Counting elliptic curves over the rationals with a 7-isogeny

Abstract

We count by height the number of elliptic curves over the rationals, both up to isomorphism over the rationals and over an algebraic closure thereof, that admit a cyclic isogeny of degree 7.

1 Introduction

1.1 Motivation and setup

Number theorists have an enduring, and recently renewed, interest in the arithmetic statistics of elliptic curves: broadly speaking, we study asymptotically the number of elliptic curves of bounded size with a given property. More precisely, every elliptic curve E over $\mathbb {Q}$ is defined uniquely up to isomorphism by a Weierstrass equation of the form

$$\begin{aligned} E :y^2 = x^3 + Ax + B \end{aligned}$$

(1.1.1)

with $A,B \in \mathbb {Z}$ satisfying $4A^3+27B^2 \ne 0$ and such that no prime $\ell$ has $\ell ^4 \mid A$ and $\ell ^6 \mid B$. Let $\mathscr {E}$ be the set of elliptic curves of this form: we define the height of $E \in \mathscr {E}$ by

$$\begin{aligned} {{\,\textrm{ht}\,}}(E) \,{:=}\,\max (\left| 4A^3\right| ,\left| 27B^2\right| ). \end{aligned}$$

(1.1.2)

For $X \ge 1$, let $\mathscr {E}_{\le X} \,{:=}\,\{E \in \mathscr {E}: {{\,\textrm{ht}\,}}(E) \le X\}$. Mathematicians have studied the count of those $E \in \mathscr {E}_{\le X}$ which admit (or are equipped with) additional level structure as $X \rightarrow \infty$, and they have done so more generally over global fields.

In recent work, many instances of this problem have been resolved. For example, Harron–Snowden [11] and Cullinan–Kenney–Voight [6] (see also previous work of Duke [9] and Grant [10]) produced asymptotics for counting those elliptic curves E for which the torsion subgroup $E(\mathbb {Q})_{tors }$ of the Mordell–Weil group is isomorphic to a given finite abelian group T, i.e., they estimated $\#\left\{ E \in \mathscr {E}_{\le X} : E(\mathbb {Q})_{{{\,\textrm{tors}\,}}} \simeq T\right\}$ as $X \rightarrow \infty$ for each of the fifteen groups T indicated in Mazur’s theorem on torsion. These cases correspond to genus zero modular curves with infinitely many rational points. For such T, they established an asymptotic with an effectively computable constant and a power-saving error term. Moreover, satisfactory interpretations of the exponent of X and the constants appearing in these asymptotics are provided. The main ingredients in the proof are the Principle of Lipschitz (also called Davenport’s Lemma [7]) and an elementary sieve.

Moving on, we consider asymptotics for

$$\begin{aligned} \# \left\{ E \in \mathscr {E}_{\le X} : E \text { admits a cyclic } N\text {-isogeny}\right\} \end{aligned}$$

(1.1.3)

(where we mean that the N-isogeny is defined over $\mathbb {Q}$). Our attention is again first drawn to the cases where the modular curve $Y_0(N)$, parametrizing elliptic curves with a cyclic N-isogeny, has genus zero: namely, $N=1, \dots , 10, 12, 13, 16, 18, 25$. For $N \le 4$, we again have an explicit power-saving asymptotic, with the case $N=3$ due to Pizzo–Pomerance–Voight [17] and the case $N=4$ due to Pomerance–Schaefer [18]. For all but four of the remaining values, namely $N=7,10,13,25$, Boggess–Sankar [3] provide at least the correct growth rate. For both torsion and isogenies, work of Bruin–Najman [5] and Phillips [16] extend these counts to a general number field K.

However, the remaining four cases have quite stubbornly resisted these methods. The obstacle can be seen in quite elementary terms. Although there is no universal elliptic curve with a cyclic N-isogeny, every such elliptic curve is of the form $dy^2 = x^3 + f(t)x + g(t)$ with $f(t),g(t) \in \mathbb {Q}[t]$ (for $t \in \mathbb {Q}$ away from a finite set and $d \in \mathbb {Z}$ a squarefree twisting parameter). For these four values of N, we have $\gcd (f(t),g(t)) \ne 1$. Phrased geometrically, the elliptic surface over $\mathbb {P}^1$ defined by $y^2=x^3+f(t)x+g(t)$ has places of additive reduction (more precisely, type II). Either way, this breaks the sieve—and new techniques are required.

1.2 Results

For $X \ge 1$, let

$$\begin{aligned} N_{}(X) \,{:=}\,\# \left\{ E \in \mathscr {E}_{\le X} : E \text { admits a (cyclic)} \ 7\text {-isogeny}\right\} . \end{aligned}$$

(1.2.1)

Our main result is as follows (Theorem 5.2.4).

Theorem 1.2.2

There exist effectively computable $c_1,c_2 \in \mathbb {R}_{>0}$ such that for every $\epsilon > 0$, we have

$$\begin{aligned} N_{}(X) = c_1 X^{1/6} \log X + c_2 X^{1/6} + O(X^{3/20 + \epsilon }) \end{aligned}$$

as $X \rightarrow \infty$, where the implied constant depends on $\epsilon$.

The constants $c_1,c_2$ in Theorem 1.2.2 are explicitly given, and estimated numerically in Sect. 6 as $c_1 = 0.09285536\ldots$ and $c_2 \approx -0.16405$. As 7 is prime, every 7-isogeny is cyclic, so we omit this adjective for the remainder of our paper. It turns out that no elliptic curve over $\mathbb {Q}$ admits two 7-isogenies with distinct kernels (Proposition 2.2.6), so $N_{}(X)$ also counts elliptic curves equipped with a 7-isogeny.

The first step in our strategy to prove Theorem 1.2.2 diverges from the methods of Boggess–Sankar [3] and Phillips [16], where the twists are resolved by use of a certain modular curve (denoted by $X_{1/2}(N)$). Instead, we first count twist classes directly, as follows. Let $\mathbb {Q}^{al }$ be an algebraic closure of $\mathbb {Q}$. Up to isomorphism over $\mathbb {Q}^{al }$, every elliptic curve E over $\mathbb {Q}$ with $j(E) \ne 0,1728$ has a unique Weierstrass model (1.1.1) with the additional property that $B>0$ and no prime $\ell$ has $\ell ^2 \mid A$ and $\ell ^3 \mid B$; such a model is called twist minimal. (See Sect. 2.1 for $j(E)=0,1728$.) Let $\mathscr {E}^{tw }\subset \mathscr {E}$ be the set of twist minimal elliptic curves, and let $\mathscr {E}^{\textrm{tw}}_{\le X}\,{:=}\,\mathscr {E}^{tw }\cap \mathscr {E}_{\le X}$ be those with height at most X. Accordingly, we obtain asymptotics for

$$\begin{aligned} N_{}^{tw }(X) \,{:=}\,\# \{E \in \mathscr {E}^{\textrm{tw}}_{\le X}: E \text { admits a } 7\text {-isogeny}\} \end{aligned}$$

(1.2.3)

as follows (Theorem 4.2.19).

Theorem 1.2.4

We have

$$\begin{aligned} N_{}^{tw }(X) = 3\zeta (2)c_1 X^{1/6} + O(X^{2/15} \log ^{17/5} X) \end{aligned}$$

as $X \rightarrow \infty$, with $c_1$ as in Theorem 1.2.2.

For an outline of the proof, see Sect. 4.1. The use of the Principle of Lipschitz remains fundamental, but the sieving is more involved: we decompose the function into progressively simpler pieces that can be estimated. (See Remark 2.2.14 for a stacky interpretation.) We then deduce Theorem 1.2.2 from Theorem 1.2.4 by counting twists using a Tauberian theorem (attributed to Landau). The techniques of this paper can be adapted to handle the cases $N = 10, 13, 25$, which have places of type III additive reduction; these will be treated in upcoming work.

1.3 Contents

In Sect. 2, we set up basic notation and investigate minimal twists. In Sect. 3, we tersely review some needed facts from analytic number theory. In Sect. 4, we pull together material from the earlier sections to prove Theorem 1.2.4. In Sect. 5, we use Landau’s Tauberian theorem and Theorem 1.2.4 to obtain Theorem 1.2.2. In Sect. 6, we describe algorithms to compute the various quantities we study in this paper, and report on their outputs.

2 Elliptic curves and isogenies

In this section, we set up what we need from the theory of elliptic curves.

2.1 Height, minimality, and defect

We begin with some notation and terminology (repeating and elaborating upon the introduction); we refer to Silverman [20, Chapter III] for background.

Let E be an elliptic curve over $\mathbb {Q}$. Recall that a (simplified) integral Weierstrass equation for E is an affine model of the form

$$\begin{aligned} y^2 = x^3 + Ax + B \end{aligned}$$

(2.1.1)

with $A,B \in \mathbb {Z}$. Let

$$\begin{aligned} H(A,B) \,{:=}\,\max (\left| 4A^3\right| ,\left| 27B^2\right| ). \end{aligned}$$

(2.1.2)

The largest $d \in \mathbb {Z}_{>0}$ such that $d^4 \mid A$ and $d^6 \mid B$ is called the minimality defect ${{\,\textrm{md}\,}}(A,B)$ of the model. We then define the height of E to be

$$\begin{aligned} {{\,\textrm{ht}\,}}(E)={{\,\textrm{ht}\,}}(A,B) \,{:=}\,\frac{H(A,B)}{{{\,\textrm{md}\,}}(A,B)^{12}}, \end{aligned}$$

(2.1.3)

well-defined up to isomorphism. In fact, E (up to isomorphism over $\mathbb {Q}$) has unique minimal model

$$\begin{aligned} y^2=x^3+(A/d^4)x+(B/d^6) \end{aligned}$$

with minimality defect $d=1$. Let $\mathscr {E}$ be the set of elliptic curves over $\mathbb {Q}$ in their minimal model, and let

$$\begin{aligned} \mathscr {E}_{\le X} \,{:=}\,\{E \in \mathscr {E}: {{\,\textrm{ht}\,}}(E) \le X\}. \end{aligned}$$

(2.1.4)

Let $\mathbb {Q}^{al }$ be an algebraic closure of $\mathbb {Q}$. We may similarly consider all integral Weierstrass equations for E which define a curve isomorphic to E over $\mathbb {Q}^{al }$—these are the twists of E (defined over $\mathbb {Q}$). Let E have $j(E) \ne 0,1728$. We call the largest $e\in \mathbb {Z}_{>0}$ such that $e^2 \mid A$ and $e^3 \mid B$ the twist minimality defect of a model (2.1.1), denoted ${{\,\textrm{tmd}\,}}(A,B)$. Explicitly, we have

$$\begin{aligned} {{\,\textrm{tmd}\,}}(E) = {{\,\textrm{tmd}\,}}(A,B) \,{:=}\,\prod _{\ell } \ell ^{v_\ell }, \quad \text {where} \ v_\ell \,{:=}\,\lfloor \min ({{\,\textrm{ord}\,}}_\ell (A)/2,{{\,\textrm{ord}\,}}_\ell (B)/3) \rfloor , \end{aligned}$$

(2.1.5)

with the product over all primes $\ell$. As above, we then define the twist height of E to be

$$\begin{aligned} {{\,\textrm{twht}\,}}(E)={{\,\textrm{twht}\,}}(A,B) \,{:=}\,\frac{H(A,B)}{{{\,\textrm{tmd}\,}}(A,B)^{6}}, \end{aligned}$$

(2.1.6)

well-defined on the $\mathbb {Q}^{al }$-isomorphism class of E; and E has a unique model over $\mathbb {Q}$ up to isomorphism over $\mathbb {Q}^{al }$ with twist minimality defect ${{\,\textrm{tmd}\,}}(E) = e = 1$ and $B > 0$, which we call twist minimal, namely,

$$\begin{aligned} y^2 = x^3 + (A/e^2)x + |B|/e^3. \end{aligned}$$

(2.1.7)

For $j=0,1728$, we choose twist minimal models as follows:

• If $j(E)=0$ (equivalently, $A=0$), then we take $y^2=x^3 + 1$ of twist height 27.

• If $j(E)=1728$ (equivalently, $B=0$), then we take $y^2=x^3+x$ of twist height 4.

Let $\mathscr {E}^{tw }\subset \mathscr {E}$ be the set of twist minimal elliptic curves, and let $\mathscr {E}^{\textrm{tw}}_{\le X}\,{:=}\,\mathscr {E}^{tw }\cap \mathscr {E}_{\le X}$ be those with twist height at most X. If $E \in \mathscr {E}$ has $j(E) \ne 0,1728$, then the set of twists of E in $\mathscr {E}$ are precisely those of the form $E^{(c)} :y^2 = x^3 + c^2 A x + c^3 B$ for $c \in \mathbb {Z}$ squarefree, and

$$\begin{aligned} {{\,\textrm{ht}\,}}(E^{(c)})=c^6 {{\,\textrm{twht}\,}}(E). \end{aligned}$$

(2.1.8)

If further $E \in \mathscr {E}^{tw }$, then of course ${{\,\textrm{twht}\,}}(E)={{\,\textrm{ht}\,}}(E)$. (For $j(E)=0,1728$, we instead have sextic and quartic twists, but these will not figure here: see Proposition 2.2.6.)

Remark 2.1.9

This setup records in a direct manner the more intrinsic notions of height coming from moduli stacks. The moduli stack $Y(1)_\mathbb {Q}$ of elliptic curves admits an open immersion into a weighted projective line $Y(1) \hookrightarrow \mathbb {P}(4,6)_\mathbb {Q}$ by $E \mapsto (A:B)$ for any choice of model (2.1.1), and the height of E is the height of the point $(A:B) \in \mathbb {P}(4,6)(\mathbb {Q})$ associated to $\mathscr {O}_{\mathbb {P}(4,6)}(12)$ (with coordinates harmlessly scaled by 4, 27): see Bruin–Najman [5, Sects. 2, 7] and Phillips [16, Sect. 2.2]. Similarly, the height of the twist minimal model is given by the height of the point $(A:B) \in \mathbb {P}(2,3)(\mathbb {Q})$ associated to $\mathscr {O}_{\mathbb {P}(2,3)}(6)$, which is almost but not quite the height of the j-invariant (in the usual sense).

2.2 Isogenies of degree 7

Next, we gather the necessary input from modular curves. Recall that the modular curve $Y_0(7)$, defined over $\mathbb {Q}$, parametrizes pairs $(E,\phi )$ of elliptic curves E equipped with a 7-isogeny $\phi$ up to isomorphism, or equivalently, a cyclic subgroup of order 7 stable under the absolute Galois group ${{\,\textrm{Gal}\,}}_\mathbb {Q}\,{:=}\,{{\,\textrm{Gal}\,}}(\mathbb {Q}^{al }\,|\,\mathbb {Q})$. For further reference on the basic facts on modular curves used in this section, see e.g. Diamond–Shurman [8] and Rouse–Sutherland–Zureick-Brown [19, Sect. 2]. We compute that the coarse space of $Y_0(7)$ is an affine open in $\mathbb {P}^1$, so the objects of interest are parametrized by its coordinate $t \ne -7,\infty$ (see Lemma 2.2.2).

More precisely, define

$$\begin{aligned} f_0(t)&\,{:=}\,-3 (t^2 - 231 t + 735) \nonumber \\&= -3 (t^2 - (3 \cdot 7 \cdot 11)t + (3 \cdot 5 \cdot 7^2)), \nonumber \\ g_0(t)&\,{:=}\,2 (t^4 + 518 t^3 - 11025 t^2 + 6174 t - 64827) \nonumber \\&= 2 (t^4 + (2 \cdot 7 \cdot 31)t^3 - (3^2 \cdot 5^2 \cdot 7^2)t^2 + (2 \cdot 3^2 \cdot 7^3)t - (3^3 \cdot 7^4)), \nonumber \\ h(t)&\,{:=}\,t^2 + t + 7, \nonumber \\ f(t)&\,{:=}\,f_0(t) h(t), \nonumber \\ g(t)&\,{:=}\,g_0(t) h(t). \end{aligned}$$

(2.2.1)

Then $h(t) = \gcd (f(t), g(t))$.

Lemma 2.2.2

The set of elliptic curves E over $\mathbb {Q}$ that admit a 7-isogeny (defined over $\mathbb {Q}$) are precisely those of the form $E :y^2 = x^3 + c^2f(t)x + c^3g(t)$ for some $c \in \mathbb {Q}^\times$ and $t \in \mathbb {Q}$ with $t \ne -7$.

Proof

Routine calculations with q-expansions for modular forms on the group $\Gamma _0(7)$, with the cusps at $t=-7,\infty$ show that every elliptic curve E over $\mathbb {Q}$ that admits a 7-isogeny is a twist of

$$\begin{aligned} E : y^2 = x^3 + f(t) x + g(t) \end{aligned}$$

for some $t \in \mathbb {Q}$. But f(t) and g(t) have no roots in $\mathbb {Q}$, so these twists must be quadratic, as desired. See [6, Proposition 3.3.16] for a similar but more expansive argument. $\square$

Of course, for elliptic curves up to isomorphism over $\mathbb {Q}^{al }$, we can ignore the factor c in Lemma 2.2.2.

Remark 2.2.3

Let

$$\begin{aligned} f^\prime _0(t)&\,{:=}\,-3 (t^2 + 9 t + 15) \nonumber \\&= -3 (t^2 + (3^2) t + (3 \cdot 5)), \nonumber \\ g^\prime _0(t)&\,{:=}\,2 (t^4 + 14 t^3 + 63 t^2 + 126 t + 189) \nonumber \\&= 2 (t^4 + (2 \cdot 7) t^3 + (3^2 \cdot 7) t^2 + (2 \cdot 3^2 \cdot 7) t + (3^3 \cdot 7)), \nonumber \\ f^\prime (t)&\,{:=}\,f^\prime _0(t) h(t), \nonumber \\ g^\prime (t)&\,{:=}\,g^\prime _0(t) h(t), \end{aligned}$$

(2.2.4)

with h(t) as above. The elliptic curve E in Lemma 2.2.2 is 7-isogenous to

$$\begin{aligned} E^\prime : y^2 = x^3 + c^2 f^\prime (t) + c^3 g^\prime (t) \end{aligned}$$

via the marked 7-isogeny. Naturally, $E^\prime$ is also isogeneous to E via the dual 7-isogeny. We obtain (2.2.4) from (2.2.1) via the Atkin-Lehner involution, which in our coordinates is given by

$$\begin{aligned} w_7 : t \mapsto -\frac{7t}{t+7}. \end{aligned}$$

(2.2.5)

All of our arguments below could have been applied equally well using the parameterization (2.2.4) instead of the parameterization (2.2.1).

Proposition 2.2.6

No elliptic curve E over $\mathbb {Q}$ admits two 7-isogenies with distinct kernels, and no E over $\mathbb {Q}$ with $j(E)=0,1728$ admits a 7-isogeny.

Proof

For the first statement: if E admits two distinct 7-isogenies, then generators for each kernel give a basis for the 7-torsion of E in which ${{\,\textrm{Gal}\,}}_\mathbb {Q}$ acts diagonally. The corresponding compactified modular curve, $X_{\text {sp}}(7)$, has genus 1 and 2 rational cusps; it is isomorphic to $X_0(49)$ over $\mathbb {Q}$, and has Weierstrass equation $y^2+xy=x^3-x^2-2x-1$ and LMFDB label 49.a4. Its Mordell–Weil group is $\mathbb {Z}/2\mathbb {Z}$, so all rational points are cusps.

For the second statement, we simply observe that f(t) and g(t) have no roots $t \in \mathbb {Q}$. $\square$

To work with integral models, we take $t=a/b$ (in lowest terms) and homogenize, giving the following polynomials in $\mathbb {Z}[a,b]$:

$$\begin{aligned} C(a, b)&\,{:=}\,b^2 h(a/b)=a^2 + ab + 7 b^2, \nonumber \\ A_0(a, b)&\,{:=}\,b^2 f_0(a/b) = -3 (a^2 - 231 a b + 735 b^2), \nonumber \\ B_0(a, b)&\,{:=}\,b^4 g_0(a/b) = 2 (a^4 + 518 a^3 b - 11025 a^2 b^2 + 6174 a b^3 - 64827 b^4), \nonumber \\ A(a,b)&\,{:=}\,b^4 f(a/b) = C(a,b)A_0(a,b) \nonumber \\ B(a,b)&\,{:=}\,b^6 f(a/b) = C(a,b)B_0(a,b). \end{aligned}$$

(2.2.7)

We have $C(a,b) = \gcd (A(a,b),B(a,b)) \in \mathbb {Z}[a,b]$.

We say that a pair $(a, b) \in \mathbb {Z}^2$ is groomed if $\gcd (a, b) = 1$, $b > 0$, and $(a, b) \ne (-7, 1)$. Thus Lemma 2.2.2 and Proposition 2.2.6 provide that the elliptic curves $E \in \mathscr {E}$ that admit a 7-isogeny are precisely those with a model

$$\begin{aligned} y^2 = x^3 + \frac{c^2 A(a, b)}{d^4} x + \frac{c^3 B(a, b)}{d^6} \end{aligned}$$

(2.2.8)

where (a, b) is groomed, $c \in \mathbb {Z}$ is squarefree, and $d={{\,\textrm{md}\,}}(c^2A(a,b),c^3B(a,b))$. Thus the count

$$\begin{aligned} N_{}(X) \,{:=}\,\#\{E \in \mathscr {E}_{\le X} : E \text { admits a } 7\text {-isogeny}\} \end{aligned}$$

(2.2.9)

can be computed as

$$\begin{aligned} N_{}(X) = \#\left\{ (a, b, c) \in \mathbb {Z}^3 : \begin{array}{c} (a, b) \text { groomed, } c \text { squarefree, and } \\ {{{\,\textrm{ht}\,}}(c^2 A(a,b),c^3 B(a,b)) \le X} \end{array} \right\} . \end{aligned}$$

(2.2.10)

with the height defined as in (2.1.3).

Similarly, but more simply, the subset of $E \in \mathscr {E}^{tw }$ that admit a 7-isogeny are

$$\begin{aligned} E_{a,b} :y^2 = x^3 + \frac{A(a, b)}{e^2} x + \frac{|B(a, b)|}{e^3} \end{aligned}$$

(2.2.11)

with (a, b) groomed and $e={{\,\textrm{tmd}\,}}(A(a,b),B(a,b))$ the twist minimality defect (2.1.5). Accordingly, if we define

$$\begin{aligned} N_{}^{tw }(X) \,{:=}\,\# \{E \in \mathscr {E}^{\textrm{tw}}_{\le X}: E \text { admits a } 7\text {-isogeny}\} \end{aligned}$$

(2.2.12)

then

$$\begin{aligned} N_{}^{tw }(X) = \# \left\{ (a, b) \in \mathbb {Z}^2 : (a, b) \text { groomed and } {{\,\textrm{twht}\,}}(A(a,b),B(a,b)) \le X\right\} . \end{aligned}$$

(2.2.13)

Remark 2.2.14

Returning to Remark 2.1.9, we conclude that counting elliptic curves equipped with a 7-isogeny is the same as counting points on $\mathbb {P}(4,6)_\mathbb {Q}$ in the image of the natural map $Y_0(7) \rightarrow Y(1) \subseteq \mathbb {P}(4,6)_\mathbb {Q}$. Counting them up to twist replaces this with the further natural quotient by $(a : b) \sim (\lambda ^2 a : \lambda ^3 b)$ for $(a : b) \in \mathbb {P}(4, 6)_\mathbb {Q}$ and $\lambda \in \mathbb {Q}^\times$, which gives us $\mathbb {P}(2,3)_\mathbb {Q}$.

2.3 Twist minimality defect

The twist minimality defect is the main subtlety in our study of $N_{}^{tw }(X)$, so we analyze it right away.

Lemma 2.3.1

Let $(a,b) \in \mathbb {Z}^2$ be groomed, let $\ell$ be prime, and let $v \in \mathbb {Z}_{\ge 0}$. Then the following statements hold.

• (a) If $\ell \ne 3, 7$, then $\ell ^v \mid {{\,\textrm{tmd}\,}}(A(a,b),B(a,b))$ if and only if $\ell ^{3v} \mid C(a, b)$.

• (b) $\ell ^{3v} \mid C(a,b)$ if and only if $\ell \not \mid b$ and $h(a/b) \equiv 0 \pmod {\ell ^{3v}}$.

• (c) If $\ell \ne 3$, then $\ell \mid C(a,b)$ implies $\ell \not \mid (2a+b)=(\partial C/\partial a)(a,b)$.

Proof

We use the notation (2.2.7) and argue as in Cullinan–Kenney–Voight [6, Proof of Theorem 3.3.1, Step 3]. For part (a), we compute the resultants

$$\begin{aligned} {{\,\textrm{Res}\,}}(A_0(t,1),B_0(t,1))={{\,\textrm{Res}\,}}(f_0(t),g_0(t))=-2^8\cdot 3^7 \cdot 7^{14} = {{\,\textrm{Res}\,}}(A_0(1,u),B_0(1,u)). \end{aligned}$$

So if $\ell \ne 2,3,7$, then $\ell \not \mid \gcd (A_0(a,b),B_0(a,b))$; so by (2.1.5), if $\ell ^v \mid {{\,\textrm{tmd}\,}}(A(a,b),B(a,b))$ then $\ell ^{2v} \mid C(a,b)$. But also

$$\begin{aligned} {{\,\textrm{Res}\,}}(B_0(t,1),C(t,1))={{\,\textrm{Res}\,}}(g_0(t),h(t)) = 2^8 \cdot 3^3 \cdot 7^7 = {{\,\textrm{Res}\,}}(B_0(1,u),C(1,u)), \end{aligned}$$

so $\ell \not \mid \gcd (B_0(a,b), C(a, b))$ and thus $\ell ^v \mid {{\,\textrm{tmd}\,}}(A(a,b),B(a,b))$ if and only if $\ell ^{3v} \mid C(a,b)$. If $\ell = 2$, a short computation confirms that B(a, b) is twice an odd integer whenever (a, b) is groomed, so our claim also holds in this case.

For (b), by homogeneity it suffices to show that $\ell \not \mid b$, and indeed this holds since if $\ell \mid b$ then $A(a,0) \equiv -3a^4 \equiv 0 \pmod {\ell }$ and $B(b,0) \equiv 2a^6 \equiv 0 \pmod {\ell }$ so $\ell \mid a$, a contradiction.

Part (c) follows from (b) and the fact that h(t) has discriminant ${{\,\textrm{disc}\,}}(h(t))=3^3$. $\square$

For $e \ge 1$, let $\widetilde{\mathcal {T}}(e) \subseteq (\mathbb {Z}/e^3\mathbb {Z})^2$ denote the image of

$$\begin{aligned} \left\{ (a, b) \in \mathbb {Z}^2 : (a, b) \ \text {groomed}, \ e \mid {{\,\textrm{tmd}\,}}(A (a, b), B (a, b))\right\} \end{aligned}$$

under the projection

$$\begin{aligned} \mathbb {Z}^2 \rightarrow (\mathbb {Z}/ e^3 \mathbb {Z})^2, \end{aligned}$$

and let $\widetilde{T}(e) \,{:=}\,\# \widetilde{\mathcal {T}}(e)$. Similarly, let $\mathcal {T}(e) \subseteq \mathbb {Z}/e^3\mathbb {Z}$ denote the image of

$$\begin{aligned} \left\{ t \in \mathbb {Z}: e^2 \mid f(t) \ \text {and} \ e^3 \mid g(t)\right\} \end{aligned}$$

under the projection

$$\begin{aligned} \mathbb {Z}\rightarrow \mathbb {Z}/ e^3 \mathbb {Z}, \end{aligned}$$

and let $T(e) \,{:=}\,\#\mathcal {T}(e)$.

As usual, we write $\varphi (n) \,{:=}\,n \prod _{p \mid n} (1 - 1/p)$ for the Euler totient function.

Lemma 2.3.2

The following statements hold.

• (a) The functions $\widetilde{T}(e)$ and T(e) are multiplicative, and $\widetilde{T}(e) = \varphi (e^3) T(e)$.

• (b) For all $\ell \ne 3,7$ and $v \ge 1$,

  $$\begin{aligned} T(\ell ^v)&= T(\ell ) = 1 + \left( \frac{\ell }{3}\right) . \end{aligned}$$

• (c) We have

  $$\begin{aligned} T(3) = 18, \ T(3^2) = 27, \text { and }\ T(3^v) = 0 \ \text {for} \ v \ge 3, \end{aligned}$$

  and

  $$\begin{aligned} T(7) = 50, \ T(7^2) = 7^4+1=2402, \text { and }\ T(7^v) = 7^7+1=823544 \ \text {for} \ v \ge 3. \end{aligned}$$

• (d) We have $T(e) =O(2^{\omega (e)})$, where $\omega (e)$ is the number of distinct prime divisors of e.

Proof

For part (a), multiplicativity follows from the CRT (Sun Zi theorem). For the second statement, let $\ell$ be a prime, and let $e = \ell ^v$ for some $v \ge 1$. Consider the injective map

$$\begin{aligned} \mathcal {T}(\ell ^v) \times (\mathbb {Z}/ \ell ^{3 v})^\times&\rightarrow \widetilde{\mathcal {T}}(\ell ^v) \nonumber \\ (t,u)&\mapsto (tu,u) \end{aligned}$$

(2.3.3)

We observe $A(1, 0) = -3$ and $B(1, 0) = 2$ are coprime, so no pair (a, b) with $b \equiv 0 \pmod \ell$ can be a member of $\widetilde{\mathcal {T}}(\ell ^v)$. Surjectivity of the given map follows, and counting both sides gives the result.

Now part (b). For $\ell \ne 3, 7$, Lemma 2.3.1(a)–(b) yield

$$\begin{aligned} \mathcal {T}(\ell ^v) = \left\{ t \in \mathbb {Z}/\ell ^{3v} \mathbb {Z}: h(t) \equiv 0 ~(mod ~{\ell ^{3v}})\right\} . \end{aligned}$$

By Lemma 2.3.1(c), $h(t) \equiv 0 \pmod \ell$ implies $h^\prime (t) \not \equiv 0 \pmod \ell$, so Hensel’s lemma applies and we need only count roots of h(t) modulo $\ell$, which by quadratic reciprocity is

$$\begin{aligned} 1 + \left( \frac{-3}{\ell }\right) = 1 + \left( \frac{\ell }{3}\right) = {\left\{ \begin{array}{ll} 2, &{} \ \text {if} \ \ell \equiv 1 ~(mod ~{3}); \\ 0, &{} \ \text {else.} \end{array}\right. } \end{aligned}$$

Next, part (c). For $\ell = 3$, we just compute $T(3) = 18$, $T(3^2) = 27$, and $T(3^3) = 0$; then $T(3^3) = 0$ implies $T(3^v) = 0$ for all $v \ge 3$. For $\ell = 7$, we compute

$$\begin{aligned} T(7) = 50, \ T(7^2) = 2402, \ T(7^3) = \dots = T(7^6) = 823544. \end{aligned}$$

Hensel’s lemma still applies to h(t): let $t_0,t_1$ be the roots of h(t) in $\mathbb {Z}_7$ with $t_0 \,{:=}\,248044 \pmod {7^7}$ (so that $t_1=-1-t_0$). We claim that

$$\begin{aligned} \mathcal {T}(7^{3v}) = \left\{ t_0\right\} \sqcup \left\{ t_1 + 7^{3v - 7} u \in \mathbb {Z}/ 7^{3v} \mathbb {Z}:u \in \mathbb {Z}/ 7^7 \mathbb {Z}\right\} , \end{aligned}$$

(2.3.4)

for $3v \ge 7$. Indeed, $g_0(t_1) \equiv 0 \pmod {7^7}$, so we can afford to approximate $t_1$ modulo $7^{3v - 7}$. As $g(t_0) \not \equiv 0 \pmod {7}$ and $g(t_1) \not \equiv 0 \pmod {7^8}$, no other values of t suffice. Thus $T(7^{3v}) = 1 + 7^7 = 823544$.

Finally, part (d). From (a)–(c) we conclude

$$\begin{aligned} T(e) \le \frac{27 \cdot 823544}{4} \cdot \prod _{\begin{array}{c} \ell \mid e \\ \ell \ne 3,7 \end{array}} \left( 1 + \left( \frac{\ell }{3}\right) \right) \le 5558922 \cdot 2^{\omega (e)} \end{aligned}$$

(2.3.5)

so $T(e) = O(2^{\omega (e)})$ as claimed. $\square$

2.4 The common factor C(a, b)

In view of Lemma 2.3.1, the twist minimality defect away from the primes 2, 3, 7 is given by the quadratic form $C(a,b)=a^2+ab+7b^2=b^2 h(a/b)$. Fortunately, this is the norm form of a quadratic order of class number 1, so although this is ultimately more than what we need, we record some consequences of this observation which take us beyond Lemma 2.3.2.

For $m \in \mathbb {Z}_{>0}$, let

$$\begin{aligned} c(m) \,{:=}\,\#\{(a, b) \in \mathbb {Z}^2 : b > 0, \ \gcd (a, b) = 1, \ C(a, b) = m\}. \end{aligned}$$

(2.4.1)

Lemma 2.4.2

The following statements hold.

• (a) We have $c(m n) = c(m) c(n)$ for $m,n \in \mathbb {Z}_{>0}$ coprime.

• (b) We have

  $$\begin{aligned} c(3) = 0, \ c(3^2) = 2, \ c(3^3) = 3, \ \text {and} \ c(3^v) = 0 \ \text {for} \ v \ge 4; \end{aligned}$$

  for $p \ne 3$ prime and $k \ge 1$ an integer, we have

  $$\begin{aligned} c(p) = c(p^k) = 1 + \left( \frac{p}{3}\right) . \end{aligned}$$

  (2.4.3)

• (c) For m and n positive integers, we have

  $$\begin{aligned} c(n^3 m) \le 3 \cdot 2^{\omega (n)-1} c(m). \end{aligned}$$

Proof

Let $\zeta \,{:=}\,(1 + \sqrt{-3})/2$, so $\overline{\zeta } = 1 - \zeta = (1 - \sqrt{-3})/2$. The quadratic form

$$\begin{aligned} C(a, b) = a^2 + a b + 7 b^2 = (a+b\left( -1 + 3 \zeta \right) )(a+b\overline{\left( -1 + 3 \zeta \right) }) ={{\,\textrm{Nm}\,}}(a+b\left( -1 + 3 \zeta \right) ) \end{aligned}$$

is the norm on the order $\mathbb {Z}[3 \zeta ]$ in basis $\left\{ 1, -1 + 3 \zeta \right\}$. Recall that $\alpha \in \mathbb {Z}[3 \zeta ]$ is primitive if no $n \in \mathbb {Z}_{>1}$ divides $\alpha$. Thus, accounting for sign,

$$\begin{aligned} 2c(m) = \#\{\alpha \in \mathbb {Z}[3 \zeta ] \ \text {primitive} : {{\,\textrm{Nm}\,}}(\alpha ) = m\}. \end{aligned}$$

(2.4.4)

The order $\mathbb {Z}[3 \zeta ]$ is a suborder of the Euclidean domain $\mathbb {Z}[\zeta ]$ of conductor 3. It inherits from $\mathbb {Z}[\zeta ]$ the following variation on unique factorization: up to sign, every nonzero $\alpha \in \mathbb {Z}[3\zeta ]$ can be written uniquely as

$$\begin{aligned} \alpha = \beta \pi _1^{e_1} \cdots \pi _r^{e_r}, \end{aligned}$$

where ${{\,\textrm{Nm}\,}}(\beta )$ is a power of 3, $\pi _1, \dots , \pi _r$ are distinct irreducibles coprime to 3, and $e_1, \dots , e_r$ are positive integers. Note that $\alpha$ is primitive if and only if $\beta$ is primitive and for $1 \le i, j \le r$ (not necessarily distinct) we have $\pi _i \ne \overline{\pi _j}$. Thus if m and n are coprime integers, $\alpha \in \mathbb {Z}[3 \zeta ]$ is primitive, and ${{\,\textrm{Nm}\,}}(\alpha ) = mn$, then $\alpha$ may be factored uniquely (up to sign) as $\alpha = \alpha _1 \alpha _2$, where ${{\,\textrm{Nm}\,}}(\alpha _1) = m$ and ${{\,\textrm{Nm}\,}}(\alpha _2) = n$. This proves (a).

We now prove (b). If $p \ne 3$ is inert in $\mathbb {Z}[3\zeta ]$ (equivalently, in $\mathbb {Z}[\zeta ]$), then no primitive $\alpha$ satisfies ${{\,\textrm{Nm}\,}}(\alpha ) = p^v$, so $c(p^v) = 0$. If $p \ne 3$ splits in $\mathbb {Z}[3 \zeta ]$ (equivalently, in $\mathbb {Z}[\zeta ]$), then no primitive $\alpha$ is divisible by more than one of the two primes above p, so $c(p^v) = 2$. This proves (2.4.3) (compare Lemma 2.3.2). Finally, if $p = 3$, we compute $c(3) = 0,$ $c(3^2) = 2,$ and $c(3^3) = 3$. Congruence conditions show $c(3^v) = 0$ for $v \ge 4$.

Part (c) follows immediately from (a) and (b). $\square$

Remark 2.4.5

We prove Lemma 2.4.2(a) and Lemma 2.4.2(b) only as a means to proving Lemma 2.4.2(c). Although the algebraic structure of the Eisenstein integers $\mathbb {Z}[\zeta ]$ may not be available in the study of other families of elliptic curves that exhibit potential additive reduction, we expect analogues of Lemma 2.4.2(c) to hold in a general context.

The twist minimality defect measures the discrepancy between H(A, B) and ${{\,\textrm{twht}\,}}(A, B)$: this discrepancy cannot be too large compared to C(a, b), as the following theorem shows.

Theorem 2.4.6

We have the following.

• (a) For all $(a, b) \in \mathbb {R}^2$, we have

  $$\begin{aligned} 108 C(a, b)^6 \le H(A(a, b), B(a, b)) \le \kappa C(a, b)^6, \end{aligned}$$

  (2.4.7)

  where $\kappa = 311\,406\,871.990\,204\ldots$ is an explicit algebraic number.

• (b) If $C(a, b) = e_0^3 m$, with m cubefree, then ${{\,\textrm{tmd}\,}}(A(a, b), B(a, b)) = e_0 e^\prime$ for some $e^\prime \mid 3 \cdot 7^3$, and

  $$\begin{aligned} \frac{2^2}{3^3 \cdot 7^{18}} e_0^{12} m^6 \le {{\,\textrm{twht}\,}}(A(a, b), B(a, b)) \le \kappa e_0^{12} m^6. \end{aligned}$$

Proof

We wish to find the extrema of $H(A(a, b), B(a, b))/C(a, b)^6$. As this expression is homogeneous of degree 0, and C(a, b) is positive definite, we may assume without loss of generality that $C(a, b) = 1$. Using Lagrange multipliers, we verify that (2.4.7) holds: the lower bound is attained at (1, 0), and the upper bound is attained when $a = 0.450\,760\dots$ and $b=-0.371\,118\dots$ are roots of

$$\begin{aligned} 1296 a^8 - 2016 a^6 + 2107 a^4 - 1596 a^2 + 252&=0 \nonumber \\ 1067311728 b^8 - 275298660 b^6 + 43883077 b^4 - 3623648 b^2 + 1849&= 0, \end{aligned}$$

(2.4.8)

respectively.

Now write $C(a, b) = e_0^3 m$ with m cubefree, and write ${{\,\textrm{tmd}\,}}(A(a, b), B(a, b)) = e_0 e^\prime$. By Lemma 2.3.1, $e^\prime = 3^v 7^w$ for some $v, w \ge 0$; a short computation shows $v \in \left\{ 0, 1\right\}$, and (2.3.4) shows $w \le \lceil 7/3\rceil = 3$. As

$$\begin{aligned} H(A(a, b), B(a, b)) = e_0^6 \left( e^\prime \right) ^6 {{\,\textrm{twht}\,}}(A(a, b), B(a, b)), \end{aligned}$$

we see

$$\begin{aligned} \frac{108}{(e^\prime )^6} e_0^{12} m^6 \le {{\,\textrm{twht}\,}}(A(a, b), B(a, b)) < \frac{\kappa }{(e^\prime )^6} e_0^{12} m^6. \end{aligned}$$

Rounding $e^\prime$ up to $3 \cdot 7^3$ on the left and down to 1 on the right gives the desired result. $\square$

Corollary 2.4.9

Let (a, b) be a groomed pair. We have

$$\begin{aligned} {{\,\textrm{tmd}\,}}(A(a, b), B(a, b)) \le \frac{3^{5/4} \cdot 7^{9/2}}{2^{1/6}} {{\,\textrm{twht}\,}}(A(a, b), B(a, b))^{1/12} \end{aligned}$$

where $3^{5/4} \cdot 7^{9/2} / 2^{1/6} = 22\,344.5\ldots$

Proof

In the notation of Theorem 2.4.6(b),

$$\begin{aligned} e_0^{12} m^6 \le \frac{3^3 \cdot 7^{18}}{2^2} {{\,\textrm{twht}\,}}(A(a, b), B(a, b)). \end{aligned}$$

Multiplying through by $(e^\prime )^{12}$, rounding m down to 1 on the left, rounding $e^\prime$ up to $3 \cdot 7^7$ on the right, and taking 12th roots of both sides, we obtain the desired result. $\square$

3 Analytic ingredients

In this section, we record some results from analytic number theory used later.

3.1 Lattices and the principle of Lipschitz

We recall (a special case of) the Principle of Lipschitz, also known as Davenport’s Lemma.

Theorem 3.1.1

(Principle of Lipschitz) Let $\mathcal {R}\subseteq \mathbb {R}^2$ be a closed and bounded region, with rectifiable boundary $\partial \mathcal {R}$. We have

$$\begin{aligned} \#(\mathcal {R}\cap \mathbb {Z}^2) = {{\,\textrm{area}\,}}(\mathcal {R}) + O({{\,\textrm{len}\,}}(\partial \mathcal {R})), \end{aligned}$$

where the implicit constant depends on the similarity class of $\mathcal {R}$, but not on its size, orientation, or position in the plane $\mathbb {R}^2$.

Proof

See Davenport [7]. $\square$

Specializing to the case of interest, for $X > 0$ let

$$\begin{aligned} \mathcal {R}(X) \,{:=}\,\left\{ (a, b) \in \mathbb {R}^2 :H(A(a, b), B(a, b)) \le X, \ b \ge 0\right\} , \end{aligned}$$

(3.1.2)

and let $R \,{:=}\,{{\,\textrm{area}\,}}(\mathcal {R}(1))$. The region $\mathcal {R}(1)$ is the common region in Fig. 1.

Fig. 1

The region $\mathcal {R}(1)$

Lemma 3.1.3

For $X > 0$, we have ${{\,\textrm{area}\,}}(\mathcal {R}(X)) = R X^{1/6}$.

Proof

Since $f(t)=A(t,1)$ and $g(t)=B(t,1)$ have no common real root, the region $\mathcal {R}(X)$ is compact [6, Proof of Theorem 3.3.1, Step 2]. The homogeneity

$$\begin{aligned} H(A(u a, ub), B(ua, ub)) = u^{12} H(A(a, b), B(a, b)) \end{aligned}$$

implies

$$\begin{aligned} {{\,\textrm{area}\,}}(\mathcal {R}(X)) = {{\,\textrm{area}\,}}(\{(X^{1/12} a, X^{1/12} b) :(a, b) \in \mathcal {R}(1)\}) = X^{1/6} {{\,\textrm{area}\,}}(\mathcal {R}(1)) = R X^{1/6} \end{aligned}$$

as desired. $\square$

The following corollaries are immediate.

Corollary 3.1.4

For $a_0, b_0, d \in \mathbb {Z}$ with $d \ge 1$, we have

$$\begin{aligned} \#\{(a, b) \in \mathcal {R}(X) \cap \mathbb {Z}^2 : (a, b) \equiv (a_0, b_0) ~(mod ~{d})\} = \frac{R X^{1/6}}{d^2} + O\left( \frac{X^{1/{12}}}{d}\right) . \end{aligned}$$

The implied constants are independent of X, d, $a_0,$ and $b_0$. In particular,

$$\begin{aligned} \#(\mathcal {R}(X) \cap \mathbb {Z}^2) = R X^{1/6} + O(X^{1/{12}}). \end{aligned}$$

(3.1.5)

Proof

Combine Lemma 3.1.3 and Theorem 3.1.1. $\square$

Corollary 3.1.6

Let $\left( c(m)\right) _{m \ge 1}$ be as in (2.4.1). We have

$$\begin{aligned} \sum _{m \le X} c(m) = O(X). \end{aligned}$$

Proof

Immediate from Corollary 3.1.4. $\square$

3.2 Dirichlet series

The following theorem is attributed to Stieltjes.

Theorem 3.2.1

Let $\alpha , \beta : \mathbb {Z}_{> 0} \rightarrow \mathbb {R}$ be arithmetic functions. If $L_\alpha (s) \,{:=}\,\sum _{n \ge 1} \alpha (n) n^{-s}$ and $L_\beta (s) \,{:=}\,\sum _{n \ge 1} \beta (n) n^{-s}$ both converge when ${{\,\textrm{Re}\,}}(s) > \sigma$, and one of these two series converges absolutely, then

$$\begin{aligned} L_{\alpha *\beta }(s) \,{:=}\,\sum _{n \ge 1} \left( \sum _{d \mid n} \alpha (d) \beta \left( \frac{n}{d}\right) \right) n^{-s} \end{aligned}$$

converges for s with ${{\,\textrm{Re}\,}}(s) > \sigma$. If both $L_\alpha (s)$ and $L_\beta (s)$ both converge absolutely when ${{\,\textrm{Re}\,}}(s) > \sigma$, then so does $L_{\alpha *\beta }(s)$.

Proof

Widder [22, Theorems 11.5 and 11.6b] proves a more general result, or see Tenenbaum [21, proof of Theorem II.1.2, Notes on p. 204]. $\square$

Let $\gamma \,{:=}\,\lim _{y \rightarrow \infty } \bigl (\sum _{n \le y} 1/n\bigr ) - \log y$ be the Euler–Mascheroni constant.

Theorem 3.2.2

The difference

$$\begin{aligned} \zeta (s) - \left( \frac{1}{s - 1} + \gamma \right) \end{aligned}$$

is entire on $\mathbb {C}$ and vanishes at $s = 1$.

Proof

Ivić [13, p. 4] proves a more general result. $\square$

3.3 Regularly varying functions

We require a fragment of Karamata’s integral theorem for regularly varying functions.

Definition 3.3.1

Let $F :\mathbb {R}_{\ge 0} \rightarrow \mathbb {R}$ be measurable and eventually positive. We say that F is regularly varying of index $\rho \in \mathbb {R}$ if for each $\lambda > 0$ we have

$$\begin{aligned} \lim _{y \rightarrow \infty } \frac{F(\lambda y)}{F(y)} = \lambda ^\rho . \end{aligned}$$

Theorem 3.3.2

(Karamata’s integral theorem) Let $F :\mathbb {R}_{\ge 0} \rightarrow \mathbb {R}$ be locally bounded and regularly varying of index $\rho \in \mathbb {R}$. Let $\sigma \in \mathbb {R}$. Then the following statements hold.

• (a) For any $\sigma > \rho + 1$, we have

  $$\begin{aligned} \int _y^\infty t^{-\sigma } F(u) \,\textrm{d}u \sim \frac{y^{1 - \sigma } F(y)}{\left| \sigma - \rho - 1\right| } \end{aligned}$$

  as $y \rightarrow \infty$.

• (b) For any $\sigma < \rho + 1$, we have

  $$\begin{aligned} \int _0^y u^{-\sigma } F(u) \,\textrm{d}u \sim \frac{y^{1 - \sigma } F(y)}{\left| \sigma - \rho - 1\right| } \end{aligned}$$

  as $y \rightarrow \infty$.

Proof

See Bingham–Glodie–Teugels [2, Theorem 1.5.11]. (Karamata’s integral theorem also includes a converse.) $\square$

Corollary 3.3.3

Let $\alpha :\mathbb {Z}_{> 0} \rightarrow \mathbb {R}$ be an arithmetic function, and suppose that for some $\kappa , \rho , \tau \in \mathbb {R}$ with $\kappa \ne 0$ and $\rho > 0$, we have

$$\begin{aligned} F(y) \,{:=}\,\sum _{n \le y} \alpha (n) \sim \kappa y^\rho \log ^\tau y \end{aligned}$$

(3.3.4)

as $y \rightarrow \infty$. Let $\sigma > 0$. Then the following statements hold, as $y \rightarrow \infty$.

• (a) If $\sigma > \rho$, then

  $$\begin{aligned} \sum _{n > y} n^{-\sigma } \alpha (n) \sim \frac{\rho y^{-\sigma } F(y)}{\left| \sigma - \rho \right| } \sim \frac{\kappa \rho y^{\rho - \sigma } \log ^\tau y}{\left| \sigma - \rho \right| }. \end{aligned}$$

• (b) If $\rho > \sigma$, then

  $$\begin{aligned} \sum _{n \le y} n^{-\sigma } \alpha (n) \sim \frac{\rho y^{-\sigma } F(y)}{\left| \sigma - \rho \right| } \sim \frac{\kappa \rho y^{\rho - \sigma } \log ^\tau y}{\left| \sigma - \rho \right| }. \end{aligned}$$

Proof

Replacing $\alpha$ and F with $-\alpha$ and $-F$ if necessary, we may assume $\kappa > 0$. As a partial sum of an arithmetic function, F(y) is measurable and locally bounded; by (3.3.4), F(y) is eventually positive. Now for any $\lambda > 0$, we compute

$$\begin{aligned} \lim _{y \rightarrow \infty } \frac{F(\lambda y)}{F(y)} = \lim _{y \rightarrow \infty } \frac{\kappa (\lambda y)^\rho \log ^\tau (\lambda y)}{\kappa y^\rho \log ^\tau y} = \lambda ^\rho , \end{aligned}$$

so F is regularly varying of index $\rho$.

Suppose first $\sigma > \rho$. Since

$$\begin{aligned} y^{-\sigma } F(y) \sim \kappa y^{\rho - \sigma } \log ^\tau y \rightarrow 0 \end{aligned}$$

as $y \rightarrow \infty$, Abel summation yields

$$\begin{aligned} \sum _{n > y} n^{-\sigma } \alpha (n) = - y^{-\sigma } F(y) + \sigma \int _y^\infty u^{-\sigma - 1} F(u) \,\textrm{d}u. \end{aligned}$$

Clearly $\sigma + 1 > \rho + 1$, so Theorem 3.3.2(a) tells us

$$\begin{aligned} \int _y^\infty u^{-\sigma - 1} F(u) \,\textrm{d}u \sim \frac{y^{-\sigma } F(y)}{\left| \sigma - \rho \right| } \sim \frac{\kappa y^{\rho - \sigma } \log ^\tau y}{\left| \sigma - \rho \right| } \end{aligned}$$

and thus

$$\begin{aligned} \sum _{n > y} n^{-\sigma } \alpha (n) \sim \frac{\rho y^{-\sigma } F(y)}{\left| \sigma - \rho \right| } \end{aligned}$$

as $y \rightarrow \infty$.

The case $\rho > \sigma$ is similar. $\square$

3.4 Bounding Dirichlet series on vertical lines

Recall that a complex function F(s) has finite order on a domain D if there exists $\xi \in \mathbb {R}_{>0}$ such that

$$\begin{aligned} F(s) = O(1 + \left| t\right| ^\xi ) \end{aligned}$$

whenever $s = \sigma + i t \in D$. If F is of finite order on a right half-plane, we define

$$\begin{aligned} \mu _F(\sigma ) \,{:=}\,\inf \{\xi \in \mathbb {R}_{\ge 0} :F(\sigma + i t) = O(1 + \left| t\right| ^\xi )\} \end{aligned}$$

where the implicit constant depends on $\sigma$ and $\xi$.

Let L(s) be a Dirichlet series with abscissa of absolute convergence $\sigma _a$ and abscissa of convergence $\sigma _c$.

Theorem 3.4.1

We have $\mu _L(\sigma )=0$ for all $\sigma > \sigma _a$, and $\mu _L(\sigma )$ is nonincreasing (as a function of $\sigma$) on any region where L has finite order.

Proof

Tenenbaum [21, Theorem II.1.21]. $\square$

Theorem 3.4.2

Let $\sigma _c < \sigma _0 \le \sigma _c+1$ and let $\epsilon > 0$. Then uniformly on

$$\begin{aligned} \left\{ s = \sigma + i t \in \mathbb {C}: \sigma _0 \le \sigma \le \sigma _c + 1, \ \left| t\right| \ge 1\right\} , \end{aligned}$$

we have

$$\begin{aligned} L(\sigma + i t) = O(t^{1 + \sigma _c - \sigma + \epsilon }). \end{aligned}$$

Proof

Tenenbaum [21, Theorem II.1.19]. $\square$

Corollary 3.4.3

For all $\sigma > \sigma _c$, we have

$$\begin{aligned} \mu _{L}(\sigma ) \le \max (0,1 + \sigma _c - \sigma ). \end{aligned}$$

Proof

It is well-known that $\sigma _a \le \sigma _c + 1$, so the claim holds for $\sigma > \sigma _c + 1$ by Theorem 3.4.1. Now for $\sigma _c< \sigma < \sigma _c + 1$, our claim follows by letting $\epsilon \rightarrow 0$ in Theorem 3.4.2. $\square$

Theorem 3.4.4

Let $\zeta (s)$ be the Riemann zeta function, and let $\sigma \in \mathbb {R}$. We have

$$\begin{aligned} \mu _\zeta (\sigma ) \le {\left\{ \begin{array}{ll} \frac{1}{2} - \sigma , &{} \text {if} \ \sigma \le 0; \\ \frac{1}{2} - \frac{141}{205} \sigma , &{} \text {if} \ 0 \le \sigma \le \frac{1}{2}; \\ \frac{64}{205}(1 - \sigma ), &{} \text {if} \ \frac{1}{2} \le \sigma \le 1; \\ 0 &{} \text {if} \ \sigma \ge 0. \end{array}\right. } \end{aligned}$$

Moreover, equality holds if $\sigma <0$ or $\sigma >1$.

Proof

Tenenbaum [21, p. 235] proves the claim when $\sigma <0$ or $\sigma >1$. Now $\mu _\zeta (1/2) \le 32/205$ by Huxley [12, Theorem 1], and our result follows from the subconvexity of $\mu _\zeta$ [21, Theorem II.1.20]. $\square$

3.5 A Tauberian theorem

We now present a Tauberian theorem, due in essence to Landau [14].

Definition 3.5.1

Let $\left( \alpha (n)\right) _{n \ge 1}$ be a sequence with $\alpha (n) \in \mathbb {R}_{\ge 0}$ for all n, and let $L_\alpha (s) \,{:=}\,\sum _{n \ge 1} \alpha (n) n^{-s}$. We say the sequence $\left( \alpha (n)\right) _{n \ge 1}$ is admissible with (real) parameters $\left( \sigma _a, \delta , \xi \right)$ if the following hypotheses hold:

• (i) $L_\alpha (s)$ has abscissa of absolute convergence $\sigma _a$.

• (ii) The function $L_\alpha (s)/s$ has meromorphic continuation to $\left\{ s = \sigma + i t \in \mathbb {C}: \sigma > \sigma _a - \delta \right\}$ and only finitely many poles in this region.

• (iii) For $\sigma > \sigma _a - \delta$, we have $\mu _{L_\alpha }(\sigma ) \le \xi$.

If $\left( \alpha (n)\right) _n$ is admissible, let $s_1, \dots , s_r$ denote the poles of $L_\alpha (s)/s$ with real part greater than $\sigma _a - \delta /(\xi + 2)$.

The following theorem is essentially an application of Perron’s formula, which is itself an inverse Mellin transform.

Theorem 3.5.2

(Landau’s Tauberian Theorem) Let $\left( \alpha (n)\right) _{n \ge 1}$ be an admissible sequence (Definition 3.5.1), and write $N_\alpha (X) \,{:=}\,\sum _{n \le X} \alpha (n)$. Then for all $\epsilon >0$,

$$\begin{aligned} N_\alpha (X) = \sum _{j = 1}^r {{\,\textrm{res}\,}}_{s=s_j}\left( \frac{L_\alpha (s) X^{s}}{s}\right) + O\!\left( X^{\sigma _a - \frac{\delta }{\left\lfloor \xi \right\rfloor + 2} + \epsilon }\right) , \end{aligned}$$

where the main term is a sum of residues and the implicit constant depends on $\epsilon$.

Proof

See Roux [15, Theorem 13.3, Remark 13.4]. $\square$

Remark 3.5.3

Landau’s original theorem [14] was fitted to a more general context, and allowed sums of the form

$$\begin{aligned} \sum _{n \ge 1} \alpha (n) \ell (n)^{-s} \end{aligned}$$

as long as $\left( \ell (n)\right) _{n \ge 1}$ was increasing and tended to $\infty$. Landau also gave an explicit expansion of

$$\begin{aligned} {{\,\textrm{res}\,}}_{s=s_j}\left( \frac{L_\alpha (s) X^{s}}{s}\right) \end{aligned}$$

in terms of the Laurent series expansion for $L_\alpha (s)$ around $s = s_j$. However, Landau also required that $L_\alpha (s)$ has a meromorphic continuation to all of $\mathbb {C}$, and Roux [15, Theorem 13.3, Remark 13.4] relaxes this assumption.

Let d(n) denote the number of divisors of n, and let $\omega (n)$ denote the number of distinct prime divisors of n. Theorem 3.5.2 has the following easy corollary.

Corollary 3.5.4

We have

$$\begin{aligned} \sum _{n \le y} 2^{\omega (n)} = \frac{y \log y}{\zeta (2)} + O(y) \quad \text {and} \quad \sum _{n \le y} d(n)^2 = \frac{y \log ^3 y}{6 \zeta (2)} + O(y \log ^2 y). \end{aligned}$$

as $y \rightarrow \infty$.

Proof

Recall that

$$\begin{aligned} \frac{\zeta (s)^2}{\zeta (2s)} = \sum _{n \ge 1} \frac{2^{\omega (n)}}{n^s} \ \text {and} \ \frac{\zeta (s)^4}{\zeta (2s)} = \sum _{n \ge 1} \frac{d(n)^2}{n^s}. \end{aligned}$$

It is straightforward to verify that $\left( 2^{\omega (n)}\right) _{n \ge 1}$ and $\left( d(n)^2\right) _{n \ge 1}$ are both admissible with parameters (1, 1/2, 1/3). We apply Theorem 3.5.2 and discard lower-order terms to obtain the result. $\square$

Remark 3.5.5

Theorem 3.5.2 furnishes lower order terms for the sums $\sum _{n \le y} 2^{\omega (n)}$ and $\sum _{n \le y} d(n)^2$, and even better estimates are known (e.g. Tenenbaum [21, Exercise I.3.54] and Zhai [23, Corollary 4]), but Corollary 3.5.4 suffices for our purposes and illustrates the use of Theorem 3.5.2.

4 Estimates for twist classes

In this section, we decompose $N_{}^{tw }(X)$, counting the number of twist minimal elliptic curves over $\mathbb {Q}$ admitting a 7-isogeny (2.2.12) in terms of progressively simpler functions. We then estimate those simple functions, and piece these estimates together until we arrive at an estimate for $N_{}^{tw }(X)$; the main result is Theorem 4.2.19, which proves Theorem 1.2.4.

4.1 Decomposition and outline

We establish some notation for brevity and ease of exposition. Suppose $\left( \alpha (X; n)\right) _{n \ge 1}$ is a sequence of real-valued functions, and $\phi : \mathbb {R}_{>0} \rightarrow \mathbb {R}_{>0}$. We write

$$\begin{aligned} \sum _{n \ge 1} \alpha (X; n) = \sum _{n \ll \phi (X)} \alpha (X; n) \end{aligned}$$

if there is a positive constant $\kappa$ such that for all $X \in \mathbb {R}_{> 0}$ and all $n > \kappa \phi (X)$, we have $\alpha (X; n) = 0$.

The function $N_{}^{tw }(X)$ is difficult to understand chiefly because of the twist minimality defect. Fortunately, the twist minimality defect cannot get too large relative to X (see Corollary 2.4.9). So we partition our sum based on the value of ${{\,\textrm{tmd}\,}}(A(a,b), B(a,b))$ in terms of the parametrization provided in Sect. 2.2.

For $e \ge 1$, let $N_{}^{tw }(X; e)$ denote the number of pairs $(a, b) \in \mathbb {Z}^2$ with

• (a, b) groomed,

• ${{\,\textrm{twht}\,}}(A(a, b), B(a, b)) \le X$, and

• ${{\,\textrm{tmd}\,}}(A(a, b), B(a, b)) = e$.

By (2.2.13) and Corollary 2.4.9, we have

$$\begin{aligned} N_{}^{tw }(X) = \sum _{e \ll X^{1/12}} N_{}^{tw }(X; e); \end{aligned}$$

(4.1.1)

more precisely, we can restrict our sum to

$$\begin{aligned} e \le \frac{3^{5/4} \cdot 7^{9/2}}{2^{1/6}} \cdot X^{1/12}. \end{aligned}$$

Determining when an integer e divides ${{\,\textrm{tmd}\,}}(A, B)$ is easier than determining when e equals ${{\,\textrm{tmd}\,}}(A, B)$, so we also let $M(X; e)$ denote the number of pairs $(a, b) \in \mathbb {Z}^2$ with

• (a, b) groomed,

• $H(A(a,b),B(a,b)) \le X$;

• $e \mid {{\,\textrm{tmd}\,}}(A(a, b), B(a, b))$;

Note that the points counted by $N_{}^{tw }(X; e)$ have twist height bounded by X, but the points counted by $M(X; e)$ have only the function H bounded by X.

Theorem 2.4.6 and the Möbius sieve yield

$$\begin{aligned} N_{}^{tw }(X; e) = \sum _{f \ll \frac{X^{1/18}}{e^{2/3}}} \mu (f) M(e^6 X; ef); \end{aligned}$$

(4.1.2)

more precisely, we can restrict our sum to

$$\begin{aligned} f \le \frac{3^{1/2} 7^2}{2^{1/9}} \cdot \frac{X^{1/18}}{e^{2/3}}. \end{aligned}$$

In order to estimate $M(X; e)$, we further unpack the groomed condition on pairs (a, b). We therefore let $M(X; d, e)$ denote the number of pairs $(a, b) \in \mathbb {Z}^2$ with

• $\gcd (da, db, e) = 1$ and $b > 0$;

• $H(A(d a, d b), B(d a, db)) \le X$;

• $e \mid {{\,\textrm{tmd}\,}}(A(d a, d b), B(da, db))$;

• $(a, b) \ne (-7, 1)$.

By Theorem 2.4.6, and because $H(A(a, b), B(a, b))$ is homogeneous of degree 12, another Möbius sieve yields

$$\begin{aligned} M(X; e) = \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \mu (d) M(X; d, e); \end{aligned}$$

(4.1.3)

more precisely, we can restrict our sum to

$$\begin{aligned} d \le \frac{1}{2^{1/6} \cdot 3^{1/4}} \cdot X^{1/12}. \end{aligned}$$

Before proceeding, we now give an outline of the argument used in this section. In Lemma 4.2.1, we use the Principle of Lipschitz to estimate $M(X; d, e)$, then piece these estimates together using (4.1.3) to estimate $M(X; e)$. Heuristically,

$$\begin{aligned} M(X; d, e) \sim \frac{R T(e) X^{1/6}}{d^2 e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) \end{aligned}$$

(4.1.4)

(where R is the area of (3.1.2) and T is the arithmetic function investigated in Lemma 2.3.2) by summing over the congruence classes modulo $e^3$ that satisfy $e \mid {{\,\textrm{tmd}\,}}(A(d a, d b), B(da, db))$. Then (4.1.3) suggests

$$\begin{aligned} M(X; e) \sim \frac{R T(e) X^{1/6}}{\zeta (2) e^3 \prod _{\ell \mid e} \left( 1 + \frac{1}{\ell }\right) }. \end{aligned}$$

(4.1.5)

To go further, we substitute (4.1.2) into (4.1.1), and let $n = e f$ to obtain

$$\begin{aligned} N_{}^{tw }(X) = \sum _{n \ll X^{1/12}} \sum _{e \mid n} \mu \left( n/e\right) M(e^6 X; n). \end{aligned}$$

(4.1.6)

This is the core identity that, in concert with the Principle of Lipschitz, enables us to estimate $N_{}^{tw }(X)$.

Substituting (4.1.5) into (4.1.6), and recalling $\varphi (n) = \sum _{e \mid n} \mu (n/e) e$, we obtain the heuristic estimate

$$\begin{aligned} N_{}^{tw }(X) \sim \frac{Q R X^{1/6}}{\zeta (2)}, \end{aligned}$$

(4.1.7)

where

$$\begin{aligned} Q \,{:=}\,\sum _{n \ge 1} \frac{T(n) \varphi (n) }{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) }. \end{aligned}$$

(4.1.8)

To make this estimate for $N_{}^{tw }(X)$ rigorous, and to get a better handle on the size of order of growth for its error term, we now decompose (4.1.6) based on the size of n into two pieces:

$$\begin{aligned} N_{\le y}^{\textrm{tw}}(X)&\,{:=}\,\sum _{n \le y} \sum _{e \mid n} \mu \left( \frac{n}{e}\right) M(e^6 X; n), \nonumber \\ N_{> y}^{\textrm{tw}}(X)&\,{:=}\,\sum _{n > y} \sum _{e \mid n} \mu \left( n/e\right) M(e^6 X; n). \end{aligned}$$

(4.1.9)

By definition, we have

$$\begin{aligned} N_{}^{tw }(X) = N_{\le y}^{\textrm{tw}}(X) + N_{> y}^{\textrm{tw}}(X). \end{aligned}$$

We then estimate $N_{\le y}^{\textrm{tw}}(X)$ in Proposition 4.2.6, and treat $N_{> y}^{\textrm{tw}}(X)$ as an error term which we bound in Lemma 4.2.14. Setting the error from our estimate equal to the error arising from $N_{> y}^{\textrm{tw}}(X)$, we obtain Theorem 4.2.19.

In the remainder of this section, we follow the outline suggested here by successively estimating $M(X; d, e)$, $M(X; e)$, $N_{\le y}^{\textrm{tw}}(X)$, $N_{> y}^{\textrm{tw}}(X)$, and finally $N_{}^{tw }(X)$.

4.2 Asymptotic estimates

We first estimate $M(X; d, e)$ and $M(X; e)$.

Lemma 4.2.1

The following statements hold.

• (a) If $\gcd (d, e) > 1$, then $M(X; d, e) = 0$. Otherwise, we have

  $$\begin{aligned} M(X; d, e) = \frac{R T(e) X^{1/6}}{d^2 e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) + O\left( \frac{T(e) X^{1/12}}{d}\right) . \end{aligned}$$

  where R is the area of (3.1.2).

• (b) We have

  $$\begin{aligned} M(X; e) = \frac{R T(e) X^{1/6}}{\zeta (2) e^3 \prod _{\ell \mid e} \left( 1 + \frac{1}{\ell }\right) } + O(T(e) X^{1/12} \log X). \end{aligned}$$

In both cases, the implied constants are independent of d, e, and X.

Proof

We begin with (a) and examine the summands $M(X; d, e)$. If d and e are not coprime, then $M(X; d, e) = 0$ because $\gcd (da, db, e) \ge \gcd (d, e) > 1$. On the other hand, if $\gcd (d, e) = 1$, we have a bijection from the pairs counted by $M(X; 1, e)$ to the pairs counted by $M(d^{12} X; d, e)$ given by $(a, b) \mapsto (d a, d b)$.

Combining Lemma 2.3.2(a) and Corollary 3.1.4, we have

$$\begin{aligned} M(X; 1, e)= & {} \sum _{(a_0, b_0) \in \widetilde{\mathcal {T}}(e)} \#\{(a, b) \in \mathcal {R}(X) \cap \mathbb {Z}^2 : (a, b) \equiv (a_0, b_0) ~(mod ~{e^3}), (a, b) \ne (-7, 1) \}\nonumber \\= & {} \varphi (e^3) T(e) \left( \frac{R X^{1/6}}{e^6} + O\left( \frac{X^{1/12}}{e^3}\right) \right) \nonumber \\= & {} \frac{R T(e) X^{1/6}}{e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) + O(T(e) X^{1/12}), \end{aligned}$$

(4.2.2)

and thus

$$\begin{aligned} M(X; d, e) = \frac{R T(e) X^{1/6}}{d^2 e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) + O\left( \frac{T(e) X^{1/12}}{d}\right) . \end{aligned}$$

For part (b), we compute

$$\begin{aligned} M(X; e)= & {} \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \mu (d) M(X; d, e) \nonumber \\= & {} \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \mu (d) \left( \frac{T(e) R X^{1/6}}{d^2 e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) + O\left( T(e) \frac{X^{1/12}}{d}\right) \right) \nonumber \\= & {} \frac{R T(e) X^{1/6}}{e^3} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell }\right) \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \frac{ \mu (d)}{d^2} + O\left( T(e) X^{1/12} \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \frac{1}{d}\right) . \end{aligned}$$

(4.2.3)

Plugging the straightforward estimates

$$\begin{aligned} \sum _{\begin{array}{c} d \ll X^{1/12} \\ \gcd (d, e) = 1 \end{array}} \frac{ \mu (d)}{d^2} = \frac{1}{\zeta (2)} \prod _{\ell \mid e} \left( 1 - \frac{1}{\ell ^2}\right) ^{-1}+ O(X^{-1/12}) \end{aligned}$$

(4.2.4)

and

$$\begin{aligned} \sum _{\begin{array}{c} d \le X^{1/12} \end{array}} \frac{1}{d} = \frac{1}{12}\log X+ O(1) \end{aligned}$$

into (4.2.3) then simplifies to give

$$\begin{aligned} M(X;e)= & {} \frac{R T(e) X^{1/6}}{\zeta (2) e^3 \prod _{\ell \mid e} \left( 1 + \frac{1}{\ell }\right) } + O(T(e) X^{1/12} \log X) \end{aligned}$$

(4.2.5)

proving (b). $\square$

We are now in a position to estimate $N_{\le y}^{\textrm{tw}}(X)$.

Proposition 4.2.6

Suppose $y \ll X^{\frac{1}{12}}$. Then

$$\begin{aligned} N_{\le y}^{\textrm{tw}}(X) = \frac{Q R X^{1/6}}{\zeta (2)} + O\left( \max \left( \frac{X^{1/6} \log y}{y}, X^{1/12} y^{3/2} \log X \log ^3 y \right) \right) \end{aligned}$$

where

$$\begin{aligned} Q \,{:=}\,\sum _{n \ge 1} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } = Q_3 Q_7 \prod _{\begin{array}{c} p \ne 7 \ \text {prime} \\ p \equiv 1 ~(mod ~{3}) \end{array}} \left( 1 + \frac{2}{(p+1)^2}\right) , \end{aligned}$$

and $Q_3 = 13/6$, $Q_7=63/8$.

Proof

Substituting the asymptotic for $M(X; e)$ from Lemma 4.2.1 into the defining series for $N_{\le y}^{\textrm{tw}}(X)$, we have

$$\begin{aligned} N_{\le y}^{\textrm{tw}}(X) = \sum _{n \le y} \sum _{e \mid n} \mu \left( n/e\right) \left( \frac{R T(n) e X^{1/6}}{\zeta (2) n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } + O\left( T(n) e^{1/2} X^{1/12} \log (e^6 X)\right) \right) . \end{aligned}$$

We handle the main term and the error of this expression separately. For the main term, we have

$$\begin{aligned} \sum _{n \le y} \sum _{e \mid n} \mu \left( n/e\right) \left( \frac{R T(n) e X^{1/6}}{\zeta (2) n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) }\right)= & {} \frac{R X^{1/6}}{\zeta (2)} \sum _{n \le y} \frac{T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } \sum _{e \mid n} \mu \left( n/e\right) e \nonumber \\= & {} \frac{R X^{1/6}}{\zeta (2)} \sum _{n \le y} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) }. \end{aligned}$$

(4.2.7)

By Lemma 2.3.2(d), we see

$$\begin{aligned} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } = O\left( \frac{2^{\omega (n)}}{n^2}\right) . \end{aligned}$$

By Corollary 3.3.3 and Corollary 3.5.4, we have

$$\begin{aligned} \sum _{n > y} \frac{2^{\omega (n)}}{n^2} \sim \frac{\log y}{\zeta (2) y} \end{aligned}$$

as $y \rightarrow \infty$. A fortiori,

$$\begin{aligned} \sum _{n> y} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } = O\left( \sum _{n > y} \frac{2^{\omega (n)}}{n^2}\right) = O\left( \frac{\log y}{y}\right) , \end{aligned}$$

so the series

$$\begin{aligned} \sum _{n \ge 1} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) } = Q \end{aligned}$$

(4.2.8)

is absolutely convergent, and

$$\begin{aligned} \sum _{n \le y} \sum _{e \mid n} \mu \left( n/e\right) \left( \frac{R T(n) e X^{1/6}}{\zeta (2) n^3 \prod _{\ell \mid n} \left( 1 + \frac{1}{\ell }\right) }\right)= & {} \frac{R X^{1/6}}{\zeta (2)} \left( Q - O\left( \frac{\log y}{y}\right) \right) \nonumber \\= & {} \frac{Q R X^{1/6}}{\zeta (2)} + O\left( \frac{X^{1/6} \log y}{y}\right) . \end{aligned}$$

(4.2.9)

As the summands of (4.2.8) constitute a nonnegative multiplicative arithmetic function, we can factor Q as an Euler product. For p prime, Lemma 2.3.2 yields

$$\begin{aligned} Q_p \,{:=}\,\sum _{a \ge 0} \frac{\varphi (p^a) T(p^a)}{p^{3a} \prod _{\ell \mid p} \left( 1 + \frac{1}{\ell }\right) } = {\left\{ \begin{array}{ll} 1 + \displaystyle {\frac{2}{p^2 + 1}}, &{} \text {if } p \equiv 1 ~(mod ~{3}) \text { and } p \ne 7; \\ 13/6, &{} \text {if } p=3; \\ 63/8, &{} \text {if } p=7; \\ 1 &{} \text {else}. \end{array}\right. } \end{aligned}$$

(4.2.10)

Thus

$$\begin{aligned} Q = \prod _{p \text { prime}} Q_p = Q_3 Q_7 \prod _{\begin{array}{c} p \ne 7 \ \text {prime} \\ p \equiv 1 ~(mod ~{3}) \end{array}} \left( 1 + \frac{2}{p^2 + 1}\right) . \end{aligned}$$

(4.2.11)

We now turn to the error term. Since $y \ll X^{1/12}$, for $e \le y$ we have $\log (e^6 X) \ll \log X$. Applying Lemma 2.3.2(d), we obtain

$$\begin{aligned}&\sum _{n \le y} \sum _{e \mid n} \mu \left( n/e\right) O\left( T(n) e^{1/2} X^{1/12} \log \left( e^6 X\right) \right) \nonumber \\&\quad = O\left( X^{1/12} \log X \sum _{n \le y} T(n) \sum _{e \mid n} \left| \mu \left( \frac{n}{e}\right) \right| e^{1/2} \right) \nonumber \\&\quad = O\left( X^{1/12} \log X \sum _{n \le y} 2^{2 \omega (n)} n^{1/2} \right) . \end{aligned}$$

(4.2.12)

Corollary 3.3.3 and Corollary 3.5.4, together with the trivial inequality $2^{2 \omega (n)} \le d(n)^2$, yield

$$\begin{aligned} \sum _{n \le y} 2^{2 \omega (n)} n^{1/2} = O(y^{3/2} \log ^3 y). \end{aligned}$$

(4.2.13)

Substituting (4.2.13) into (4.2.12) gives our desired result. $\square$

We now bound $N_{> y}^{\textrm{tw}}(X)$.

Lemma 4.2.14

We have

$$\begin{aligned} N_{> y}^{\textrm{tw}}(X) = O\left( \frac{X^{1/6} \log ^3 y}{y}\right) . \end{aligned}$$

Proof

We have

$$\begin{aligned} N_{> y}^{\textrm{tw}}(X)&= \sum _{n> y} \sum _{e \mid n} \mu \left( n/e\right) M(e^6 X; n) \le \sum _{n > y} 2^{\omega (n)} M(n^6 X; n). \end{aligned}$$

(4.2.15)

Write $n = 3^v 7^w n^\prime$ where $\gcd (n^\prime , 3) = \gcd (n^\prime , 7) = 1$. We define

$$\begin{aligned} n_0 \,{:=}\,3^{\max (v - 1, 0)} 7^{\max (w - 3, 0)} n^\prime , \end{aligned}$$

so

$$\begin{aligned} \frac{n}{3 \cdot 7^3} \le n_0 \le n. \end{aligned}$$

Let $(a, b) \in \mathbb {Z}^2$ be a groomed pair. By Theorem 2.4.6(a), $H(A(a, b), B(a, b)) \le n^6 X$ implies $108 C(a, b)^6 \le n^6 X$, and by Theorem 2.4.6(b), $n \mid {{\,\textrm{tmd}\,}}(A(a, b), B(a, b))$ implies $n_0^3 \mid C(a, b)^3$. Thus

$$\begin{aligned} M(n^6 X; n) \le \#\left\{ (a, b) \in \mathbb {Z}^2 \ \text {groomed} : 108\, C(a, b)^6 \le n^6 X, \ n_0^3 \mid C(a, b)\right\} . \end{aligned}$$

(4.2.16)

Recalling (2.4.1) and Lemma 2.4.2(c), we deduce

$$\begin{aligned} M(n^6 X; n) \le \sum _{m \ll X^{1/6}/n^2} c(n_0^3 m) \le 3 \cdot 2^{\omega (n_0) - 1} \sum _{m \ll X^{1/6}/n^2} c(m). \end{aligned}$$

But $2^{\omega (n)} \le 4 \cdot 2^{\omega (n_0)}$, so by Corollary 3.1.6, we have

$$\begin{aligned} M(n^6 X; n) = O\left( \frac{2^{\omega (n)} X^{1/6}}{n^2}\right) , \end{aligned}$$

and substituting this expression into (4.2.15) yields

$$\begin{aligned} N_{> y}^{\textrm{tw}}(X) = O\left( \sum _{n> y} \frac{\left( 2^{\omega (n)}\right) ^2 X^{1/6}}{n^2}\right) = O\left( X^{1/6} \sum _{n > y} \frac{2^{2 \omega (n)}}{n^2}\right) . \end{aligned}$$

(4.2.17)

As in the proof of Proposition 4.2.6, combining Corollary 3.3.3 and Corollary 3.5.4 together with the trivial inequality $2^{2 \omega (n)} \le d(n)^2$ yields

$$\begin{aligned} \sum _{n > y} \frac{2^{2 \omega (n)}}{n^2} = O\left( \frac{\log ^3 y}{y}\right) . \end{aligned}$$

(4.2.18)

Substituting (4.2.18) into (4.2.17) gives our desired result. $\square$

We are now in a position to prove Theorem 1.2.4, which we restate here with the notations we have established.

Theorem 4.2.19

We have

$$\begin{aligned} N_{}^{tw }(X) = \frac{Q R X^{1/6}}{\zeta (2)} + O(X^{2/15} \log ^{17/5} X), \end{aligned}$$

where

$$\begin{aligned} Q = \sum _{n \ge 1} \frac{\varphi (n) T(n)}{n^3 \prod _{\ell \mid n} \left( 1 + 1/\ell \right) }, \end{aligned}$$

and R is the area of the region

$$\begin{aligned} \mathcal {R}(1) = \left\{ (a, b) \in \mathbb {R}^2 : H(A (a, b), B (a, b)) \le 1, b \ge 0\right\} . \end{aligned}$$

Proof

Let y be a positive quantity with $y \ll X^{1/12}$; in particular, $\log y \ll \log X$. Proposition 4.2.6 and Lemma 4.2.14 together tell us

$$\begin{aligned} N_{}^{tw }(X) = \frac{Q R X^{1/6}}{\zeta (2)} + O\left( \max \left( \frac{X^{1/6} \log ^3 y}{y}, X^{1/12} y^{3/2} \log X \log ^3 y\right) \right) . \end{aligned}$$

(4.2.20)

We let $y = X^{1/30}/\log ^{2/5} X$, so

$$\begin{aligned} \frac{X^{1/6} \log ^3 y}{y} \asymp X^{1/12} y^{3/2} \log X \log ^3 y \asymp X^{2/15} \log ^{17/5} X, \end{aligned}$$

(4.2.21)

and we conclude

$$\begin{aligned} N_{}^{tw }(X) = \frac{Q R X^{1/6}}{\zeta (2)} + O(X^{2/15} \log ^{17/5} X) \end{aligned}$$

as desired. $\square$

4.3 L-series

To conclude, we set up the next section by interpreting Theorem 4.2.19 in terms of Dirichlet series. Let

$$\begin{aligned} h^{\textrm{tw}}(n) \,{:=}\,\# \left\{ (a, b) \in \mathbb {Z}^2 \text { groomed } : {{\,\textrm{twht}\,}}(A(a,b),B(a,b)) = n\right\} \end{aligned}$$

(4.3.1)

and define

$$\begin{aligned} L^{\textrm{tw}}(s) \,{:=}\,\sum _{n \ge 1} \frac{h^{\textrm{tw}}(n)}{n^s} \end{aligned}$$

(4.3.2)

wherever this series converges. Then $N_{}^{tw }(X) = \sum _{n \le X} h^{\textrm{tw}}(n)$, and conversely we have $L^{\textrm{tw}}(s) = \int _0^\infty u^{-s} \,\textrm{d}N_{}^{tw }(u)$.

Corollary 4.3.3

The Dirichlet series $L^{\textrm{tw}}(s)$ has abscissa of (absolute) convergence $\sigma _a=\sigma _c = 1/6$ and has a meromorphic continuation to the region

$$\begin{aligned} \left\{ s = \sigma + i t \in \mathbb {C}: \sigma > 2/15\right\} . \end{aligned}$$

(4.3.4)

Moreover, $L^{\textrm{tw}}(s)$ has a simple pole at $s = 1/6$ with residue

$$\begin{aligned} {{\,\textrm{res}\,}}_{s=\frac{1}{6}} L^{\textrm{tw}}(s) = \frac{QR}{6\zeta (2)} \end{aligned}$$

and is holomorphic elsewhere on the region (4.3.4).

Proof

Let $s = \sigma + i t \in \mathbb {C}$ be given with $\sigma > 1/6$. Abel summation yields

$$\begin{aligned} \begin{aligned} \sum _{n \le X} h^{\textrm{tw}}(n) n^{-s}&= N_{}^{tw }(X) X^{-s} + s \int _1^X N_{}^{tw }(u) u^{-s-1} \,\textrm{d}u \\&= O\left( X^{1/6 - \sigma } + s \int _1^X u^{- 5/6 - \sigma } \,\textrm{d}u\right) ; \end{aligned} \end{aligned}$$

(4.3.5)

as $X \rightarrow \infty$ the first term vanishes and the integral converges. Thus, when $\sigma > 1/6$,

$$\begin{aligned} \sum _{n \ge 1} h^{\textrm{tw}}(n) n^{-s} = s \int _1^\infty N_{}^{tw }(u) u^{-1-s}\,\textrm{d}u \end{aligned}$$

and this integral converges. A similar argument shows that the sum defining $L^{\textrm{tw}}(s)$ diverges when $\sigma < 1/6$. We have shown $\sigma _c = 1/6$ is the abscissa of convergence for $L^{\textrm{tw}}(s)$, but as $h^{\textrm{tw}}(n) \ge 0$ for all n, it is also the abscissa of absolute convergence $\sigma _a=\sigma _c$.

Now define $L_{\mathrm{{rem}}}^{\textrm{tw}}(s)$ so that

$$\begin{aligned} L^{\textrm{tw}}(s) = \frac{QR}{\zeta (2)} \zeta (6s) + L_{\mathrm{{rem}}}^{\textrm{tw}}(s). \end{aligned}$$

(4.3.6)

Abel summation and the substitution $u \mapsto u^{1/6}$ yields for $\sigma >1$

$$\begin{aligned} \zeta (6s) = s \int _1^\infty \left\lfloor u^{1/6} \right\rfloor u^{- 1 - s}\,\text {d}u = s \int _1^\infty \left( u^{1/6} + O(1)\right) u^{- 1 - s} \, \text {d}u. \end{aligned}$$

Let

$$\begin{aligned} \delta (n) \,{:=}\,{\left\{ \begin{array}{ll} 1, &{} \text {if} \ n = k^6 \ \text {for some} \ k \in \mathbb {Z}; \\ 0, &{} \text {else.} \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} L_{\mathrm{{rem}}}^{\textrm{tw}}(s)&= \sum _{n \ge 1} \left( h^{\textrm{tw}}(n) - \frac{QR}{\zeta (2)}\delta (n)\right) n^{-s}\\&= s \int _1^\infty \left( N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right) u^{-1-s} \, \text {d}u \end{aligned} \end{aligned}$$

(4.3.7)

when $\sigma > 1/6$. But then for any $\epsilon > 0$,

$$\begin{aligned} N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor = O(u^{2/15 + \epsilon }) \end{aligned}$$

(4.3.8)

by Theorem 4.2.19. Substituting (4.3.8) into (4.3.7), we obtain

$$\begin{aligned} L_{\mathrm{{rem}}}^{\textrm{tw}}(s) = s \int _1^\infty \left( N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right) u^{-1-s}\, \text {d}u&= O\left( s \int _1^\infty u^{-13/15 - \sigma + \epsilon } \,\text {d}u\right) \end{aligned}$$

(4.3.9)

where the integral converges whenever $\sigma > 2/15 + \epsilon$. Letting $\epsilon \rightarrow 0$, we obtain an analytic continuation of $L_{\mathrm{{rem}}}^{\textrm{tw}}(s)$ to the region (4.3.4).

At the same time, $\zeta (6s)$ has meromorphic continuation to $\mathbb {C}$ with a simple pole at $s=1/6$ with residue 1/6. Thus looking back at (4.3.6), we find that

$$\begin{aligned} L^{\textrm{tw}}(s) = \frac{QR}{\zeta (2)} \zeta (6s) + s \int _1^\infty \left( N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right) u^{-1-s} \, \text {d}u \end{aligned}$$

when $\sigma > 1/6$, but in fact the right-hand side of this equality defines a meromorphic function on the region (4.3.4) with a simple pole at $s = 1/6$ and no other poles. Our claim follows. $\square$

5 Estimates for rational isomorphism classes

In Sect. 4, we counted the number of elliptic curves over $\mathbb {Q}$ with a 7-isogeny up to isomorphism over $\mathbb {Q}^{al }$ (Theorem 4.2.19). In this section, we count all isomorphism classes over $\mathbb {Q}$ by enumerating over twists using a Tauberian theorem ( Theorem 3.5.2).

5.1 Setup

Breaking up the sum (2.2.10), let

$$\begin{aligned} h(n) \,{:=}\,\#\{(a, b, c) \in \mathbb {Z}^3 : (a, b) \text { groomed, } c \text { squarefree, } {{\,\textrm{ht}\,}}(c^2 A(a,b),c^3 B(a,b)) = n\} \nonumber \\. \end{aligned}$$

(5.1.1)

Then $h(n)$ counts the number of elliptic curves $E \in \mathscr {E}$ of height n that admit a 7-isogeny (2.2.9) and

$$\begin{aligned} N_{}(X) = \sum _{n \le X} h(n). \end{aligned}$$

(5.1.2)

We also let

$$\begin{aligned} L(s) \,{:=}\,\sum _{n \ge 1} \frac{h(n)}{n^s} \end{aligned}$$

(5.1.3)

wherever this sum converges.

Theorem 5.1.4

The following statements hold.

• (a) We have

  $$\begin{aligned} h(n) = 2 \sum _{c^6 \mid n} \left| \mu (c)\right| h^{\textrm{tw}}(n/c^6) \end{aligned}$$

• (b) For $s = \sigma + i t \in \mathbb {C}$ with $\sigma > 1/6$ we have

  $$\begin{aligned} L(s) = \frac{2 \zeta (6s) L^{\textrm{tw}}(s)}{\zeta (12s)} \end{aligned}$$

  (5.1.5)

  with absolute convergence on this region.

• (c) The Dirichlet series $L(s)$ has a meromorphic continuation to the region (4.3.4) with a double pole at $s = 1/6$ and no other singularities on this region.

• (d) The Laurent expansion for $L(s)$ at $s = 1/6$ begins

  $$\begin{aligned} L(s)= & {} \frac{1}{3 \zeta (2)^2}\nonumber \\{} & {} \left( \frac{QR}{6} \left( s - \frac{1}{6}\right) ^{-2} + \left( \zeta (2) \ell _0 + Q R \left( \gamma - \frac{2 \zeta ^\prime (2)}{\zeta (2)}\right) \right) \left( s - \frac{1}{6}\right) ^{-1}+ O(1)\right) ,\nonumber \\ \end{aligned}$$

  (5.1.6)

  where

  $$\begin{aligned} \ell _0 \,{:=}\,\frac{Q R \gamma }{\zeta (2)} + \frac{1}{6} \int _1^\infty \left( N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right) u^{-7/6} \,\textrm{d}u \end{aligned}$$

  (5.1.7)

  is the constant term of the Laurent expansion for $L^{\textrm{tw}}(s)$ around $s = 1/6$.

Proof

Part (a) follows directly from Lemma 2.2.2 and (2.1.8) and is something true independent of the parametrization: to count all elliptic curves up to isomorphism by height, it suffices to count them as twists of only the twist minimal curves. More precisely, from (2.1.8) we have ${{\,\textrm{ht}\,}}(c^2 A(a,b), c^3B(a,b))=n$ with c squarefree if and only if ${{\,\textrm{twht}\,}}(A(a,b),B(a,b))=n/(c')^6$ where $c' \,{:=}\,ce/\gcd (c,e)^2$ and $e \,{:=}\,{{\,\textrm{tmd}\,}}(A(a,b),B(a,b))$ and $c'$ squarefree. Thus

$$\begin{aligned} h(n) = \sum _{\begin{array}{c} c' \text { squarefree} \\ (c')^6 \mid n \end{array}} h^{\textrm{tw}}(n/(c')^6) \end{aligned}$$

which of course gives

$$\begin{aligned} h(n) = 2 \sum _{(c')^6 \mid n} \left| \mu (c')\right| h^{\textrm{tw}}(n/(c')^6) \end{aligned}$$

proving (a).

For (b), we see that $h_7(n)$ is the the nth coefficient of the Dirichlet convolution of $L^{\textrm{tw}}(s)$ and

$$\begin{aligned} 2\sum _{n \ge 1} \left| \mu (n)\right| n^{-6s} = \frac{2\zeta (6s)}{\zeta (12s)}. \end{aligned}$$

Write $s = \sigma + i t$. As both $L^{\textrm{tw}}(s)$ and $\zeta (6s)/\zeta (12s)$ are absolutely convergent when $\sigma > 1/6$, we see

$$\begin{aligned} L(s) = \frac{2 \zeta (6s) L^{\textrm{tw}}(s)}{\zeta (12s)} \end{aligned}$$

when $\sigma > 1/6$, and $L(s)$ converges absolutely in this half-plane.

For (c), since $\zeta (s)$ is nonvanishing when $\sigma > 1$, the ratio $\zeta (6s)/\zeta (12s)$ is meromorphic for $\sigma > 1/12$. But Corollary 4.3.3 gives a meromorphic continuation of $L^{\textrm{tw}}(s)$ to the region (4.3.4). The function $L(s)$ is a product of these two meromorphic functions on (4.3.4), and so it is a meromorphic function on this region. The holomorphy and singularity for $L(s)$ then follow from those of $L^{\textrm{tw}}(s)$ and $\zeta (s)$.

We conclude (d) by computing Laurent expansions. We readily verify

$$\begin{aligned} \frac{\zeta (6s)}{\zeta (12s)} = \frac{1}{\zeta (2)}\left( \frac{1}{6}\left( s - \frac{1}{6}\right) ^{-1}+ \left( \gamma - \frac{2 \zeta ^\prime (2)}{\zeta (2)}\right) + O\left( s - \frac{1}{6}\right) \right) , \end{aligned}$$

(5.1.8)

whereas the Laurent expansion for $L^{\textrm{tw}}(s)$ at $s = 1/6$ begins

$$\begin{aligned} L^{\textrm{tw}}(s) = \frac{1}{\zeta (2)} \left( \frac{QR}{6}\left( s - \frac{1}{6}\right) ^{-1}+ \zeta (2) \ell _0 + \dots \right) , \end{aligned}$$

(5.1.9)

with $\ell _0$ given by (5.1.7). Multiplying the Laurent series tails gives the desired result. $\square$

5.2 Proof of main result

We are now poised to finish off the proof of our main result.

Lemma 5.2.1

The sequence $\left( h(n)\right) _{n \ge 1}$ is admissible ((Definition 3.5.1) with parameters (1/6, 1/30, 128/1025).

Proof

We check each condition in (Definition 3.5.1. Since $h(n)$ counts objects, we indeed have $h(n) \in \mathbb {Z}_{\ge 0}$.

For (i), Corollary 4.3.3 tells us that $L^{\textrm{tw}}(s)$ has 1/6 as its abscissa of absolute convergence. Likewise, $\displaystyle {\frac{\zeta (6s)}{\zeta (12s)}}$ has 1/6 as its abscissa of absolute convergence. By Theorem 5.1.4(b),

$$\begin{aligned} L(s) = \frac{2\zeta (6s) L^{\textrm{tw}}(s)}{\zeta (12s)}, \end{aligned}$$

and by Theorem 3.2.1 this series converges absolutely for $\sigma > \sigma _a$, so the abscissa of absolute convergence for $L(s)$ is at most 1/6. But for $\sigma < 1/6$, $L(\sigma ) > L^{\textrm{tw}}(\sigma )$ by termwise comparison of coefficients, so the Dirichlet series for $L(s)$ diverges when $\sigma < 1/6$, and (i) holds with $\sigma _a = 1/6$.

For (ii), Corollary 4.3.3 tells us that $L^{\textrm{tw}}(s)$ has a meromorphic continuation when $\sigma ={{\,\textrm{Re}\,}}(s)>2/15$; on the other hand, as $\zeta (12s)$ is nonvanishing for $\sigma > 1/12$, we see that $\zeta (6s)/\zeta (12s)$ has a meromorphic contintuation to $\sigma >1/12$, and so (ii) holds with

$$\begin{aligned} \delta = 1/6 - 2/15=1/30. \end{aligned}$$

(The only pole of $L(s)/s$ with $\sigma > 2/15$ is the double pole at $s = 1/6$ indicated in Theorem 5.1.4(e).)

For (iii), let $\sigma > 2/15$. Let $\zeta _a(s)=\zeta (as)$. Applying Theorem 3.4.4, we have

$$\begin{aligned} \mu _{\zeta _6}(\sigma ) = \mu _{\zeta }(6\sigma ) < \frac{64}{205}\left( 1 - \frac{6 \cdot 2}{15}\right) = \frac{64}{1025} \end{aligned}$$

(5.2.2)

if $\sigma \le 1/6$, and by Theorem 3.4.1, $\mu _{\zeta _6}(\sigma ) = 0$ if $\sigma > 1/6$. We recall (4.3.6); applying Theorem 3.4.1 again we see $\mu _{L_{\mathrm{{rem}}}^{\textrm{tw}}}(\sigma ) = 0$ if $\sigma > 2/15$, so (5.2.2) implies $\mu _{L^{\textrm{tw}}}(\sigma ) < 64/1025$ if $\sigma > 2/15$. Finally, as $\zeta (12s)^{-1}$ is absolutely convergent for $s > 1/12$, Theorem 3.4.1 tells us $\mu _{{\zeta _{12}}^{-1}}(\sigma ) = 0$. Taken together, we see

$$\begin{aligned} \mu _{L}(\sigma ) < \frac{64}{1025} + \frac{64}{1025} + 0 = \frac{128}{1025}, \end{aligned}$$

(5.2.3)

so the sequence $\left( h(n)\right) _{n \ge 1}$ is admissible with final parameter $\xi = 128/1025$. $\square$

We now prove Theorem 1.2.2, which we restate here for ease of reference in our established notation.

Theorem 5.2.4

For all $\epsilon > 0$,

$$\begin{aligned} N_{}(X)= & {} \frac{QR}{3 \zeta (2)^2} X^{1/6} \log X + \frac{2}{\zeta (2)^2}\left( \zeta (2) \ell _0 + Q R \left( \gamma - 1 - \displaystyle {\frac{2 \zeta ^\prime (2)}{\zeta (2)}}\right) \right) X^{1/6}\\{} & {} + O\left( X^{3/20 + \epsilon }\right) \end{aligned}$$

as $X \rightarrow \infty$, where the implicit constant depends on $\epsilon$. The constants Q, R are defined in Theorem 4.2.19, and $\ell _0$ is defined in (5.1.7).

Proof

By Lemma 5.2.1, $\left( h(n)\right) _{n \ge 1}$ is admissible with parameters $\left( 1/6, 1/30, 128/1025\right)$. We now apply Theorem 3.5.2 to the Dirichlet series $L(s)$, and our claim follows. $\square$

Remark 5.2.5

We suspect that the true error on both $N_{}(X)$ and $N_{}^{tw }(X)$ is $O(X^{1/12 + \epsilon })$. Some improvements to our error term are possible. Improvements to the error term for $N_{}^{tw }(X)$ will directly improve the error term for $N_{}(X)$. In addition, we believe that (with appropriate hypotheses) the denominator $\left\lfloor \xi \right\rfloor + 2$ in the exponent of the error for Theorem 3.5.2 can be replaced with $\xi + 1$. If so, the exponent $3/20 + \epsilon$ in the error term may be replaced with $158/1153 + \epsilon$. Under this assumption, improvements in the estimate of $\mu _\zeta (\sigma )$ will translate directly to improvements in the error term of $N_{}(X)$ (see Bourgain [4, Theorem 5]). In the most optimistic scenario, if the Lindelöf hypothesis holds, the exponent of our error term would be the same as that of $N_{}^{tw }(X)$.

Remark 5.2.6

Here we combine Landau’s Tauberian theorem (Theorem 3.5.2) with Theorem 5.1.4(b) in order to obtain asymptotics for $L(s)$. In doing so, we implicitly invoke the apparatus of complex analysis, which is used in the proof of Perron’s formula and of Landau’s Tauberian theorem. Indeed, this suggests a general strategy. However, we believe an elementary argument applying Dirichlet’s hyperbola method [21, Theorem I.3.1] to Theorem 5.1.4(a) could achieve similar asymptotics, and perhaps even modestly improve on the error term.

6 Computations

In this section, we conclude by describing some computations which make our main theorems completely explicit.

6.1 Computing elliptic curves with 7-isogeny

We begin by outlining an algorithm for computing all elliptic curves that admit a 7-isogeny up to twist height X. In a nutshell, we iterate over possible factorizations $e^3 m$ with m cubefree to find all groomed pairs (a, b) for which $C(a, b) = e^3 m$, then check if ${{\,\textrm{twht}\,}}(A(a, b), B(a, b)) \le X$.

In detail, our algorithm proceeds as follows.

• 1. We list all primes $p \equiv 1 \pmod 3$ up to $(X/108)^{1/6}$ (this bound arises from Theorem 2.4.6(a)).

• 2. For each pair $(a, b) \in \mathbb {Z}^2$ with $b > 0$, $\gcd (a, b) = 1$, $b > 0$, and C(a, b) coprime to 3 and less than Y, we compute C(a, b). We organize the results into a lookup table, so that for each c we can find all pairs (a, b) with $b > 0$, $\gcd (a, b) = 1$, $b > 0$, and $C(a, b) = c$. We append 1 to our table with lookup value (1, 0). For each c in our lookup table, we record whether c is cubefree by sieving against the primes we previously computed.

• 3. For positive integer pairs $(e_0, m)$, $e_0^{12} m^6 \le X/108$, and m cubefree, we find all groomed pairs $(a, b) \in \mathbb {Z}^2$ with $C(a, b) = e_0^3 m$. If $\gcd (e_0, 3) = \gcd (m, 3) = 1$, we can do this as follows. If $e_0^3 < Y$, we iterate over groomed pairs $(a_e, b_e)$ and $(a_m, b_m)$ yielding $C(a_e, b_e) = e_0^3$ and $C(a_m, b_m) = m$ respectively, and taking the product

  $$\begin{aligned} (a_e + b_e\left( -1+3\zeta \right) ) (a_m + b_m\left( -1+3\zeta \right) ) = a + b \left( -1+3\zeta \right) \in \mathbb {Z}[3\zeta ] \end{aligned}$$

  as in the proof of Lemma 2.4.2. If $e_0^3 > Y$, we iterate over groomed pairs $(a_e^\prime , b_e^\prime )$ with $C(a_e^\prime , b_e^\prime ) = e_0$ instead of over groomed pairs $(a_e, b_e)$, and compute

  $$\begin{aligned} (a_e^\prime + b_e\left( -1+3\zeta \right) )^3 (a_m + b_m\left( -1+3\zeta \right) ) = a + b \left( -1+3\zeta \right) \in \mathbb {Z}[3\zeta ]. \end{aligned}$$

  If $\gcd (e_0, 3) > 1$ or $\gcd (m, 3) > 1$, we perform the steps above for the components of $e_0$ and m coprime to 3, and then postmultiply by those groomed pairs $(a_3, b_3) \in \mathbb {Z}^2$ with $C(a_3, b_3)$ an appropriate power of 3 (which is necessarily 9 or 27 by Lemma 2.4.2(b).

• 4. For each pair (a, b) with $C(a, b) = e_0^3 m$, obtained in the previous step, we compute $H(A(a, b), B(a, b))$. We compute the 3-component of the twist minimality defect $e_3$, the 7-component of the twice minimality defect $e_7$, and thereby compute the twist minimality defect $e = {{\,\textrm{lcm}\,}}(e_0, e_3, e_7)$. We compute the twist height using the reduced pairs $(A(a, b) / e^2, \left| B(a, b)\right| / e^3)$. If this result is less than or equal to X, we report (a, b), together with their twist height and any auxiliary information we care to record.

We list the first few twist minimal elliptic curves admitting a 7-isogeny in Table 1.

Table 1 $E \in \mathscr {E}^{tw }$ with 7-isogeny and ${{\,\textrm{twht}\,}}E \le 10^{9}$

Running this algorithm out to $X = 10^{42}$ took us approximately 34 CPU hours on a single core, producing 4 582 079 elliptic curves admitting a 7-isogeny in $\mathscr {E}^{tw }_{\le 10^{42}}$. To check the accuracy of our code, we confirmed that the j-invariants of these curves are distinct. We also confirmed that the 7-division polynomial of each curve has a linear or cubic factor over $\mathbb {Q}$; this took 3.5 CPU hours. For $X = 10^{42}$, we have

$$\begin{aligned} \frac{\zeta (2)}{Q R}\frac{ N_{}^{tw }(10^{42})}{(10^{42})^{1/6}} = 0.99996\ldots , \end{aligned}$$

which is close to 1.

Substituting Theorem 5.1.4(a) into (5.1.2) and reorganizing the resulting sum, we find

$$\begin{aligned} N_{}(X) = 2 \sum _{n \le X} h^{\textrm{tw}}\left( n/c^6\right) \sum _{c \le (X/n)^{1/6}} \left| \mu (c)\right| . \end{aligned}$$

(6.1.1)

Letting $X = 10^{42}$ and using our list of 4 582 079 elliptic curves admitting a 7-isogeny, we compute that there are $88\,157\,174$ elliptic curves admitting a 7-isogeny in $\mathscr {E}_{\le 10^{42}}$.

6.2 Computing constants

We also estimate the constants in our main theorems. First and easiest among these is Q, given by (4.2.11). Truncating the Euler product as a product over $p \le Y$ gives us a lower bound

$$\begin{aligned} Q_{\le Y} \,{:=}\,\frac{273}{16} \prod _{\begin{array}{c} 7 < p \le Y \\ p \equiv 1 ~(mod ~{3}) \end{array}} \left( 1 + \frac{2}{p^2+1}\right) \end{aligned}$$

for Q. To obtain an upper bound, we compute

$$\begin{aligned} Q < Q_{\le Y} \exp \left( 2 \sum _{\begin{array}{c} p > Y \\ p \equiv 1 ~(mod ~{3}) \end{array}} \frac{1}{p^2+1}\right) . \end{aligned}$$

For $a, b \in \mathbb {Z}$ coprime integers and $X \in \mathbb {R}_{>0}$, write

$$\begin{aligned} \pi (X; a, b) \,{:=}\,\# \left\{ p \ \text {prime} : p \equiv a \pmod b\right\} . \end{aligned}$$

(6.2.1)

Suppose $Y \ge 8 \cdot 10^9$. Using Abel summation and Bennett–Martin–O’Bryant–Rechnitzer [1, Theorem 1.4], we obtain

$$\begin{aligned} \sum _{\begin{array}{c} p > Y \\ p \equiv 1 ~(mod ~{3}) \end{array}} \frac{1}{p^2+1}&= -\frac{\pi (Y;3,1)}{Y^2 + 1} + 2 \int _Y^\infty \frac{\pi (u; 3, 1) u}{(u^2 + 1)^2} \,\textrm{d}u \\&< -\frac{Y}{2\left( Y^2 + 1\right) \log Y} + \left( \frac{1}{\log Y} + \frac{5}{2 \log ^2 Y}\right) \int _Y^\infty \frac{u^2}{(u^2 + 1)^2} \,\textrm{d}u \\&= \frac{1}{2} \left( \frac{5Y}{2 (Y^2 + 1) \log Y} + \left( \frac{1}{\log Y} + \frac{5}{2 \log ^2 Y}\right) \left( \frac{\pi }{2} - \tan ^{-1}(Y)\right) \right) \end{aligned}$$

so

$$\begin{aligned} Q <Q_{\le Y} \cdot \exp \left( \frac{5Y}{2 (Y^2 + 1) \log Y} + \left( \frac{1}{\log Y} + \frac{5}{2 \log ^2 Y}\right) \left( \frac{\pi }{2} - \tan ^{-1}(Y)\right) \right) . \end{aligned}$$

In particular, letting $Y = 10^{12}$, we compute

$$\begin{aligned} 17.46040523112662< Q < 17.460405231134835 \end{aligned}$$

This computation took approximately 9 CPU days, although an estimate nearly as good could be computed much more quickly.

We now turn our attention to R, given in (3.1.2). We observe

$$\begin{aligned} \mathcal {R}(1) \subseteq [-0.677, 0.677] \times [0, 0.078], \end{aligned}$$

so we can estimate $\mathcal {R}(1)$ by performing rejection sampling on the rectangle $[-0.677, 0.677] \times [0, 0.078]$, which has area 0.105612. Of our $s \,{:=}\,595\,055\,000\,000$ samples, $r \,{:=}\,243\,228\,665\,9$65 lie in R, so

$$\begin{aligned} R \approx 0.105612 \cdot \frac{r}{s} = 0.04316889\ldots \end{aligned}$$

with standard error

$$\begin{aligned} 0.105612 \cdot \sqrt{\frac{r(s - r)}{s^3}} < 6.8 \cdot 10^{-8}. \end{aligned}$$

This took 11 CPU weeks to compute, although an estimate nearly as good could be computed much more quickly. With a little more care, we believe that R could be estimated via numerical integration with a provable error bound.

We can approximate $\ell _0$ by truncating the integral (5.1.7) and using our approximations for Q and R. This yields $\ell _0 \approx -1.62334$. In Theorem 4.2.19, we have shown that for some $M > 0$ and for all $u > X$, we have

$$\begin{aligned} \left| N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right| < M u^{2/15} \log ^{17/5} u. \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned}&\left| \int _X^\infty \left( N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor \right) u^{-7/6} \,\textrm{d}u\right| \\&\qquad \qquad< M \int _X^\infty u^{-31/30} \log ^{17/5} u\,\textrm{d}u \\&\qquad \qquad < M \int _X^\infty u^{-31/30} \log ^4 u\,\textrm{d}u \\&\qquad \qquad = 30 M X^{-1/30} \\&\qquad \qquad \left( \log ^4 X + 120 \log ^3 X + 10800 \log ^2 X + 648000 \log X + 19440000\right) ; \end{aligned} \end{aligned}$$

this gives us a bound on our truncation error. We do not know the exact value for M, but empirically, we find that for $1 \le u \le 10^{42}$,

$$\begin{aligned} -3.3119 \cdot 10^{-5} \le \frac{N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor }{u^{2/15} \log ^{17/5} u} \le 4.3226 \cdot 10^{-6}. \end{aligned}$$

If we assume $M \approx 3.3119 \cdot 10^{-5}$, we find the truncation error for $\ell _0$ is bounded by 253.23, which catastrophically dwarfs our initial estimate.

We can do better with stronger assumptions. Suppose for the moment that $N_{}^{tw }(X) - \frac{QR}{\zeta (2)} X^{1/6} = O(X^{1/12 + \epsilon })$, as we guessed in Remark 5.2.5. We let $\epsilon \,{:=}\,10^{-4}$, and find that for $1 \le u \le 10^{42}$,

$$\begin{aligned} -1.2174 \le \frac{N_{}^{tw }(u) - \frac{QR}{\zeta (2)} \left\lfloor u^{1/6} \right\rfloor }{u^{1/12 + \epsilon }} \le 0.52272. \end{aligned}$$

If

$$\begin{aligned} \left| N_{}^{tw }(X) - \frac{QR}{\zeta (2)} X^{1/6}\right| \le M X^{1/12 + \epsilon } \end{aligned}$$

for $M \approx 1.2174$, we get an estimated truncation error of $2.43 \cdot 10^{-5}$, which is much more manageable.

Our estimate of $\ell _0$ is also skewed by our estimates of QR. An error of $\epsilon$ in our estimate for QR induces an error of

$$\begin{aligned} \frac{\epsilon }{6\zeta (2)} \int _1^X \left\lfloor u^{1/6} \right\rfloor u^{-7/6} \,\textrm{d}u < \frac{\epsilon }{6\zeta (2)} \int _1^X u^{-1}\,\textrm{d}u = \frac{\epsilon \log X}{6\zeta (2)} \end{aligned}$$

in our estimate of $\ell _0$. When $X = 10^{42}$, this gives an additional error of $1.15 \cdot 10^{-5}$, for an aggregate error of 253.23 or $2.43 \cdot 10^{-5}$, depending on our assumptions.

Given Q, R, and $\ell _0$, it is straightforward to compute $c_1$ and $c_2$ using the expressions given for them in Theorem 5.2.4. We find $c_1 = 0.09285536\ldots$ with an error of $6.02 \cdot 10^{-8}$, and $c_2 \approx -0.16405$ with an error of 307.89 or of $2.98 \cdot 10^{-5}$, depending on the assumptions made above. Note that both of these estimates for $c_2$ depended on empirical rather than theoretical estimates for the implicit constant in the error term of Theorem 5.2.4. As a sanity check, we also verify that

$$\begin{aligned} \frac{N_{}^{tw }(10^{42})}{10^7} - 42 c_1 \log 10 = -0.1641924\ldots \approx c_2, \end{aligned}$$

which agrees to three decimal places with the estimate for $c_2$ we gave above.