# Two torsion, the endomorphism field, and the finitude of endomorphism
algebras for a class of abelian varieties

Pip Goodman

###### Abstract

A conjecture attributed to Robert Coleman states there should only be finitely
many isomorphism classes for endomorphism algebras of abelian varieties of
fixed dimension over a fixed number field.

In this paper, we establish new cases of this conjecture, given conditions on
the 2-torsion field. In fact, for the abelian varieties considered, we provide
an explicit list of possible endomorphism algebras. This list is then refined
by studying the endomorphism field. In particular, we find conditions which
guarantee it is contained in the 2-torsion field.

##  1 Introduction

A conjecture attributed to Robert Coleman states111Here we have presented a
weaker version than usually stated. For the full-strength version see, for
example, [BFGR06, pg 384]. that for any positive integer $g\geq 1$ there
should only be finitely many isomorphism classes for $\operatorname{{\rm
End}}^{0}(A)$ among all abelian varieties $A/\mathbb{Q}$ of dimension $g$.
Here $\operatorname{{\rm End}}^{0}(A)=\operatorname{{\rm
End}}(A)\otimes\mathbb{Q}$ and $\operatorname{{\rm End}}(A)$ denotes the ring
of all endomorphisms of $A$ defined over a fixed algebraic closure of the
ground field.

This conjecture is currently only known to hold in full generality for $g=1$.
For $g=2$, recent work of Fité and Guitart [FG20] shows the conjecture holds
when $A$ is geometrically split. However the case of geometrically simple
abelian surfaces over $\mathbb{Q}$ remains completely open.

The theorem below presents new evidence towards this conjecture.

######  Theorem 1.1 ($=$ Corollary 2.6).

Let $A/\mathbb{Q}$ be an abelian variety of dimension $g\geq 1$ with $p=2g+1$
prime. Suppose $\operatorname{{\rm Gal}}(\mathbb{Q}(A[2])/\mathbb{Q})\cong
C_{p}$. Then either

  * •

$\operatorname{{\rm End}}^{0}(A)$ is a proper subfield of
$\mathbb{Q}(\zeta_{p})$; or

  * •

$p\equiv 3\mod{4}$, $g\geq 3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to
the power of an elliptic curve with complex multiplication by
$\mathbb{Q}(\sqrt{-p})$.

In particular there are only finitely many possibilities for
$\operatorname{{\rm End}}^{0}(A)$.

Let $K$ be a number field. In the proof of the above theorem we study the
minimal extension $L/K$ over which all endomorphisms of an abelian variety
$A/K$ are defined. Following Guralnick and Kedlaya [GK17], we call $L$ the
_endomorphism field_.

By exploiting the connection of $L/K$ to Sato-Tate groups, strong restrictions
on the degree $[L:K]$ (for $g$ fixed, $A$ and $K$ varying) have been found
[FKRS12], [GK17], [FKS21, top of page 3]. See also [Goo21, Theorem 2.5] for a
partial converse to the main theorem of [GK17].

The results we present on $L/K$ here are in the same vein as those of
Silverberg, who showed $L$ is contained in $K(A[n])$ for $n\geq 3$ [Sil92,
Theorem 2.4]. The elliptic curve $y^{2}=x^{3}-x$, which has CM by
$\mathbb{Z}[i]$, shows this result is not true in general for $n=2$. However,
we have the following:

######  Theorem 1.2 ($=$ Theorem 2.1).

Suppose $E=\operatorname{{\rm End}}^{0}(A)$ is a (finite) Galois extension of
$\mathbb{Q}$ and $L\nsubseteq K(A[2])$. The following hold:

  * •

$\operatorname{{\rm Gal}}(E/\mathbb{Q})$ has a non-trivial normal elementary
abelian 2-subgroup;

  * •

if $\operatorname{{\rm End}}(A)$ is 2-maximal in $E$, then $2$ is wildly
ramified in $E/\mathbb{Q}$.

In particular, if $E/\mathbb{Q}$ is Galois, $\operatorname{{\rm End}}(A)$ is a
2-maximal order and $2$ is not wildly ramified, then $L\subseteq K(A[2])$.

An order $\mathcal{O}\subseteq\operatorname{{\rm End}}^{0}(A)$ is said to be
_2-maximal_ if for any other order
$\mathcal{O}^{\prime}\subseteq\operatorname{{\rm End}}^{0}(A)$ satisfying
$\mathcal{O}\otimes\mathbb{Z}_{2}\subseteq\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}$
we have
$\mathcal{O}\otimes\mathbb{Z}_{2}=\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}$.

######  Example 1.3.

The condition that $\operatorname{{\rm End}}(A)$ is 2-maximal cannot be
removed. Indeed, the elliptic curve $y^{2}=(x+2)(x^{2}-2x-11)$ has CM by
$\mathbb{Z}[\sqrt{-3}]$ and its 2-torsion field is $\mathbb{Q}(\sqrt{3})$, see
[Sil94, Appendix A §3].

######  Remark 1.4.

By Proposition 2.2 of [Goo21] $\operatorname{{\rm End}}(A)$ is a 2-maximal
order and 2 is totally inert in $E$ when the absolute Galois group $G_{L}$
acts irreducibly on $A[2]$. For a more explicit criterion, see [Goo21, Theorem
2.9].

Theorem 1.2 along with Remark 1.4 facilitates finding the endomorphism field
in explicit examples. However, our main motivation comes from questions of the
form “How does $\operatorname{{\rm Gal}}(K(A[2])/K)$ influence
$\operatorname{{\rm End}}(A)=\operatorname{{\rm End}}_{L}(A)$?”. We note this
question may be made explicit for hyperelliptic jacobians. Indeed, if $J/K$ is
the jacobian of a hyperelliptic curve determined by $y^{2}=f(x)$, $f\in K[x]$,
then $\operatorname{{\rm Gal}}(f)\coloneqq\operatorname{{\rm
Gal}}(K(f)/K)\cong\operatorname{{\rm Gal}}(K(J[2])/K)$, see for example
[Zar01, Theorem 4.1]. Let us list a few further corollaries of Theorems 1.1
and 1.2 in this direction.

######  Corollary 1.5 ($=$ Corollary 2.7).

Let $C\colon y^{2}=f(x)$ be an elliptic curve defined over a number field with
a real embedding. If $\operatorname{{\rm Gal}}(f)\cong C_{3}$, then C does not
have complex multiplication.

We let $\operatorname{{\rm End}}_{F}(A)$ denote the endomorphisms of $A/K$
defined over a field $F/K$ and write $\operatorname{{\rm
End}}^{0}_{F}(A)=\operatorname{{\rm End}}_{F}(A)\otimes\mathbb{Q}$.

######  Corollary 1.6 ($=$ Corollary 2.8).

Let $A/\mathbb{Q}$ be an abelian surface. Suppose $\operatorname{{\rm
Gal}}(\mathbb{Q}(A[2])/\mathbb{Q})\cong C_{5}$. Then either
$\operatorname{{\rm End}}(A)=\mathbb{Z}$ or $\operatorname{{\rm
End}}_{\mathbb{Q}}^{0}(A)=\operatorname{{\rm
End}}^{0}(A)=\mathbb{Q}(\sqrt{5})$.

######  Example 1.7.

It is well known the first case occurs and examples are easily found. The
jacobian $J$ of the hyperelliptic curve
$y^{2}=x(x^{5}-4x^{4}+2x^{3}+5x^{2}-2x-1)$ has $\operatorname{{\rm
End}}_{\mathbb{Q}}(J)=\operatorname{{\rm
End}}(J)\cong\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ by [Wil00,
Proposition 1] and $x^{5}-4x^{4}+2x^{3}+5x^{2}-2x-1$ has Galois group
isomorphic to $C_{5}$. This shows the second case may also occur.

Combining Theorems 1.2, 2.12 with Theorems 1.5 and 2.9 of [Goo21] along with
the main theorem of [Zar00] (for the final statement). We obtain the
following:

######  Theorem 1.8.

Let $f(x)\in K[x]$ be a polynomial of degree 5.

  1. 1.

Suppose $\operatorname{{\rm Gal}}(f)\cong F_{5}$. The following hold:

    1. (i)

If $\operatorname{{\rm End}}^{0}(J)$ is a (real) quadratic field, then
$L=K(\sqrt{\Delta_{f}})$, the unique degree 2 extension of $K$ contained in
$K(f)$.

    2. (ii)

If $\operatorname{{\rm End}}^{0}(J)\cong E$ is a degree 4 CM field, then it is
cyclic and $L$ is the unique degree 4 extension of $K$ contained in $K(f)$.
Moreover, $L=EK$.

  2. 2.

Suppose $\operatorname{{\rm Gal}}(f)\cong D_{5}$. If $\operatorname{{\rm
End}}^{0}(J)$ is a degree 4 CM field, then $L$ is the unique degree 2
extension of $K$ contained in $K(f)$.

  3. 3.

Suppose $\operatorname{{\rm Gal}}(f)\cong C_{5}$. Then $L=K$. Furthermore, if
$\operatorname{{\rm End}}^{0}(J)$ is a degree 4 CM field, then $K$ contains a
real quadratic field with discriminant congruent to 5 modulo 8.

In particular, if $K=\mathbb{Q}$, $J$ has CM and $\operatorname{{\rm Gal}}(f)$
contains an element of order 5, then $\operatorname{{\rm Gal}}(f)\cong F_{5}$.

Motivated by a question of Ciaran Schembri and John Voight, linked to their
work on an analogue of Mazur’s Theorem for abelian surfaces defined over
$\mathbb{Q}$ with QM (see §3 for a definition), we prove analogues of Theorem
1.2 for certain quaternion algebras, see Theorems 3.3 and 3.5.

######  Acknowledgements.

The author would like to thank Ciaran Schembri and John Voight for their
question and communicating examples in Section 3. Likewise he expresses his
gratitude to Víctor Rotger for clarifying details on his paper with Luis
Dieulefait. He also thanks Tim Dokchitser, Jeremy Rickard and David Zywina for
reading an earlier draft of this article.

The author thanks the University of Bristol and the Université Clermont
Auvergne for their hospitality during the period this paper was written. The
author was funded by the EPSRC (grant number EP/N509619/1) and an early career
fellowship from the LMS.

##  2 Number fields

Silverberg has shown the endomorphism field $L/K$ of an abelian variety $A/K$
is contained in the $n$-torsion field for every $n\geq 3$ [Sil92, Theorem
2.4]. The case $n=4$ implies $\operatorname{{\rm Gal}}(L/L\cap K(A[2]))$ is an
elementary abelian 2-group, since $\operatorname{{\rm Gal}}(K(A[4])/K(A[2]))$
is too, for details see [Goo21, Proposition 3.9]. Below we provide conditions
on $\operatorname{{\rm End}}(A)$ which guarantee $L\subseteq K(A[2])$.

######  Theorem 2.1.

Suppose $E=\operatorname{{\rm End}}^{0}(A)$ is a (finite) Galois extension of
$\mathbb{Q}$ and $L\nsubseteq K(A[2])$. The following hold:

  * •

$\operatorname{{\rm Gal}}(E/\mathbb{Q})$ has a non-trivial normal elementary
abelian 2-subgroup;

  * •

if $\operatorname{{\rm End}}(A)$ is 2-maximal in $E$, then $2$ is wildly
ramified in $E/\mathbb{Q}$.

In particular, if $E/\mathbb{Q}$ is Galois, $\operatorname{{\rm End}}(A)$ is a
2-maximal order and $2$ is not wildly ramified, then $L\subseteq K(A[2])$.

###### Proof.

If $\operatorname{{\rm Gal}}(L/K)$ acts faithfully on $\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$, then we may recuperate Lemma 2.1 in
[Sil92] with $N=2$, $\Lambda=\operatorname{{\rm End}}(A)$ and $G$ the image of
$\operatorname{{\rm Gal}}(L/K)$ in $\operatorname{{\rm Aut}}(\Lambda)$.
Following through the proofs of Propositions 2.2, 2.3 and Theorem 2.4 in
[Sil92] shows $L\subseteq K(A[2])$. Hence we may suppose the action of
$\operatorname{{\rm Gal}}(L/K)$ on $\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$ has a non-trivial element $\gamma$ in
its kernel.

Since $\operatorname{{\rm Gal}}(L/K)$ acts on $\operatorname{{\rm End}}(A)$ by
$\mathbb{Z}$-linear automorphisms of $\operatorname{{\rm End}}(A)$, we may
view its image as a subgroup of $\operatorname{{\rm Gal}}(E/\mathbb{Q})$.
Identify $\gamma$ with its image in $\operatorname{{\rm Gal}}(E/\mathbb{Q})$.
Let $d$ be the rank of $\operatorname{{\rm End}}(A)$ as a free
$\mathbb{Z}$-module and view $\operatorname{{\rm Gal}}(E/\mathbb{Q})$ as a
(finite) subgroup of $\operatorname{{\rm GL}}_{d}(\mathbb{Z})$ via its action
on $\operatorname{{\rm End}}(A)$. The reduction map $\operatorname{{\rm
GL}}_{d}(\mathbb{Z})\rightarrow\operatorname{{\rm
GL}}_{d}(\mathbb{Z}/4\mathbb{Z})$ is injective on finite groups and the kernel
of $\operatorname{{\rm
GL}}_{d}(\mathbb{Z}/4\mathbb{Z})\rightarrow\operatorname{{\rm
GL}}_{d}(\mathbb{Z}/2\mathbb{Z})$ is an elementary abelian $2$-group (for more
details see [Goo21, Prop. 3.9]). Thus the kernel of the map
$\operatorname{{\rm Gal}}(E/\mathbb{Q})\rightarrow\operatorname{{\rm
Aut}}(\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z})\cong\operatorname{{\rm
GL}}_{d}(\mathbb{Z}/2\mathbb{Z})$ is an elementary abelian $2$-group which is
non-trivial since it contains $\gamma$.

Assume now $\operatorname{{\rm End}}(A)$ is a 2-maximal order in $E$. As
$\gamma$ acts trivially on the order $\operatorname{{\rm End}}(A)$ modulo $2$,
it also acts trivially on $\operatorname{{\rm End}}(A)$ modulo every prime
above $2$ in $E$. Since $\operatorname{{\rm End}}(A)$ is 2-maximal, we find
$\gamma$ belongs to the inertia group $I_{\mathfrak{p}}$ for every prime
$\mathfrak{p}$ above 2 in $E$. By the above $\gamma$ has order two, and hence
$2$ is wildly ramified in $E/\mathbb{Q}$. ∎

######  Remark 2.2.

The condition that $\operatorname{{\rm End}}(A)$ is 2-maximal cannot be
removed. Indeed, the elliptic curve $y^{2}=x^{3}-15x-22$ has CM by
$\mathbb{Z}[\sqrt{-3}]$ and its 2-torsion field is $\mathbb{Q}(\sqrt{3})$, see
[Sil94, Appendix A §3].

We now prove finitude results for endomorphism algebras of certain abelian
varieties and apply the above theorem to restrict the list of possible
endomorphism algebras.

######  Proposition 2.3.

Let $A/K$ be an abelian variety of dimension $g\geq 1$ with $p=2g+1$ prime.
Suppose $\mathfrak{q}$ is a prime of bad reduction for $A$ and there is an
element of order $p$ in the image of the inertia group $I_{\mathfrak{q}}$ on
$A[\ell]$ for some $\mathfrak{q}\nmid\ell$. Then either

  * •

$p$ does not divide $[L:K]$ and $\operatorname{{\rm End}}^{0}(A)$ is a
subfield of $\mathbb{Q}(\zeta_{p})$; or

  * •

$p$ divides $[L:K]$, $g\geq 3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to
the power of an absolutely simple abelian variety with complex multiplication
by a proper subfield of $\mathbb{Q}(\zeta_{p})$.

In particular there are only finitely many possibilities for
$\operatorname{{\rm End}}^{0}(A)$.

###### Proof.

Suppose $p$ divides $[L:\mathbb{Q}]$. Then by Theorem 2.5 of [Goo21], $g\geq
3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to the power of an absolutely
simple abelian variety with complex multiplication by a proper subfield $F$ of
$\mathbb{Q}(\zeta_{p})$. Hence we may suppose $p$ does not divide $[L:K]$.

The criterion of Néron-Ogg-Shafarevich implies that the image of the inertia
group $I_{\mathfrak{q}}$ on the Tate module $T_{\ell}(A)$ for any prime
$\ell$, not divisible by $\mathfrak{q}$, contains an element $\tau$ of order
$p$. Furthermore, as the trace of $\tau$ is an integer [ST68, Thm. 2] its
eigenvalues are the primitive $p$-th roots of unity.

By Dirichlet’s theorem on arithmetic progressions, we can find a prime $\ell$,
not divisible by $\mathfrak{q}$, which is a primitive root modulo $p$. The
reduction of $\tau$ modulo $\ell$ lands in $\operatorname{{\rm
Gal}}(K(A[\ell])/K)$ and has order $p$. This allows us to apply [Goo21,
Theorem 2.9] and deduce $E\coloneqq\operatorname{{\rm End}}^{0}(A)$ is a
field.

As $p$ does not divide $[L:K]$, the element $\tau$ lies in the image of
$G_{L}$. The decomposition
$E\otimes\mathbb{Q}_{\ell}=\prod_{\lambda|\ell}E_{\lambda}$ induces a
decomposition $T_{\ell}(A)=\prod_{\lambda|\ell}T_{\lambda}(A)$ giving
representations $G_{L}\rightarrow\operatorname{{\rm GL}}_{n}(E_{\lambda})$
where $n=\frac{2g}{[E:\mathbb{Q}]}$, see §2 of [Rib76] for further details.
Let $\tau_{\lambda}\in\operatorname{{\rm GL}}_{n}(E_{\lambda})$ be the
projection of $\tau$ onto $T_{\lambda}(A)$. By the above, the eigenvalues of
$\tau_{\lambda}$ are distinct primitive $p$-th roots of unity. Taking the
trace of $\tau_{\lambda}$ we deduce $E$ contains a subfield of
$\mathbb{Q}(\zeta_{p})$ of degree $[E:\mathbb{Q}]$. In other words,
$E\subseteq\mathbb{Q}(\zeta_{p})$. ∎

By finding a condition on the ray class groups of primes above $2$ in $K$, we
deduce an explicit version of the above:

######  Theorem 2.4.

Let $A/K$ be an abelian variety of dimension $g\geq 1$ with $p=2g+1$ prime.
Suppose $\operatorname{{\rm Gal}}(K(A[2])/K)\cong C_{p}$ and $p$ divides
neither the class number of $K$, nor the multiplicative order of the residue
field of any prime above 2. Then either

  * •

$p$ does not divide $[L:K]$ and $\operatorname{{\rm End}}^{0}(A)$ is a
subfield of $\mathbb{Q}(\zeta_{p})$; or

  * •

$p$ divides $[L:K]$, $g\geq 3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to
the power of an absolutely simple abelian variety with complex multiplication
by a proper subfield of $\mathbb{Q}(\zeta_{p})$.

In particular there are only finitely many possibilities for
$\operatorname{{\rm End}}^{0}(A)$.

###### Proof.

Let $F/K$ be an odd degree extension unramified outside of 2. Then by class
field theory, any prime dividing $[F\colon K]$ divides either the class number
of $K$, or the multiplicative order of the residue field of a prime above 2 in
$K$.

It follows that $K(A[2])/K$ is ramified at some prime $\mathfrak{q}\nmid 2$.
This allows us to apply the above proposition and conclude. ∎

The following lemma is well-known.

######  Lemma 2.5.

Let $A/K$ be an absolutely simple abelian variety with CM by a Galois
extension $E/\mathbb{Q}$. Then $L=E^{*}K$ and $E\supseteq E^{*}$, where
$E^{*}$ is the reflex field of $E$. Furthermore if $E/\mathbb{Q}$ is abelian,
then $E=E^{*}$.

###### Proof.

As $E/\mathbb{Q}$ is Galois, the reflex field $E^{*}$ is a subfield of $E$
[Shi98, Prop. 28, pg 62]. Moreover, since $A$ is absolutely simple its CM type
is primitive. Proposition 30 on page 65 of [Shi98] applies to tell us the
endomorphism field $L$ equals $E^{*}K$. Finally, if $E/\mathbb{Q}$ is an
abelian extension, then $E=E^{*}$ by Example (1) on page 63 of [Shi98]. ∎

######  Corollary 2.6.

Let $A/\mathbb{Q}$ be an abelian variety of dimension $g\geq 1$ with $p=2g+1$
prime. Suppose $\operatorname{{\rm Gal}}(\mathbb{Q}(A[2])/\mathbb{Q})\cong
C_{p}$. Then either

  * •

$\operatorname{{\rm End}}^{0}(A)$ is a proper subfield of
$\mathbb{Q}(\zeta_{p})$; or

  * •

$p\equiv 3\mod{4}$, $g\geq 3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to
the power of an elliptic curve with complex multiplication by
$\mathbb{Q}(\sqrt{-p})$.

In particular there are only finitely many possibilities for
$\operatorname{{\rm End}}^{0}(A)$.

###### Proof.

Clearly, we may apply Theorem 2.4. Suppose first $p$ divides $[L:\mathbb{Q}]$.
Then $g\geq 3$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to the power of an
absolutely simple abelian variety $B$ with complex multiplication by a proper
subfield $F$ of $\mathbb{Q}(\zeta_{p})$. The above lemma applies to $B$ and
shows $L$, the endomorphism field of $A$, contains $F$.

Proposition 3.9 of [Goo21] (recalled at the beginning of this section) shows
$\operatorname{{\rm Gal}}(L/\mathbb{Q})$ has order $2^{m}p$ for some $m$.
Moreover, as the quotient $\operatorname{{\rm
Gal}}(\mathbb{Q}(A[2])/\mathbb{Q})\cong\operatorname{{\rm
Gal}}(L/\mathbb{Q})/C_{2}^{m}$ has odd order $p$, any element of even order in
$\operatorname{{\rm Gal}}(L/\mathbb{Q})$ has either order two or $2p$.

Since $p$ does not divide $[F:\mathbb{Q}]$, the cyclic quotient
$\operatorname{{\rm Gal}}(F/\mathbb{Q})$ of $\operatorname{{\rm
Gal}}(L/\mathbb{Q})$ has order two (it is not trivial since $F$ is a CM
field). Using $F$ is a CM field only ramified at $p$, we find
$F=\mathbb{Q}(\sqrt{-p})$ where $p\equiv 3\mod{4}$.

We now suppose $p$ does not divide $[L:\mathbb{Q}]$ and show
$E\neq\mathbb{Q}(\zeta_{p})$. Suppose we had equality, then applying Lemma 2.5
we find $L=\mathbb{Q}(\zeta_{p})$. If $p=3$, then we apply Theorem 2.9 of
[Goo21] and Theorem 2.1 to obtain a contradiction. Else, $p\geq 5$ and
Proposition 3.9 of [Goo21] provides us with a contradiction, since
$\operatorname{{\rm Gal}}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ is not an
elementary abelian $2$-group. ∎

Let us record some easy corollaries of Theorem 2.9 of [Goo21], Theorems 2.1,
2.6 and Lemma 2.5.

######  Corollary 2.7.

Let $A\colon y^{2}=f(x)$ be an elliptic curve defined over a number field with
a real embedding. If $\operatorname{{\rm Gal}}(f)\cong C_{3}$, then $A$ does
not have complex multiplication.

######  Corollary 2.8.

Let $A/\mathbb{Q}$ be an abelian surface. Suppose $\operatorname{{\rm
Gal}}(\mathbb{Q}(A[2])/\mathbb{Q})\cong C_{5}$. Then either
$\operatorname{{\rm End}}(A)=\mathbb{Z}$ or $\operatorname{{\rm
End}}_{\mathbb{Q}}^{0}(A)=\operatorname{{\rm
End}}^{0}(A)=\mathbb{Q}(\sqrt{5})$.

######  Remark 2.9.

The above result is the best possible. Indeed, it is well known the first case
occurs and examples are easily found. The jacobian $J$ of the hyperelliptic
curve $y^{2}=x(x^{5}-4x^{4}+2x^{3}+5x^{2}-2x-1)$ has $\operatorname{{\rm
End}}_{\mathbb{Q}}(J)=\operatorname{{\rm
End}}(J)\cong\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ by [Wil00,
Proposition 1] and $x^{5}-4x^{4}+2x^{3}+5x^{2}-2x-1$ has Galois group
isomorphic to $C_{5}$. This provides an example for the second case.

We present a variant of Theorem 2.4 for abelian varieties over imaginary
quadratic fields.

######  Theorem 2.10.

Let $g\geq 2$ be an integer and suppose $p=2g+1$ is prime. Let $K$ be an
imaginary quadratic field of class number coprime to $p$. Let $A/K$ be an
abelian variety of dimension $g$. Suppose $\operatorname{{\rm
Gal}}(K(A[2])/K)\cong C_{p}$. Then either

  * •

$\operatorname{{\rm End}}^{0}(A)$ is a proper subfield of
$\mathbb{Q}(\zeta_{p})$; or

  * •

$p\equiv 3\mod{4}$ and $A$ is isogenous over $\bar{\mathbb{Q}}$ to the power
of an elliptic curve with complex multiplication by $\mathbb{Q}(\sqrt{-p})$.

In particular there are only finitely many possibilities for
$\operatorname{{\rm End}}^{0}(A)$.

###### Proof.

Clearly, we may apply Theorem 2.4. Suppose first $p$ divides $[L:K]$. Then $A$
is isogenous over $\bar{K}$ to the power of an absolutely simple abelian
variety $B$ with complex multiplication by a proper subfield $F$ of
$\mathbb{Q}(\zeta_{p})$. The above lemma applies to $B$ and shows $L$, the
endomorphism field of $A$, contains $F$.

Proposition 3.9 of [Goo21] (recalled at the beginning of this section) shows
$\operatorname{{\rm Gal}}(L/K)$ has order $2^{m}p$ for some $m$. Moreover, as
the quotient $\operatorname{{\rm Gal}}(K(A[2])/K)\cong\operatorname{{\rm
Gal}}(L/K)/C_{2}^{m}$ has odd order $p$, any element of even order in
$\operatorname{{\rm Gal}}(L/K)$ has either order two or $2p$. Since $p$ does
not divide $[F:\mathbb{Q}]$, the cyclic quotient $\operatorname{{\rm
Gal}}(FK/K)$ of $\operatorname{{\rm Gal}}(L/K)$ has order dividing two.

In turn we deduce $[F:\mathbb{Q}]$ divides $4$ and is equal to 4 only if
$F\supsetneq K$. Being a CM field, $[F:\mathbb{Q}]=4$ would imply $F$ also
contains a real quadratic field, whence $\operatorname{{\rm
Gal}}(F/\mathbb{Q})\cong C_{2}\times C_{2}$ giving a contradiction. Thus
$[F:\mathbb{Q}]=2$. Using again that $F$ is a CM field only ramified at $p$,
we deduce $F=\mathbb{Q}(\sqrt{-p})$ where $p\equiv 3\mod{4}$.

We now suppose $p$ does not divide $[L:K]$ and show
$E\neq\mathbb{Q}(\zeta_{p})$. Suppose we had equality, then by Lemma 2.5,
$L=K(\zeta_{p})$. Proposition 3.9 of [Goo21] then implies $\operatorname{{\rm
Gal}}(L/K)=\operatorname{{\rm Gal}}(K(\zeta_{p})/K)\cong\operatorname{{\rm
Gal}}(\mathbb{Q}(\zeta_{p})/\mathbb{Q}(\zeta_{p})\cap K)$ has order at most
$2$. Hence $[\mathbb{Q}(\zeta_{p}):\mathbb{Q}]$ divides $4$, being equal to 4
only if $K$ is contained in $\mathbb{Q}(\zeta_{p})$. But
$[\mathbb{Q}(\zeta_{p}):\mathbb{Q}]=4$ implies $p=5$ and
$\mathbb{Q}(\zeta_{5})$ does not have an imaginary subfield. Whence
$[\mathbb{Q}(\zeta_{p}):\mathbb{Q}]=2$ and $p=3$. This in turn implies $g=1$
which we have ruled out by assumption. ∎

######  Remark 2.11.

The condition on the class number of $K$ cannot be removed. Indeed, the
polynomial $f(x)=x^{5}-19x^{4}+107x^{3}+95x^{2}+88x-16$ has Galois group
$D_{5}$ and its splitting field is the Hilbert class field of
$K\coloneqq\mathbb{Q}(\sqrt{-131})$. The jacobian $J_{f}$ of the hyperelliptic
curve defined by $y^{2}=f(x)$, has endomorphism algebra isomorphic to
$\mathbb{Q}(\sqrt{13})$ and the group $\operatorname{{\rm
Gal}}(K(J_{f}[2])/K)$ is isomorphic to $C_{5}$.

However, notice $13\equiv 5\mod{8}$ as predicted by [Goo21, Theorem 2.10].
Also note $J_{f}$ could not have had CM by combining Theorem 2.1 and Theorem
2.10 of [Goo21].

Before proving Theorem 1.8, we give the following extension of [Goo21, Theorem
3.5].

######  Theorem 2.12.

Let $f\in K[x]$ be a polynomial of odd degree $n$ with Galois group isomorphic
to a Frobenius group $G$ of order $n(n-1)$. Let $J_{f}$ be the jacobian
associated to the hyperelliptic curve defined by $y^{2}=f(x)$.

Suppose $E={\rm End}^{0}(J_{f})$ is a number field of dimension $s$ over
$\mathbb{Q}$. Then $E/\mathbb{Q}$ is Galois with ${\rm Gal}(E/\mathbb{Q})$
isomorphic to a quotient of $H$ the Frobenius complement of $G$.

Furthermore, $L/K$ is an extension of degree $s$ contained in $K(f)$, and as
abstract groups, ${\rm Gal}(L/K)\cong{\rm Gal}(E/\mathbb{Q})$. Moreover, if
$s=n-1$, then $L=EK$.

Finally, if $\operatorname{{\rm End}}(J_{f})$ is $2$-maximal, then $E$ is
unramified at 2.

###### Proof.

This result is [Goo21, Theorem 3.5] with the extra assertion that if $s=n-1$,
then $L=EK$. If $s=n-1$, then $J_{f}$ is an absolutely simple abelian variety
with CM by $E$. We may therefore apply Lemma 2.5 to find $L=E^{*}K$ and
$E^{*}\subseteq E$. Hence $[E^{*}:\mathbb{Q}]\geq[L:K]=s=[E:\mathbb{Q}]$. It
follows we have equality $E^{*}=E$. ∎

######  Example 2.13.

For $f(x)=x^{5}-2$, it is well known the endomorphism algebra of $J_{f}$ is
$\mathbb{Q}(\zeta_{5})$ and $\mathbb{Q}(\zeta_{5})$ is the unique degree 4
extension contained in the splitting field $\mathbb{Q}(f)$.

A more interesting example is given by the genus 2 curve with LMFDB label
28561.a.371293.1 first found in [BSS+16] (though we use an odd degree model
computed using Magma [BCP97]). Here
$f(x)=52x^{5}+104x^{4}+104x^{3}+52x^{2}+12x+1$, $\operatorname{{\rm
Gal}}(f)\cong F_{5}$ and $J_{f}$ has CM by the number field $K$ defined by
$x^{4}-x^{3}+2x^{2}+4x+3$. The unique degree 4 extension contained in
$\mathbb{Q}(f)$ is given by $K$.

This field is totally inert at 2 (as predicted by [Goo21, Theorem 2.9]) and
unramified outside 13. We note its class number is one, but in line with the
theorems presented above, $\mathbb{Q}(f)/K$ is a degree 5 extension ramified
only at 2.

Theorem 1.8 is a direct consequence of Theorems 2.1, 2.12 and Theorems 1.5,
2.9 of [Goo21] with the final statement then following from the main theorem
of [Zar00].

##  3 Quaternion algebras

Let $B=\mathbb{Q}+i\mathbb{Q}+j\mathbb{Q}+k\mathbb{Q}$ be an indefinite
quaternion algebra generated by $i,j$ satisfying $i^{2}=D/m$, $j^{2}=m$ and
$k=ij$, where $D$ is a positive squarefree integer and $m|D$. We note
$0>-m=\mathrm{n}(j)$, the (reduced) norm of $j$.

An order of $B$ is said to be hereditary if it has squarefree (reduced)
discriminant (for alternate, equivalent, definitions see [DR04, Voi21]). We
shall take $\mathcal{O}$ to be a (hereditary) order in $B$ of discriminant
$D$.

An abelian surface $A/K$ is said to have quaternion multiplication by
$\mathcal{O}$ if there is an isomorphism
$\iota\colon\mathcal{O}\xrightarrow{\sim}\operatorname{{\rm End}}(A)$. Any
such abelian surface admits a principal polarisation over $\bar{K}$ [Voi21,
43.6.6. pg 818]. Equivalently, there is an element $\mu\in\mathcal{O}$
satisfying $\mu^{2}+D=0$. Fix such an $A$, $\mu$ and $\iota$.

Following [DR04, Definition 3.3], we say $\chi\in B$ is a _twist_ of
$(\mathcal{O},\mu)$ if it lies in both $\mathcal{O}$ and the normaliser
$N_{B^{*}}(\mathcal{O})$, has trace zero, $\mathrm{n}(\chi)$ divides $D$, and
$\mu\chi=-\chi\mu$. The existence of such an element can be verified by a
finite computation, we refer the reader to the remark at the bottom of page 9
of [DR04] for further details.

Owing in part to the fact $(\mathcal{O},\mu)$ has been fixed, there are only
two possible values222This follows from [Rot04, Lemmas 3.5 and 3.7]. Indeed,
in the notation of the paper, we have $F=\mathbb{Q}$ and $D\neq 3$, as $D$ is
divisible by an even number of primes. This forces $\omega_{odd}=1$. The cited
lemmas in turn show $C_{2}\cong U_{0}\leq V_{0}\cong C_{2}\times C_{2}$. The
non-trivial element of $U_{0}$ can be represented by $\mu=\sqrt{-D}$. This
allows us to write $V_{0}=\langle[\mu],[\chi]\rangle$ where the representative
$\chi$ may be taken to have reduced norm $m|D$. Moreover, any representative
of $[\chi]$ has reduced norm $m$ up to a rational square. Likewise,
representatives of the class $[\mu\chi]$ have reduced norm $D/m$ up to
rational squares.  $\mathrm{n}(\chi)$ can take and the product of these values
is equal to $D$.

We call $(\mathcal{O},\mu,\chi)$ a _twisted principally polarised order_ , and
say $(\mathcal{O},\mu,\chi)$ is of discriminant $D$ and norm
$-\mathrm{n}(\chi)$.

As before, we let $L/K$ denote the endomorphism field of $A$. Dieulefait and
Rotger [DR04, Theorem 1.3] showed $\operatorname{{\rm Gal}}(L/K)$ is
isomorphic to one of the trivial group, $C_{2}$, or $C_{2}\times C_{2}$.
Moreover, they proved in each case $\operatorname{{\rm End}}_{K}^{0}(A)$ is
respectively isomorphic to $B$; one of
$\mathbb{Q}(\mu),\mathbb{Q}(\chi),\mathbb{Q}(\mu\chi)$; or $\mathbb{Q}$.

This determines (and heavily restricts) the possible images of

$\operatorname{{\rm Gal}}(L/K)\rightarrow\operatorname{{\rm
Aut}}(\operatorname{{\rm End}}(A))\cong N_{B^{*}}(\mathcal{O})/\mathbb{Q}^{*}$

arising from the natural action of $\operatorname{{\rm Gal}}(L/K)$ on
$\operatorname{{\rm End}}(A)$.

To gain information on the intersection $L\cap K(A[2])$, we will use a
description of orders of discriminant $D$ in $B$.

######  Lemma 3.1.

Suppose $2|D$ and $m\equiv 3\mod{4}$. Then

$\mathcal{O}=\mathbb{Z}+\frac{1}{2}(1+j+k)\mathbb{Z}+\frac{1}{2}(1+j-k)\mathbb{Z}+\frac{1}{2}(i+k)\mathbb{Z}$

is an order of discriminant $D$ in $B$. Moreover, any order
$\mathcal{O}^{\prime}$ of discriminant $D$ which contains
$\mathbb{Z}[1,mi,j,k]$, satisfies
$\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}=\mathcal{O}\otimes\mathbb{Z}_{2}$.

###### Proof.

It is a routine calculation to show $\mathcal{O}$ has discriminant $D$ (for an
example, see [Vig80, pg. 85 - 86]).

We now show the final statement. As $m$ is odd we have
$\mathbb{Z}[1,mi,j,k]\otimes\mathbb{Z}_{2}=\mathbb{Z}[1,i,j,k]\otimes\mathbb{Z}_{2}$.
Thus we may assume $\mathcal{O}^{\prime}$ contains $\mathbb{Z}[1,i,j,k]$. The
discriminant of $\mathbb{Z}[1,i,j,k]$ equals $4D$ and thus any element
contained in $\mathcal{O}^{\prime}$ but not in $\mathbb{Z}[1,i,j,k]$ is of the
form $\frac{1}{4}\alpha$ with $\alpha\in\mathbb{Z}[1,i,j,k]$.

By considering the (monic) characteristic polynomial of $\frac{1}{4}\alpha$,
we find $\alpha\in 2\mathbb{Z}[1,i,j,k]$. Hence it suffices to check elements
of the form $\frac{1}{2}(a+bi+cj+dk)$ with $a,b,c,d\in\\{0,1\\}$. One finds
$\frac{1}{2}(1+j+k)$, $\frac{1}{2}(1+i+j)$, $\frac{1}{2}(i+k)$ are the only
integral such and that they must all belong to $\mathcal{O}^{\prime}$. ∎

######  Lemma 3.2.

Suppose $m\equiv 1\mod{4}$. Then

$\mathcal{O}=\mathbb{Z}+\frac{1}{2}(1+j)\mathbb{Z}+k\mathbb{Z}+\frac{1}{2}(i+k)\mathbb{Z}$

is an order of discriminant $D$ in $B$. Furthermore, if $D\equiv 1\mod{4}$
then

$\mathcal{O}_{1}=\mathbb{Z}+\frac{1}{2}(1+i)\mathbb{Z}+j\mathbb{Z}+\frac{1}{2}(j+k)\mathbb{Z}$

is an order of discriminant $D$ in $B$. Likewise, if $D\equiv 3\mod{4}$ then

$\mathcal{O}_{3}=\mathbb{Z}+\frac{1}{2}(1+k)\mathbb{Z}+j\mathbb{Z}+\frac{1}{2}(i+j)\mathbb{Z}$

is an order of discriminant $D$ in $B$.

Moreover, any order $\mathcal{O}^{\prime}$ of even discriminant $D$ which
contains $\mathbb{Z}[1,mi,j,k]$ satisfies
$\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}=\mathcal{O}\otimes\mathbb{Z}_{2}$.
Any order $\mathcal{O}^{\prime}$ of discriminant $D\equiv t\mod{4}$ with
$t\in\\{1,3\\}$, which contains $\mathbb{Z}[1,mi,j,k]$, satisfies either
$\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}=\mathcal{O}\otimes\mathbb{Z}_{2}$
or
$\mathcal{O}^{\prime}\otimes\mathbb{Z}_{2}=\mathcal{O}_{t}\otimes\mathbb{Z}_{2}$.

###### Proof.

The proof follows the same strategy as for Lemma 3.1. ∎

######  Theorem 3.3.

Let $A/K$ be an abelian surface with QM by a twisted principally polarised
order $(\mathcal{O},\mu,\chi)$ of discriminant $D$ and norm $m$, where
$\mathcal{O}\subseteq B$.

If $2|D$ and $m\equiv 3\mod{4}$, then $L\subseteq K(A[2])$.

###### Proof.

As $\operatorname{{\rm End}}(A)$ is a hereditary order, the results of [DR04,
Theorem 1.3] apply to show a non-trivial element of $\operatorname{{\rm
Gal}}(L/K)$ acts on $\operatorname{{\rm End}}(A)$ (possibly after scaling) by
conjugation as one of $\mu,\chi$ or $\mu\chi$.

We look to determine the action of these elements on $\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$. To do so we reduce a
$\mathbb{Z}$-basis of $\operatorname{{\rm End}}(A)$ modulo $2$, thus we may
work with $\operatorname{{\rm End}}(A)\otimes\mathbb{Z}_{2}$ in place of
$\operatorname{{\rm End}}(A)$. By considering the algebraic relations they
satisfy, we may assume $\mu=k$, $\chi=j$ and $\mu\chi=mi$. Hence Lemma 3.1
allows us to take $\operatorname{{\rm
End}}(A)=\mathbb{Z}+\frac{1}{2}(1+j+k)\mathbb{Z}+\frac{1}{2}(1+j-k)\mathbb{Z}+\frac{1}{2}(i+k)\mathbb{Z}$.

Let $X=\frac{1}{2}(1+j+k)$, $Y=\frac{1}{2}(1+j-k)$ and $Z=\frac{1}{2}(i+k)$.
Let us examine the action of $i,j$ and $k$ on the basis of $\operatorname{{\rm
End}}(A)$ given by $1,X,Y,Z$. Each of $i,j$ and $k$ fix $1$. For $i$ we have
$iXi^{-1}=1-X$, $iYi^{-1}=1-Y$ and $iZi^{-1}=Z+Y-X$. For $j$ we have
$jXj^{-1}=Y$, $jYj^{-1}=X$, $jZj^{-1}=-Z$. Looking at the coefficients, we see
the action remains faithful on $\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$. By proceeding as in the proof of
Theorem 2.1 we are now done. ∎

######  Example 3.4.

The hereditary assumption is necessary. The following example shows not only
$L$ need not be contained in $\mathbb{Q}(A[2])$ for a non-hereditary order,
but also the result of Dieulefait and Rotger fails.

Let $J$ be the jacobian defined by the hyperelliptic curve associated to
$y^{2}+y=6x^{5}+9x^{4}−x^{3}−3x^{2}$ with LMFDB label 20736.l.373248.1. This
surface has QM by an order of (reduced) discriminant $6^{2}$ in
$\left(\frac{2,3}{\mathbb{Q}}\right)$. In particular, the order is not
hereditary. The endomorphism field has defining polynomial
$x^{8}+4x^{6}+10x^{4}+24x^{2}+36$ and the two torsion field of $J$ is
$\mathbb{Q}(\sqrt{2},\sqrt{3})$.

This curve, made easily available on the LMFDB [LMF], was first found in
[BSS+16]. Computations linked to its endomorphism algebra were carried out in
Magma using code from [CMSV19].

For ease of notation let $F=K(A[2])$.

######  Theorem 3.5.

Let $A/K$ be an abelian surface with QM by a twisted principally polarised
order $(\mathcal{O},\mu,\chi)$ of discriminant $D$ and norm $m$, where
$\mathcal{O}\subseteq B$.

If $m\equiv 1\mod{4}$, then $\operatorname{{\rm End}}^{0}_{F}(A)$ contains at
least one of $\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{-D})$ and
$\mathbb{Q}(\sqrt{D/m})$. Moreover, if $D$ is even, then $\operatorname{{\rm
End}}^{0}_{F}(A)$ contains at least one of $\mathbb{Q}(\sqrt{-D})$ and
$\mathbb{Q}(\sqrt{D/m})$.

###### Proof.

As $\operatorname{{\rm End}}(A)$ is a hereditary order, the results of [DR04,
Theorem 1.3] apply to show a non-trivial element of $\operatorname{{\rm
Gal}}(L/K)$ acts on $\operatorname{{\rm End}}(A)$ (possibly after scaling) by
conjugation as one of $\mu,\chi$ or $\mu\chi$.

We look to determine the action of these elements on $\operatorname{{\rm
End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$. To do so we reduce a
$\mathbb{Z}$-basis of $\operatorname{{\rm End}}(A)$ modulo $2$, thus we may
work with $\operatorname{{\rm End}}(A)\otimes\mathbb{Z}_{2}$ in place of
$\operatorname{{\rm End}}(A)$. By considering the algebraic relations they
satisfy, we may assume $\mu=k$, $\chi=j$ and $\mu\chi=mi$. Lemma 3.2 allows us
to take $\operatorname{{\rm End}}(A)$ equal to one of the three given orders.
As these orders differ by permuting $i,j$ and $k$, we only give details for
the case $\operatorname{{\rm
End}}(A)=\mathbb{Z}+\frac{1}{2}(1+j)\mathbb{Z}+k\mathbb{Z}+\frac{1}{2}(i+k)\mathbb{Z}$.

Let $X=\frac{1}{2}(1+j)$, $Y=k$ and $Z=\frac{1}{2}(i+k)$. Let us examine the
action of $i,j$ and $k$ on the basis of $\operatorname{{\rm End}}(A)$ given by
$1,X,Y,Z$. Each of $i,j$ and $k$ fix $1$. For $i$ we have $iXi^{-1}=1-X$,
$iYi^{-1}=-Y$ and $iZi^{-1}=Z-Y$. For $j$ we have $jXj^{-1}=X$, $jYj^{-1}=-Y$,
$jZj^{-1}=-Z$. Looking at the coefficients, we see $j$ acts trivially on
$\operatorname{{\rm End}}(A)\otimes\mathbb{Z}/2\mathbb{Z}$ whereas $i$ and $k$
act by the same involution. Proceeding as in the proof of Theorem 2.1, we find
at least one of $\mathbb{Q}(i)\cong\mathbb{Q}(\sqrt{D/m})$ and
$\mathbb{Q}(k)\cong\mathbb{Q}(\sqrt{-D})$ is contained in $\operatorname{{\rm
End}}_{F}^{0}(A)$. ∎

######  Example 3.6.

The jacobian $J$ of the hyperelliptic curve
$y^{2}=-2x^{6}-12x^{5}-21x^{4}-10x^{3}-3x^{2}+6x+1$ has QM by the maximal
order of $B=\left(\frac{3,5}{\mathbb{Q}}\right)$ which has discriminant 15.
The endomorphism field of $J$ is $L=\mathbb{Q}(\sqrt{3},\sqrt{-3})$, we have
$L\cap\mathbb{Q}(A[2])=\mathbb{Q}(\sqrt{-3})$ and $\operatorname{{\rm
End}}^{0}_{\mathbb{Q}(\sqrt{-3})}(J)=\mathbb{Q}(\sqrt{-15})$. These
calculations were performed in Magma using code from [CMSV19]. The curve comes
from [LY20].

## References

  * [BCP97] Wieb Bosma, John Cannon, and Catherine Playoust.  The Magma algebra system. I. The user language.  J. Symbolic Comput., 24(3-4):235–265, 1997.  Computational algebra and number theory (London, 1993).
  * [BFGR06] Nils Bruin, E. Victor Flynn, Josep González, and Víctor Rotger.  On finiteness conjectures for endomorphism algebras of abelian surfaces.  Math. Proc. Cambridge Philos. Soc., 141(3):383–408, 2006.
  * [BSS+16] Andrew R. Booker, Jeroen Sijsling, Andrew V. Sutherland, John Voight, and Dan Yasaki.  A database of genus-2 curves over the rational numbers.  LMS J. Comput. Math., 19(suppl. A):235–254, 2016.
  * [CMSV19] Edgar Costa, Nicolas Mascot, Jeroen Sijsling, and John Voight.  Rigorous computation of the endomorphism ring of a Jacobian.  Math. Comp., 88(317):1303–1339, 2019.
  * [DR04] Luis V. Dieulefait and Víctor Rotger.  The arithmetic of QM-abelian surfaces through their Galois representations.  J. Algebra, 281(1):124–143, 2004.  For the current version see https://web.mat.upc.edu/victor.rotger/docs/ArithmeticQM.pdf.
  * [FG20] Francesc Fité and Xavier Guitart.  Endomorphism algebras of geometrically split abelian surfaces over $\mathbb{Q}$.  Algebra Number Theory, 14(6):1399–1421, 2020.
  * [FKRS12] Francesc Fité, Kiran S. Kedlaya, Víctor Rotger, and Andrew V. Sutherland.  Sato-Tate distributions and Galois endomorphism modules in genus 2\.  Compos. Math., 148(5):1390–1442, 2012.
  * [FKS21] Francesc Fité, Kiran S. Kedlaya, and Andrew V. Sutherland.  Sato-Tate groups of abelian threefolds.  ArXiv preprint https://arxiv.org/abs/2106.13759, 2021.
  * [GK17] Robert Guralnick and Kiran S. Kedlaya.  Endomorphism fields of abelian varieties.  Res. Number Theory, 3:Art. 22, 10, 2017.
  * [Goo21] Pip Goodman.  Restrictions on endomorphism rings of Jacobians and their minimal fields of definition.  Trans. Amer. Math. Soc., 374(7):4639–4654, 2021.
  * [LMF] The LMFDB Collaboration.  The L-functions and Modular Forms Database.  http://www.lmfdb.org.
  * [LY20] Yi-Hsuan Lin and Yifan Yang.  Quaternionic loci in Siegel’s modular threefold.  Math. Z., 295(1-2):775–819, 2020.
  * [Rib76] Kenneth A. Ribet.  Galois action on division points of Abelian varieties with real multiplications.  Amer. J. Math., 98(3):751–804, 1976.
  * [Rot04] Víctor Rotger.  The field of moduli of quaternionic multiplication on abelian varieties.  Int. J. Math. Math. Sci., (49-52):2795–2808, 2004.
  * [Shi98] Goro Shimura.  Abelian varieties with complex multiplication and modular functions, volume 46 of Princeton Mathematical Series.  Princeton University Press, Princeton, NJ, 1998.
  * [Sil92] Alice Silverberg.  Fields of definition for homomorphisms of abelian varieties.  J. Pure Appl. Algebra, 77(3):253–262, 1992.
  * [Sil94] Joseph H. Silverman.  Advanced topics in the arithmetic of elliptic curves, volume 151 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 1994.
  * [ST68] Jean-Pierre Serre and John Tate.  Good reduction of abelian varieties.  Ann. of Math. (2), 88:492–517, 1968.
  * [Vig80] Marie-France Vignéras.  Arithmétique des algèbres de quaternions, volume 800 of Lecture Notes in Mathematics.  Springer, Berlin, 1980.
  * [Voi21] John Voight.  Quaternion algebras, volume 288 of Graduate Texts in Mathematics.  Springer, Cham, [2021] ©2021.
  * [Wil00] John Wilson.  Explicit moduli for curves of genus 2 with real multiplication by ${\mathbb{Q}}(\sqrt{5})$.  Acta Arith., 93(2):121–138, 2000.
  * [Zar00] Yuri G. Zarhin.  Hyperelliptic Jacobians without complex multiplication.  Math. Res. Lett., 7(1):123–132, 2000.
  * [Zar01] Yuri G. Zarhin.  Hyperelliptic Jacobians and modular representations.  In Moduli of abelian varieties (Texel Island, 1999), volume 195 of Progr. Math., pages 473–490. Birkhäuser, Basel, 2001.