above classification implies the following crucial fact, used in the main theorem of the paper, and which was an important motivation for this section.

Corollary 5.13.Every isogeny class of elliptic curves defined over $K$ with prime conductor and a $K$‐rational 2‐torsion point contains a curve whose discriminant has odd valuation.

Proof.If $E$ is a curve over $K$ of prime conductor $\mathfrak{p}$ and a $K$‐rational 2‐torsion point, we are in the multiplicative case. By Theorem 5.12 $E$ is either a sporadic curve of conductor norm 17, 241 over $\mathbb{Q}(\sqrt{-1})$ or 257 over $\mathbb{Q}(\sqrt{-3})$ (all of these have a curve with prime discriminant in their isogeny class) or is isogenous to $E_{a,b}$ with $b=16\epsilon$ and $a^2=u{\pi }^r+64\epsilon$ with $r$ odd. Such curves have discriminant $2^{8}u{\epsilon }^2{\pi }^r$, so odd valuation. $\square$

Remark 5.14.The computations done in this section could be generalized to other number fields of class number 1, as the number of units modulo squares is always finite. The case of real quadratic fields is of particular interest, requiring almost no modification except to allow for ${\mathcal{O}}_K^{\ast }/{({\mathcal{O}}_K^{\ast })}^2$ having order 4. In this case, Proposition 5.11 gives a possibly infinite family of elliptic curves with bad reduction only at the primes dividing $u^k+16$, where $u$ is a fundamental unit. For example, over $\mathbb{Q}(\sqrt{5})$, we get curves of prime conductor with norms $1009,35569,1659169,\dots ,$ but to our knowledge it is not known whether we can get infinitely many curves of prime conductor in this way.

We end this section with an interesting phenomenon concerning curves of prime conductor and rational 2‐torsion.

Theorem 5.15.Let $K$ be an imaginary quadratic field with class number 1, and $E/K$ be an elliptic curve of prime power conductor with a $K$‐rational 2‐torsion point. Then $E$ has rank 0.

Proof.A simple 2‐descent computation shows that this is the case for curves in the Setzer–Neumann family (this phenomenon also occurs for rational elliptic curves, and the proof is the same). The remaining sporadic cases can be handled by looking at tables 20 or computing the rank of the curve 27.a1 over the different fields $K$ directly, for example using SageMath 36 (using a Pari/GP implementation due to Denis Simon based on the article 35). $\square$

6 On some irreducible finite flat group schemes over $Spec({\mathcal{O}}_K)$

Let $E/K$ be a modular elliptic curve of odd prime conductor, whose discriminant is a square. Then by Theorem 2.2 $E[2]$ is a finite flat group scheme over $Spec({\mathcal{O}}_K)$ of type (2,2). It is either reducible or irreducible. In the reducible case, it contains a factor isomorphic to the multiplicative or the additive group ([37, Corollary p. 21]), hence the curve has a point of order 2 as studied in the previous section. The group ${Aut}_{G_K}(E[2])$ cannot be isomorphic to the whole of ${GL}_2({\mathbb{F}}_2)$ as proved in Theorem 2.3, so the only remaining possibility for it is the cyclic group of order 3. Note that such a group scheme $E[2]$ does not occur over $Spec(\mathbb{Z})$ as there are no cubic extensions of $\mathbb{Q}$ unramified outside 2. The same is true for four of the nine imaginary quadratic fields under consideration here.

Lemma 6.1.There are no elliptic curves $E/K$ of odd prime conductor and even discriminant valuation whose residual 2‐adic Galois representation is cyclic of order 3 for $K=\mathbb{Q}(\sqrt{-d})$, $d=1,2,3$, or 7.

Proof.By the aforementioned result of Mazur, the extension $L/K$ obtained by adjoining the 2‐torsion points of $E$ is a cyclic cubic extension unramified outside 2. It is easy to verify (for example using explicit class field theory as implemented in 24) that there is no such extension for these particular fields $K$. $\square$

For the remaining fields $K=\mathbb{Q}(\sqrt{-d})$ for $d\in \{11,19,43,67,163\}$, there is a unique cyclic cubic extension $L/K$ unramified outside 2, namely the ring class field associated to the order $\mathbb{Z}[\sqrt{-d}]$ of index 2, which has class number 3 (since 2 is inert). These are the splitting fields of the following polynomials $p(x)$ of discriminant $-4d\mathit{\text{:}}$

• (i) $p(x)=x^3-x^2+x+1$ over $\mathbb{Q}(\sqrt{-11})$,
• (ii) $p(x)=x^3-2x-2$ over $\mathbb{Q}(\sqrt{-19})$,
• (iii) $p(x)=x^3-x^2-x+3$ over $\mathbb{Q}(\sqrt{-43})$,
• (iv) $p(x)=x^3-x^2-3x+5$ over $\mathbb{Q}(\sqrt{-67})$,
• (v) $p(x)=x^3-8x+10$ over $\mathbb{Q}(\sqrt{-163})$.

In particular the splitting field of $E[2]$ must be one of these fields $L$. Moreover, for each of these five values of $d$ there is an elliptic curve $E$ defined over $\mathbb{Q}$ with prime conductor $d$ and discriminant $-d$, namely 11.a3, 19.a3, 43.a1, 67.a1, and 163.a1. In each case the base change to $\mathbb{Q}(\sqrt{-d})$ has prime conductor $(\sqrt{-d})$ and square discriminant $-d={\sqrt{-d}}^2$, and the 2‐division field is the splitting field of the corresponding cubic $p(x)$.

We can construct further examples over each field as follows. Rubin and Silverberg showed in [26, Theorem 1] how to parametrize all elliptic curves with given level 2 structure: given one curve $E:y^2=x^3+ax+b$, all curves with residual 2‐adic representation isomorphic to that of $E$ are obtained by specializing the family of curves

$$y^2=x^3+A_4(u,v)x+A_6(u,v)$$(6.1)

with

$$\begin{array}{cc}\hfill A_4(u,v)& =3(3a{v}^2+9buv-a^2{u}^2),\hfill \\ \hfill A_6(u,v)& =27b{v}^3-18a^{2}u{v}^2-27ab{u}^{2}v-(2a^3+27b^2)u^3\hfill \end{array}$$

at a pair of elements $(u,v)$ of $K$. Note that scaling $u,v$ by $c\in K^{\ast }$ gives the quadratic twist by $c$; hence, up to isomorphism, we may assume that $u,v\in {\mathcal{O}}_K$ with square‐free gcd. The discriminant of 6.1 is $2^4{3}^6\mathrm{\Delta }(F)F^2$ where $F(u,v)=v^3+av{u}^2+b{u}^3$ with discriminant $\mathrm{\Delta }(F)=-(4a^3+27b^2)$.

Theorem 6.2.Let $K$ be an imaginary quadratic field of class number 1 and $E/K$ be a semistable elliptic curve whose residual 2‐adic representation has image cyclic of order 3. Then $K=\mathbb{Q}(\sqrt{-d})$ for $d=11,19,43,67,163$ and the valuation of $\mathcal{D}(E)$ at each prime of bad reduction is exactly 2.

Proof.The first condition comes from Lemma 6.1. Secondly, by the proof of Theorem 2.3 the valuation of $\mathcal{D}(E)$ is a power of 2. It remains to show that the discriminant cannot be a fourth power: to see this, we use the parametrized families 6.1, recalling that 2 is inert in each case.

From 6.1 for each field, after a little simplification we get $A_4=-6P(u,v)$ and $A_6=2Q(u,v)$, with discriminant $1728(8P{(u,v)}^3-Q{(u,v)}^2)$ where

• (i) $P=2u^2-17uv-v^2$, $Q=586u^3+102u^{2}v+12u{v}^2-17v^3$ for $d=11$,
• (ii) $P=2u^2+9uv+3v^2$, $Q=46u^3+54u^{2}v+36u{v}^2+27v^3$ for $d=19$,
• (iii) $P=8u^2-35uv+2v^2$, $Q=-2386u^3+420u^{2}v-48u{v}^2+35v^3$ for $d=43$,
• (iv) $P=50u^2-53uv+5v^2$, $Q=-4618u^3+1590u^{2}v-300u{v}^2+53v^3$ for $d=67$,
• (v) $P=32u^2+45uv+12v^2$, $Q=838u^3+1080u^{2}v+576u{v}^2+135v^3$ for $d=163$.

If $E$ has good reduction at 2, since $\mathcal{D}(E)$ is even, it must be divisible by $2^{12}$, so $4\mid Q(u,v)$. Since $Q(u,v)\equiv v^3(mod2)$, $v$ must be even, and the condition $4\mid Q(u,v)$ implies that $u$ is even as well. After the substitution $(u,v)\to (u/2,v/2)$, the new $Q(u,v)$ is odd (from the minimality condition) with invariants $c_4=72P(u,v)$ and $c_6=-216Q(u,v)$. Clearly $v_2(c_4)⩾3$, but if it equals 3, we cannot get good reduction at 2 by Kraus's criterion since the conditions

$$\begin{array}{ccc}\hfill d=-a_1^6+3a_1^2{c}_4+2c_6& \equiv & 0(mod16)\hfill \\ \hfill 4a_1^{2}d& \equiv & {(a_1^4-c_4)}^2(mod256),\hfill \end{array}$$

are not compatible. The first one implies that $a_1$ is even, hence the left‐hand side of the second equation is zero, while the right‐hand side is not. Then $2\mid P(u,v)$ and $v_2(c_4)⩾4$. Kraus's criterion now implies that there exists $a_1$ such that $a_1^2\equiv c_6/8=-27Q(u,v)\equiv Q(u,v)(mod4)$. The discriminant is now $27(8P^3-Q^2)\equiv 5Q^2\equiv 5a_1^4(mod8)$, which cannot be a fourth power since 5 is not a fourth power modulo 8. $\square$

A natural question is whether there are infinitely many curves of prime conductor, whose discriminant is a prime square. They are all obtained by evaluating the previous equations at suitable pairs $(u,v)$. The model described above has discriminant $-2^6{3}^{6}dF{(u,v)}^2$, where $F(u,v)$ is an explicit cubic form of discriminant $-4d$. The values of $(u,v)$ to get good reduction at 2, $3$, and $\sqrt{-d}$ are given by congruence conditions, each one giving a potentially infinite family, where one expects the cubic $F(u,v)$ to attain infinitely many prime values.

To end the paper we give an example of an elliptic curve of this type over each field, in addition to the base change examples given above.

• (1) $K=\mathbb{Q}(\sqrt{-11})=\mathbb{Q}(\alpha )$ where ${\alpha }^2-\alpha +3=0$:
  $$E:y^2+y=x^3+\alpha x^2-x$$
  (with LMFDB label22 The other curves here do not yet have LMFDB labels.
  2.0.11.1‐47.1‐a1) has prime conductor $\mathfrak{p}=(\pi )$ with $\pi =7-2\alpha$ of norm 47, and discriminant ${\pi }^2$.
• (2) $K=\mathbb{Q}(\sqrt{-19})=\mathbb{Q}(\alpha )$ where ${\alpha }^2-\alpha +5=0$:
  $$E:y^2+y=x^3+(-\alpha -1)x^2+(2\alpha )x+(-\alpha -1)$$
  has prime conductor $\mathfrak{p}=(\pi )$ with $\pi =18\alpha -7$ of norm 1543, and discriminant ${\pi }^2$.
• (3) $K=\mathbb{Q}(\sqrt{-43})=\mathbb{Q}(\alpha )$ where ${\alpha }^2-\alpha +11=0$:
  $$E:y^2+y=x^3+(\alpha -1)x^2+(-\alpha -2)x+2$$
  has prime conductor $\mathfrak{p}=(\pi )$ with $\pi =29-2\alpha$ of norm 827, and discriminant ${\pi }^2$.
• (4) $K=\mathbb{Q}(\sqrt{-67})=\mathbb{Q}(\alpha )$ where ${\alpha }^2-\alpha +17=0$:
  $$E:y^2+y=x^3+(\alpha +1)x^2+2\alpha x+(\alpha -1)$$
  has prime conductor $\mathfrak{p}=(\pi )$ with $\pi =6\alpha -65$ of norm 4447, and discriminant ${\pi }^2$.
• (5) $K=\mathbb{Q}(\sqrt{-163})=\mathbb{Q}(\alpha )$ where ${\alpha }^2-\alpha +41=0$:
  $$E:y^2+y=x^3+(\alpha +1)x^2+(\alpha -18)x+(-3\alpha -4)$$
  has prime conductor $\mathfrak{p}=(\pi )$ with $\pi =47+6\alpha$ of norm 3967, and discriminant ${\pi }^2$.

Acknowledgements

We would like to thank Nicolas Vescovo for participating in some discussions of the present article, to Luis Dieulefait, for explaining to us some technicalities used in the proof of Theorem 2.5, to Angelos Koutsianas for his computations used in the proof of Theorem 2.3, and to Samir Siksek and Haluk Şengün for many useful conversations. The second author would like to thank the University of Warwick for its hospitality during his visit as a Leverhulme Visiting Professor. Last but not least, we would like to thank the anonymous referee for many useful comments and corrections.

1 This can be avoided over $\mathbb{Q}$, since for every odd prime ideal $(p)$, either $\pm p\equiv 1(mod4)$, and $\mathbb{Q}(\sqrt{\pm p})$ is unramified at 2.

2 The other curves here do not yet have LMFDB labels.