A note on 8-division fields of elliptic curves

TL;DR

This paper explicitly describes the generators and Galois action on the 8-division fields of elliptic curves over fields of characteristic not 2, using roots of defining polynomials.

Contribution

It provides explicit formulas for generators of the 8-division field and describes the Galois automorphism action in terms of polynomial roots.

Findings

Explicit generators for the 8-division field are given.

The Galois automorphism action is explicitly described.

Results apply to elliptic curves defined by degree 3 or 4 polynomials.

Abstract

Let $K$ be a field of characteristic different from $2$ and let $E$ be an elliptic curve over $K$, defined either by an equation of the form $y^{2} = f(x)$ with degree $3$ or as the Jacobian of a curve defined by an equation of the form $y^{2} = f(x)$ with degree $4$. We obtain generators over $K$ of the $8$-division field $K(E[8])$ of $E$ given as formulas in terms of the roots of the polynomial $f$, and we explicitly describe the action of a particular automorphism in $\mathrm{Gal}(K(E[8]) / K)$.

Equations

$$A_1^2+A_2^2+A_3^2=0,$$

$${\varphi }_{2^n}:\mathrm{Gal}(K(E[2^n])/K)\hookrightarrow \mathrm{SL}(E[2^n])$$

$${\theta }_{2^n}:\mathrm{Gal}(L(E[2^n])/L)\to \mathrm{Gal}(K(E[2^n])/K)$$

$$S:=\mathrm{Spec}(\mathbb{Z}[\frac{1}{2},\{{\tilde{\alpha }}_i{\}}_{i=1}^d,\{({\tilde{\alpha }}_i-{\tilde{\alpha }}_j{)}^{-1}{\}}_{1\le i<j\le d}{]}^{S_d})$$

$${\beta }_i{\beta }_j$$

$${\beta }_i{\beta }_{i+1}{\beta }_i$$

$$⟨\sigma ,\tau |{\sigma }^4={\tau }^4=[{\sigma }^2,\tau ]=[\sigma ,{\tau }^2]=[\sigma ,\tau {]}^2=[[\sigma ,\tau ],\sigma ]=[[\sigma ,\tau ],\tau ]=1⟩.$$

$$(A_i(A_{i+1}+{\zeta }_4{A}_{i+2}))(A_i(A_{i+1}-{\zeta }_4{A}_{i+2}))=-A_i^4.$$

$$\sigma :(A_1,A_2,A_3,B_1,B_2,B_3)\mapsto (A_1,A_2,-A_3,B_1^{\prime },{\zeta }_4{B}_2^{\prime },{\zeta }_4{B}_3),$$

$$\tau :(A_1,A_2,A_3,B_1,B_2,B_3)\mapsto (-A_1,A_2,A_3,{\zeta }_4{B}_1,B_2^{\prime },{\zeta }_4{B}_3^{\prime }).$$

$$\mathrm{Gal}(K^{\prime }/K(E[2]))=⟨\sigma ,\tau ⟩\times ⟨\mu ⟩\cong \Gamma (2{)}^{\prime }/\Gamma (8)\times \{\pm 1\}\cong \Gamma (2)/\Gamma (8).$$

$$(B_i\pm B_i^{\prime }{)}^2=2A_i{A}_{i+1}\pm 2{\zeta }_4{A}_i^2;(B_i\pm {\zeta }_4{B}_i^{\prime }{)}^2=2{\zeta }_4{A}_i{A}_{i+2}\mp 2A_i^2.$$

$$\sqrt{-A_i^2\pm {\zeta }_4{A}_i{A}_{i+1}}=(1\mp {\zeta }_4{)}^{-1}(B_i\pm B_i^{\prime }),\sqrt{{\zeta }_4{A}_i{A}_{i+2}\pm A_i^2}=({\zeta }_8+{\zeta }_8^{-1}{)}^{-1}(B_i\mp {\zeta }_4{B}_i^{\prime })\in K(E[8]).$$

$$2^{-1}{\zeta }_4(B_i-B_i^{\prime }{)}^2{B}_{i+2}^2/(A_i+{\zeta }_4{A}_{i+1}{)}^2=(A_i(A_i+{\zeta }_4{A}_{i+1}))(A_{i+2}(A_i+{\zeta }_4{A}_{i+1}))/(A_i+{\zeta }_4{A}_{i+1}{)}^2$$

$$=A_i{A}_{i+2}(A_i+{\zeta }_4{A}_{i+1}{)}^2/(A_i+{\zeta }_4{A}_{i+1}{)}^2=A_i{A}_{i+2}.$$

$$P:=(x_0+(r_1{r}_2+r_2{r}_3+r_3{r}_1),-y_0+(r_1+r_2+r_3)(r_1{r}_2+r_2{r}_3+r_3{r}_1))$$

$$Q=({\zeta }_4{A}_2{A}_3+(r_1{r}_2+r_2{r}_3+r_3{r}_1),-{\zeta }_4{A}_2{A}_3(A_2+{\zeta }_4{A}_3)+(r_1+r_2+r_3)(r_1{r}_2+r_2{r}_3+r_3{r}_1)),$$

$$r_1^2={\zeta }_4{A}_2{A}_3-{\alpha }_1^{\prime }={\zeta }_4{A}_2{A}_3;r_2^2={\zeta }_4{A}_2{A}_3-{\alpha }_2^{\prime }={\zeta }_4{A}_2{A}_3+A_3^2;r_3^2={\zeta }_4{A}_2{A}_3-{\alpha }_3^{\prime }={\zeta }_4{A}_2{A}_3-A_2^2;$$

$$r_1\in \{\pm ({\zeta }_8-{\zeta }_8^{-1}{)}^{-1}(B_3-B_3^{\prime })B_2/(A_3+{\zeta }_4{A}_1)\},r_2\in \{\pm ({\zeta }_8+{\zeta }_8^{-1}{)}^{-1}(B_3-{\zeta }_4{B}_3^{\prime })\},r_3\in \{\pm (1-{\zeta }_4{)}^{-1}(B_2+B_2^{\prime })\}$$

Full text

Let $K$ be any field of characteristic different from $2$, and let $E$ be an elliptic curve over $K$. For any integer $N\geq 1$, we write $E[N]$ for the $N$-torsion subgroup of $E$ and $K(E[N])$ for the (finite algebraic) extension of $K$ obtained by adjoining the coordinates of the points in $E[N]$ to $K$. Let $T_{2}(E)$ denote the $2$-adic Tate module of $E$; it is a free $\mathbb{Z}_{2}$-module of rank $2$ given by the inverse limit of the finite groups $E[2^{n}]$ with respect to the multiplication-by-$2$ map. The absolute Galois group $G_{K}=\mathrm{Gal}(\bar{K}/K)$ of $K$ acts in a natural way on each free rank-$2$ $\mathbb{Z}/2^{n}\mathbb{Z}$-module $E[2^{n}]$; we denote this action by $\bar{\rho}_{2^{n}}:G_{K}\to\mathrm{Aut}(E[2^{n}])$. This induces an action of $G_{K}$ on $T_{2}(E)$, which we denote by $\rho_{2}:G_{K}\to\mathrm{Aut}(T_{2}(E))$.

The purpose of this note is to provide formulas for generators of the $8$-division field $K(E[8])$ of an elliptic curve $E$ and to describe how a certain Galois element in $\mathrm{Gal}(K(E[8])/K)$ acts on these generators. We will consider the case where $E$ is given by a standard Weierstrass equation of the form $y^{2}=\prod_{i=1}^{3}(x-\alpha_{i})\in K[x]$ (the “degree-$3$ case”) and the case where $E$ is the Jacobian of the genus-$1$ curve given by an equation of the form $y^{2}=\prod_{i=1}^{4}(x-\alpha_{i})\in K[x]$ (the “degree-$4$ case”), where in both cases the elements $\alpha_{i}\in\bar{K}$ are distinct.

For the statement of the main theorem and the rest of this article, we fix the following algebraic elements over $K$ (for ease of notation, we will treat indices $i$ as elements of $\mathbb{Z}/3\mathbb{Z}$). In the degree-$3$ case, for each $i\in\mathbb{Z}/3\mathbb{Z}$, we choose an element $A_{i}\in\bar{K}$ whose square is $\alpha_{i+1}-\alpha_{i+2}$. In the degree-$4$ case, for each $i\in\mathbb{Z}/3\mathbb{Z}$, we choose an element $A_{i}\in\bar{K}$ whose square is $(\alpha_{i}-\alpha_{4})(\alpha_{i+1}-\alpha_{i+2})$. One checks that in either case, we have the identity

[TABLE]

which we will exploit below.

In the degree-$3$ case, it is well known that $K(E[2])=K(\alpha_{1},\alpha_{2},\alpha_{3})$. Menawhile, in the degree-$4$ case, the extension $K(E[2])/K$ is generated by polynomials in the roots $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$ which are fixed by the group of permutations in $S_{4}$ that fix all partitions of the roots into $2$-element subsets. This follows from a well-known description of the $2$-torsion points of the Jacobian of a hyperelliptic curve (see for instance the statement and proof of [7, Corollary 2.11]) which says that the points in $E[2]$ are parametrized by partitions of the set of roots $\{\alpha_{i}\}_{i=1}^{4}$ into even-cardinality subsets, and that $G_{K}$ acts on $E[2]$ via the Galois action on these partitions determined by permutation of the $\alpha_{i}$’s. In fact, it is clear (from examining, for instance, the solution to the “generic” quartic equation via the resolvent cubic) that $K(E[2])$ coincides with $K(\gamma_{1},\gamma_{2},\gamma_{3})$, where $\gamma_{i}=(\alpha_{i+1}+\alpha_{i+2})(\alpha_{i}+\alpha_{4})$ for $i\in\mathbb{Z}/3\mathbb{Z}$; note that $A_{i}^{2}=\gamma_{i+1}-\gamma_{i+2}$ for each $i$. Thus, in either case, we have $A_{1}^{2},A_{2}^{2},A_{3}^{2}\in K(E[2])$.

Now for each $i\in\mathbb{Z}/3\mathbb{Z}$, fix an element $B_{i}\in\bar{K}$ whose square is $A_{i}(A_{i+1}+\zeta_{4}A_{i+2})$. Let $\zeta_{8}\in\bar{K}$ be a primitive $8$th root of unity, and let $\zeta_{4}=\zeta_{8}^{2}$, which is a primitive $4$th root of unity. Our result is as follows.

b) If the scalar automorphism $-1\in\mathrm{Aut}(E[8])$ lies in the image under $\bar{\rho}_{8}$ of some Galois element $\sigma\in G_{K}$, then $\sigma$ acts on $K(E[8])$ by fixing $K(E[2],\zeta_{8})$ and changing the sign of each generator $A_{i},B_{i}\in\bar{K}$.

b) For certain elliptic curves, it is possible to determine using various methods that the image of $\rho_{2}$ contains $\Gamma(8)$. See [13, Example 4.3], which shows this for elliptic curves in Legendre form whose Weierstrass roots satisfy certain arithmetic conditions; e.g. $y^{2}=x(x-1)(x-10)$. One can then use our result to determine the full $2$-adic image in these cases.

The rest of this article is devoted to proving Theorem 1. We begin by justifying a simplifying assumption about the ground field $K$. From now on, the superscript “$S_{d}$” over a ring containing independent transcendental variables $\tilde{\alpha}_{1},...,\tilde{\alpha}_{d}$ indicates the subring of elements fixed under all permutations of the variables $\tilde{\alpha}_{i}$.

Step 1: We show that the statements are true for $K=k(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$, where $k$ is any subfield of $\mathbb{C}$. Due to the Galois equivariance of the Weil pairing, we have $\zeta_{4}\in K(E[4])$ and $\zeta_{8}\in K(E[8])$. We will therefore assume that $\zeta_{8}\in k$, so that the image of $\mathrm{Gal}(\bar{K}/K)$ under $\rho_{2}$ modulo $4$ (resp. modulo $8$) is contained in the group $\mathrm{SL}(E[4])$ (resp. $\mathrm{SL}(E[8])$) of automorphisms of determinant $1$. For any $n\geq 1$, write

[TABLE]

for the obvious injection induced by $\rho_{2}$, and define $\phi_{2^{n},L}:\mathrm{Gal}(L(E[2^{n}])/L)\hookrightarrow\mathrm{SL}(E[2^{n}])$ analogously. From the formulas given in Theorem 1 and elementary computations of the orders of the (finite) groups above for $n\in\{2,3\}$, we see that $\phi_{4,L}$ and $\phi_{8,L}$ are isomorphisms. Now let

[TABLE]

be the composition of the natural inclusion $\mathrm{Gal}(L(E[2^{n}])/L)\hookrightarrow\mathrm{Gal}(L(E[2^{n}])/K)$ with the natural restriction map $\mathrm{Gal}(L(E[2^{n}])/K)\twoheadrightarrow\mathrm{Gal}(K(E[2^{n}])/K)$. Note that the automorphism in $\mathrm{Gal}(L(E[8])/L(E[2]))$ which changes the sign of each generator given in Theorem 1 is sent by $\theta_{8}$ to the automorphism in $\mathrm{Gal}(K(E[8])/K)$ which changes the sign of each of these generators.

It is clear that $\phi_{2^{n},L}=\phi_{2^{n}}\circ\theta_{2^{n}}$. It will therefore suffice to show that $\theta_{4}$ and $\theta_{8}$ are isomorphisms. Indeed, they are injections due to the fact that $L(E[2^{n}])$ is the compositum of the subfields $K(E[2^{n}])$ and $\mathbb{C}$ for each $n\geq 1$, and the fact that they are surjections in the case of $n\in\{2,3\}$ follows immediately from the surjectivity of $\phi_{4,L}$ and $\phi_{8,L}$.

Step 2: We show that the statements are true for $K=\mathbb{F}_{p}(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$, where $p\neq 2$. Let $E_{0}$ be the elliptic curve defined in the obvious way over $K_{0}:=\mathbb{Q}(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$ in the degree-$d$ case. By what was shown in Step 1, the statement of Theorem 1 is true for $E_{0}$. It is easy to see that $E_{0}$ admits a model $\mathcal{E}$ over

[TABLE]

which is an abelian scheme whose fiber over the prime $(p)$ is isomorphic to $E$. For each $n\geq 1$, Proposition 20.7 of [5] implies that the kernel of the multiplication-by-$2^{n}$ map on $\mathcal{E}\to S$, which we denote by $\mathcal{E}[2^{n}]\to S$, is a finite étale group scheme over $S$. Since the morphism $\mathcal{E}[2^{n}]\to S$ is finite, $\mathcal{E}[2^{n}]$ is an affine scheme; we write $\mathcal{O}_{S,2^{n}}\supset\mathcal{O}_{S}$ for the minimal extension of scalars under which $\mathcal{E}[2^{n}]$ becomes constant. Note that the ring $\mathbb{Z}[\frac{1}{2},\{\tilde{\alpha}_{i}\}_{i=1}^{d},\{(\tilde{\alpha}_{i}-\tilde{\alpha}_{j})^{-1}\}_{1\leq i<j\leq d}]$, along with all subrings of invariants under finite groups of automorphisms, is integrally closed. It follows from the fact that $\mathcal{O}_{S}$ is integrally closed and from the finite étaleness of $\mathcal{O}_{S}[2^{n}]$ that $\mathcal{O}_{S,2^{n}}$ is also integrally closed. Moreover, the fraction field of $\mathcal{O}_{S,2^{n}}$ coincides with $K_{0}(E_{0}[2^{n}])$ while the fraction field of $\mathcal{O}_{S,2^{n}}/(p)$ coincides with $K(E[2^{n}])$ for each $n$. We further observe that $\mathcal{O}_{S}[\{\tilde{\alpha}_{i}\}_{i=1}^{3}]$ (resp. the subring of invariants in $\mathcal{O}_{S}[\{\tilde{\alpha}_{i}\}_{i=1}^{4}]$ under the permutations which fix all partitions of $\{\tilde{\alpha}_{i}\}_{i=1}^{4}$ into even-cardinality subsets) is integrally closed with fraction field equal to $K_{0}(E[2])$ and therefore coincides with $\mathcal{O}_{S,2}$ in the degree-$3$ case (resp. in the degree-$4$ case).

We now need to show that $\mathcal{O}_{S,2}[\zeta_{4},A_{1},A_{2},A_{3}]$ and $\mathcal{O}_{S,2}[\zeta_{8},A_{1},A_{2},A_{3},B_{1},B_{2},B_{3}]$ (where $A_{i},B_{i}$ are the generators given in the statement of Theorem 1) are integerally closed and therefore coincide with $\mathcal{O}_{S,4}$ and $\mathcal{O}_{S,8}$ respectively. This claim follows from the elementary observation that extensions of integrally closed rings given by adjoining square roots of squarefree integral elements are integrally closed, provided that $2$ is invertible. Moreover, the automorphism of $\mathcal{O}_{S,8}$ which fixes $\mathcal{O}_{S,2}$ and changes the signs of the $A_{i}$’s and $B_{i}$’s acts on the sections of the finite group scheme $\mathcal{E}[8]\to S$ as multiplication by $-1$, so we are done.

Step 3: We show that the statement is true for $K=k(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$, where $k$ is any field of characteristic different from $2$. Indeed, in the degree-$d$ case, the curve given by $y^{2}=\prod_{i=1}^{d}(x-\tilde{\alpha}_{i})$ is clearly defined over $\mathfrak{F}:=\mathfrak{f}(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$, where $\mathfrak{f}$ is the prime subfield of $k$, so $E$ is defined over $\mathfrak{F}$. The statements are true over $\mathfrak{F}$ by what was shown in Steps 1 and 2, and the validity of the statements over $K$ follows from observing that $K(E[2^{n}])=K\mathfrak{F}(E[2^{n}])$ for all $n\geq 1$.

Step 4: We show that the statements are true when $K$ is any field of characteristic different from $2$. Let $E_{0}$ be the elliptic curve defined in the obvious way over $K(\{\tilde{\alpha}_{i}\}_{i=1}^{d})^{S_{d}}$. Then $E_{0}$ admits a model over $\mathrm{Spec}(K[\{\tilde{\alpha}_{i}\}_{i=1}^{d},(\tilde{\alpha}_{i}-\tilde{\alpha}_{j})_{1\leq i<j\leq d}]^{S_{d}})$, of which $E$ is the fiber over the closed point given by $\{\tilde{\alpha}_{i}=\alpha_{i}\}_{i=1}^{d}$, and the claims follow by a similar argument to what was used for Step 2.

∎

In light of the above lemma, we assume from now on that the $\alpha_{i}$’s are independent transcendental variables over $\mathbb{C}$ and that $K=\mathbb{C}(\{\alpha_{i}\}_{i=1}^{d})^{S_{d}}$ with $d=3$ (resp. $d=4$) in the degree-$3$ case (resp. the degree-$4$ case). Note that again due to the equivariance of the Weil pairing, the image of $\rho_{2}$ is contained in the subgroup $\mathrm{SL}(T_{2}(E))\subset\mathrm{Aut}(T_{2}(E))$ of automorphisms of determinant $1$.

We now recall the abstract definitions of the braid group and the pure braid group on $d$ strands for any integer $d\geq 1$. The braid group on $d$ strands, denoted $B_{d}$, is generated by elements $\beta_{1},\beta_{2},...,\beta_{d-1}$, with relations given by

[TABLE]

There is a surjective homomorphism of $B_{d}$ onto the symmetric group $S_{d}$ given by mapping each generator $\beta_{i}$ to the transposition $(i,i+1)\in S_{d}$. We define the pure braid group on $d$ strands, denoted $P_{d}\lhd B_{d}$, to be the kernel of this surjection. A set of generators $\{A_{i,j}\}_{1\leq i<j\leq d}$ of $P_{d}$ with relations is given by [2, Lemma 1.8.2]. We write $\widehat{B}_{d}$ and $\widehat{P}_{d}$ for the profinite completions of $B_{d}$ and $P_{d}$ respectively. Note that since $P_{d}$ and $B_{d}$ are residually finite, they inject into their profinite completions; moreover, since $P_{d}\lhd B_{d}$ has finite index, this injection induces an inclusion of profinite completions $\widehat{P}_{d}\lhd\widehat{B}_{d}$.

Let $G_{K}^{\mathrm{unr}}$ denote the Galois group of the maximal algebraic extension $K^{\mathrm{unr}}$ of $K$ which is unramified over the discriminant locus $\Delta$ of the affine scheme $\mathrm{Spec}(\mathbb{C}[\{\alpha_{i}\}_{i=1}^{d}]^{S_{d}})$, where $d=3$ (resp. $d=4$) in the degree-$3$ case (resp. the degree-$4$ case). For any integer $n\geq 0$, we write $\Gamma(2^{n})$ for the level-$2^{n}$ congruence subgroup of $\mathrm{SL}(T_{2}(E))$ given by $\{\sigma\in\mathrm{SL}(T_{2}(E))\ |\ \sigma\equiv 1\ (\mathrm{mod}\ 2^{n})\}$.

b) The map $\rho_{2}:G_{K}\to\mathrm{SL}(T_{2}(E))$ is surjective and factors through the obvious restriction map $G_{K}\twoheadrightarrow G_{K}^{\mathrm{unr}}\cong\widehat{B}_{d}$, inducing a surjection $\rho_{2}^{\mathrm{unr}}:\widehat{B}_{d}\twoheadrightarrow\mathrm{SL}(T_{2}(E))$.

c) For each $n\geq 0$, the algebraic extension $K(E[2^{n}])/K$ is a subextension of $K^{\mathrm{unr}}/K$ and corresponds to the normal subgroup $(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2^{n}))\lhd\widehat{B}_{d}$.

d) The normal subgroup $(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2))\lhd\widehat{B}_{d}$ coincides with $\widehat{P}_{3}\lhd\widehat{B}_{3}$ in the degree-$3$ case, and it coincides with a subgroup $H\lhd\widehat{B}_{4}$ which strictly contains $\widehat{P}_{4}\lhd\widehat{B}_{4}$ and which is isomorphic to $\widehat{P}_{3}$ in the degree-$4$ case.

To prove (b), we first note that $\rho_{2}$ is surjective because it is known that there exist elliptic curves with “largest possible” $2$-adic Galois images; see also [12, Corollary 1.2(b)]. Now choose any prime $\mathfrak{p}$ of the coordinate ring of $X_{d}$ and note that $E$ has good reduction with respect to this prime. It follows from the criterion of Néron-Ogg-Shafarevich ([11, Theorem 1]) that the action $\rho_{2}$ is unramified with respect to $\mathfrak{p}$ and therefore factors through an algebraic extension of $K(\{\alpha_{i}\}_{i=1}^{d})$ which is unramified over $\mathfrak{p}$. The second claim of (b) follows.

Part (c) is immediate from the observation that the action $\bar{\rho}_{2^{n}}:G_{K}\to\mathrm{Aut}(E[2^{n}])$ is clearly the composition of $\rho_{2}$ with the quotient-by-$\Gamma(2^{n})$ map.

Finally, we investigate the subgroup $(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2))\lhd\widehat{B}_{d}$. In the degree-$3$ case, we get $(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2))=\widehat{P}_{3}\lhd\widehat{B}_{3}$ from the fact that $K(E[2])=K(\alpha_{1},\alpha_{2},\alpha_{3})$, which is the function field of the ordered configuration space $Y_{3}$ as defined above. In the degree-$4$ case, we have seen that $K(E[2])=K(\gamma_{1},\gamma_{2},\gamma_{3})\subsetneq K(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4})$, which is the function field of $Y_{4}$. Therefore, we have $H:=(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2))\supsetneq\widehat{P}_{4}\lhd\widehat{B}_{4}$. It is easy to check that the $\gamma_{i}$’s are independent and transcendental over $\mathbb{C}$, so that $K(E[2])$ and the function field of $Y_{3}$ are isomorphic as abstract $\mathbb{C}$-algebras. Thus, $H\cong\widehat{P}_{3}$, and (d) is proved.

∎

We now present several well-known group-theoretic facts which will be needed later.

b) The abelianization of $P_{d}$ is isomorphic to $\mathbb{Z}^{d(d-1)/2}$. More explicitly, it is freely generated by the images of the above generators $A_{i,j}$.

c) For each $n\geq 1$, the quotient $\Gamma(2^{n})/\Gamma(2^{n+1})$ is an elementary abelian group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{3}$.

∎

It is now easy to determine the $4$-division field of $E$ in both cases and to describe how $G_{K}$ acts on it. We note that in the degree-$3$ case, parts (a) and (b) are well known and can be deduced by straightforward calculations of order-$4$ points (for instance, in [1, Example 2.2]; see also [12, Proposition 3.1]).

b) Any Galois element $\sigma\in G_{K}$ with $\rho_{2}(\sigma)=-1\in\mathrm{SL}(T_{2}(E))$ acts on $K(E[4])$ by fixing $K(E[2])$ and changing the signs of each generator $A_{i}\in\bar{K}$.

c) In the degree-$3$ case, the scalar automorphism $-1\in\mathrm{SL}(T_{2}(E))$ is the image of the braid $\Sigma\in\widehat{P}_{3}$ under $\rho_{2}^{\mathrm{unr}}$. In the degree-$4$ case, the scalar $-1\in\mathrm{SL}(T_{2}(E))$ is the image of the braid $\Sigma\in\widehat{P}_{3}\cong H$, where $H\lhd\widehat{B}_{4}$ is the subgroup from the statement of Lemma 4(c).

Part (b) follows from checking that the automorphism of $K(E[4])$ defined by changing the signs of all the $A_{i}$’s is the only nontrivial automorphism lying in the center of $\mathrm{Gal}(K(E[2],A_{1},A_{2},A_{3})/K)$.

Now it follows from (a) and (b) of Lemma 5 that $\Sigma$ has nontrivial image in $P_{3}^{\mathrm{ab}}/2P_{3}^{\mathrm{ab}}$. It therefore has nontrivial image in $\Gamma(2)/\Gamma(4)$, so $\rho_{2}^{\mathrm{unr}}(\Sigma)$ is a nontrivial element of $\mathrm{SL}(T_{2}(E))$. We know from Lemma 5(a) that $\Sigma$ lies in the center of $\widehat{P}_{3}$. It follows from Lemma 4(d) that $\rho_{2}^{\mathrm{unr}}$ restricted to $(\rho_{2}^{\mathrm{unr}})^{-1}(\Gamma(2))\cong\widehat{P}_{3}$ is surjective onto $\Gamma(2)\lhd\mathrm{SL}(T_{2}(E))$, so it takes the center of $\widehat{P}_{3}$ to the center of $\Gamma(2)$, which is $\{\pm 1\}$. We therefore get $\rho_{2}(\sigma)=-1\in\Gamma(2)$, which is the statement of (c).

∎

We now want to find generators for the extension $K(E[8])/K(E[2])$. In order to do so, we will first prove that $\widehat{P}_{3}$ has a unique quotient isomorphic to $\Gamma(2)/\Gamma(8)$ (Lemma 8 below), and then we will show that the extension of $K(E[2])$ given in the statement of Theorem 1 has Galois group isomorphic to $\Gamma(2)/\Gamma(8)$ (Lemma 9 below). For the following, we note that after fixing a basis of the free rank-$2$ $\mathbb{Z}_{2}$-module $T_{2}(E)$, we may consider $\mathrm{SL}(T_{2}(E))$ as the matrix group $\mathrm{SL}_{2}(\mathbb{Z}_{2})$. Moreover, by applying a suitable form of the Strong Approximation Theorem (see for instance Theorem 7.12 of [8]), we have $\Gamma(2)/\Gamma(8)\cong(\Gamma(2)\cap\mathrm{SL}_{2}(\mathbb{Z}))/(\Gamma(8)\cap\mathrm{SL}_{2}(\mathbb{Z}))$. In light of this, in the proofs of the next two lemmas, we use the symbols $\Gamma(2)$ and $\Gamma(8)$ to denote principal congruence subgroups of $\mathrm{SL}_{2}(\mathbb{Z})$ rather than of $\mathrm{SL}_{2}(\mathbb{Z}_{2})\cong\mathrm{SL}(T_{2}(E))$.

[TABLE]

Let $\sigma$ (resp. $\tau$) be the image of $\tilde{\sigma}:=\begin{bmatrix}1&-2\\ 0&1\end{bmatrix}$ (resp. $\tilde{\tau}:=\begin{bmatrix}1&0\\ 2&1\end{bmatrix}$) in $\Gamma(2)^{\prime}/\Gamma(8)$. It is well known that $\tilde{\sigma}$ and $\tilde{\tau}$ generate $\Gamma(2)^{\prime}\cap\mathrm{SL}_{2}(\mathbb{Z})$ (see, for instance, Proposition A.1 of [6]), so $\sigma$ and $\tau$ generate $\Gamma(2)^{\prime}/\Gamma(8)$. It is then straightforward to check that the relations given in (3) hold. To show that these relations determine the group $\Gamma(2)^{\prime}/\Gamma(8)$, one checks that the only nontrivial element of the commutator subgroup of the group given by (3) has order $2$ and that the quotient by the commutator subgroup is isomorphic to $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$; therefore, the group has order $32$. Since $\Gamma(2)^{\prime}/\Gamma(8)$ also has order $32$, it must be fully determined by the relations in (3).

∎

Any surjection $\Gamma(2)\twoheadrightarrow\Gamma(2)/\Gamma(8)$ takes $-1\in\Gamma(2)$ to a nontrivial element $\mu\in\Gamma(2)/\Gamma(8)$ and takes $\Gamma(2)^{\prime}$ to some proper subgroup of $\Gamma(2)/\Gamma(8)$ not containing $\mu$, since $\Gamma(2)^{\prime}$ can be generated by only $2$ elements while $\Gamma(2)/\Gamma(8)$ cannot. Therefore, such a surjection takes $\Gamma(2)^{\prime}$ to a subgroup of $\Gamma(2)/\Gamma(8)$ isomorphic to $\Gamma(2)^{\prime}/\Gamma(8)$. So in fact it suffices to show that $\Gamma(8)$ is the only normal subgroup of $\Gamma(2)^{\prime}$ which induces a quotient isomorphic to $\Gamma(2)^{\prime}/\Gamma(8)$.

Let $N^{\prime}\lhd\Gamma(2)^{\prime}$ be a normal subgroup such that $\Gamma(2)^{\prime}/N^{\prime}\cong\Gamma(2)^{\prime}/\Gamma(8)$. Let $\tilde{\sigma}$ and $\tilde{\tau}$ be the matrices given in the proof of Lemma 7, and let $\phi_{N^{\prime}}:\Gamma(2)^{\prime}\twoheadrightarrow\Gamma(2)^{\prime}/N^{\prime}$ be the obvious quotient map. One checks from the presentation given in the statement of Lemma 7 that each element of $\Gamma(2)^{\prime}/N^{\prime}$ has order dividing $4$; that each square element lies in the center; and that each commutator has order dividing $2$ and lies in the center. It follows that $\phi_{N^{\prime}}(\tilde{\sigma}^{4})=\phi_{N^{\prime}}(\tilde{\tau}^{4})=\phi_{N^{\prime}}([\tilde{\sigma}^{2},\tilde{\tau}])=\phi_{N^{\prime}}([\tilde{\sigma},\tilde{\tau}^{2}])=\phi_{N^{\prime}}([\tilde{\sigma},\tilde{\tau}]^{2})=\phi_{N^{\prime}}([[\tilde{\sigma},\tilde{\tau}],\tilde{\sigma}])=\phi_{N^{\prime}}([[\tilde{\sigma},\tilde{\tau}],\tilde{\tau}])=1$. Thus, $N^{\prime}$ contains the subgroup normally generated by $\{\tilde{\sigma}^{4},\tilde{\tau}^{4},[\tilde{\sigma}^{2},\tilde{\tau}],[\tilde{\sigma},\tilde{\tau}^{2}],[\tilde{\sigma},\tilde{\tau}]^{2},[[\tilde{\sigma},\tilde{\tau}],\tilde{\sigma}],[[\tilde{\sigma},\tilde{\tau}],\tilde{\tau}]\}$. But Lemma 7 implies that $\Gamma(8)\lhd\Gamma(2)^{\prime}$ is normally generated by this subset, so $\Gamma(8)\unlhd N^{\prime}$. Since $\Gamma(2)^{\prime}/N^{\prime}$ and $\Gamma(2)^{\prime}/\Gamma(8)$ have the same (finite) order, we have $N^{\prime}=\Gamma(8)$, as desired.

∎

Using the relation (1), for each $i$, we compute

[TABLE]

In light of this, for $i\in\mathbb{Z}/3\mathbb{Z}$, we define $B_{i}^{\prime}$ to be the element of $K^{\prime}$ such that $B_{i}^{\prime 2}=A_{i}(A_{i+1}-\zeta_{4}A_{i+2})$ and $B_{i}B_{i}^{\prime}=\zeta_{4}A_{i}^{2}\in K(E[2])$. Define $\sigma\in\mathrm{Gal}(K^{\prime}/K(E[2]))$ as the automorphism which acts by

[TABLE]

and let $\tau\in\mathrm{Gal}(K^{\prime}/K(E[2]))$ be the automorphism which acts by

[TABLE]

Note that $\sigma^{2}$ and $\tau^{2}$ both act trivially on $K(E[4])$ while sending $(B_{1},B_{2},B_{3})$ to $(B_{1},B_{2},-B_{3})$ and to $(-B_{1},B_{2},B_{3})$ respectively; it is now easy to check that $\sigma^{2}$ (resp. $\tau^{2}$) has order $2$ and commutes with $\tau$ (resp. $\sigma$). One also verifies that $[\sigma,\tau]$ acts trivially on $K(E[4])$ and sends $(B_{1},B_{2},B_{3})$ to $(-B_{1},-B_{2},-B_{3})$, and that this automorphism also commutes with both $\sigma$ and $\tau$. Thus, $\sigma$ and $\tau$ satisfy all of the relations given in (3). Moreover, $\sigma$ and $\tau$ each have order $4$, while $[\sigma,\tau]$ has order $2$. It is elementary to verify that this implies that $\langle\sigma,\tau\rangle$ has order $32$, which is the order of $\Gamma(2)^{\prime}/\Gamma(8)$; therefore $\langle\sigma,\tau\rangle\cong\Gamma(2)^{\prime}/\Gamma(8)$. Note also that $\langle\sigma,\tau\rangle$ fixes $A_{2}$, whose orbit under $\mathrm{Gal}(\bar{K}/K(E[2]))$ has cardinality $2$, so if $\mu$ is any automorphism in $\mathrm{Gal}(K^{\prime}/K(E[2]))$ which does not fix $A_{2}$, then $\langle\sigma,\tau,\mu\rangle$ has order $64$ and must be all of $\mathrm{Gal}(K^{\prime}/K(E[2]))$. Let $\mu$ be the automorphism that acts by changing the sign of all $A_{i}$’s and all $B_{i}$’s. Then $\mu$ commutes with $\sigma$ and $\tau$, and

[TABLE]

∎

The next two propositions (Propositions 10 and 12 below) imply Theorem 1.

∎

a) We first compute, for $i\in\mathbb{Z}/3\mathbb{Z}$ (and with $B_{i}^{\prime}$ defined as in the proof of Lemma 9), that

[TABLE]

Therefore, for each $i$, we have (up to sign changes)

[TABLE]

b) We similarly compute, for $i\in\mathbb{Z}/3\mathbb{Z}$, that

[TABLE]

[TABLE]

Therefore, for each $i$, we have $\pm\sqrt{A_{i}A_{i+2}}=\pm(1-\zeta_{4})^{-1}(B_{i}-B_{i}^{\prime})B_{i+2}/(A_{i}+\zeta_{4}A_{i+1})\in K(E[8])$.

Given a point $(x_{0},y_{0})\in E(\bar{K})$, in [1, §2], Bekker and Zarhin describe an algorithm to find a point $P\in E(\bar{K})$ with $2P=(x_{0},y_{0})$. In order to find such a point, one chooses elements $r_{1},r_{2},r_{3}\in\bar{K}$ with $r_{i}^{2}=x_{0}-\alpha_{i}$ for $i=1,2,3$ and with $r_{1}r_{2}r_{3}=-y_{0}$. Then

[TABLE]

satisfies $2P=(x_{0},y_{0})$. Following this algorithm, we get a point $P$ of order $4$ with $2P=(0,0)$ given by $P=(\zeta_{4}A_{2}A_{3},\zeta_{4}A_{2}A_{3}(A_{2}+\zeta_{4}A_{3}))$. Similarly, we get a point $Q$ of order $8$ with $2Q=P$ given by

[TABLE]

where $r_{1},r_{2},r_{3}\in\bar{K}$ are elements satisfying

[TABLE]

and $r_{1}r_{2}r_{3}=-\zeta_{4}A_{2}A_{3}(A_{2}+\zeta_{4}A_{3})$. Using the formulas computed in Remark 11, we see that one may choose

[TABLE]

with $r_{1}r_{2}r_{3}$ as specified above. It follows that the $x$-coordinate (resp. the $y$-coordinate) of $Q$ can be written as a quotient of homogeneous polynomial functions in the $A_{i}$’s, $B_{i}$’s, and $B_{i}^{\prime}$’s with coefficients in $K$ whose degree (the degree of the numerator minus the degree of the denominator) is $2$ (resp. $3$). Therefore, since $\sigma$ changes the sign of each $B_{i}^{\prime}$ as well as each of the $A_{i}$’s and $B_{i}$’s, we see that $\sigma$ fixes the $x$-coordinate of $Q$ while changing the sign of the $y$-coordinate, so $\sigma(Q)=-Q$. Thus, $\rho_{2}(\sigma)=-1\in\mathrm{SL}(T_{2}(E))$, and we have proven the statement of the proposition for the degree-$3$ case. In particular, we see from Proposition 6(c) that if we choose $\sigma\in G_{K}$ to be an automorphism whose image under the restriction map to $G_{K}^{\mathrm{unr}}\cong\widehat{P}_{3}$ is the central element $\Sigma\in\widehat{P}_{3}$, then $\sigma$ acts on $K(E[8])$ by changing the sign of each of the $A_{i}$’s and $B_{i}$’s.

In the degree-$4$ case, similarly let $\sigma\in\mathrm{Gal}(\bar{K}/K(E[2]))$ be an automorphism whose image under the restriction map to $\mathrm{Gal}(K^{\mathrm{unr}}/K(E[2]))\cong\widehat{P}_{3}$ is the central element $\Sigma\in\widehat{P}_{3}$ as defined in the statement of Lemma 5(a). Then it follows from what was remarked at the end of the last paragraph that $\sigma$ again acts on $K(E[8])$ by changing the sign of each of the $A_{i}$’s and $B_{i}$’s, so to prove the theorem for this case it suffices to show that $\rho_{2}(\sigma)=-1\in\mathrm{SL}(T_{2}(E))$. But this is given by Proposition 6(c).

∎

The author is grateful to Yuri Zarhin for his guidance in supervising the author’s investigation of $8$-torsion of elliptic curves during his time at The Pennsylvania State University. The author would also like to thank the referee for pointing out a gap in an earlier version of this work.