DOI: 10.1515/crelle-2026-0061 ISSN: 0075-4102
Constructing Jacobians of rank 1
Peter Koymans, Adam Morgan Abstract
Let 𝐾 be a number field, let
g
≥
1
g\geq 1
be an integer and let
f
(
x
)
=
(
x
−
a
1
)
⋯
(
x
−
a
2
g
+
1
)
∈
O
K
[
x
]
see text
f(x)=(x-a_{1})\cdots(x-a_{2g+1})\in O_{K}[x]
be a polynomial that splits into
2
g
+
1
2g+1
distinct linear factors.
Write 𝐶 for the hyperelliptic curve given by
C
:
y
2
=
f
(
x
)
C:y^{2}=f(x)
and write
J
=
Jac
(
C
)
J=\operatorname{Jac}(C)
for its Jacobian.
Under mild technical assumptions on 𝑓 that are satisfied almost always, we prove that there exists some
d
∈
K
×
d\in K^{\times}
such that the quadratic twist
J
d
J^{d}
has rank exactly equal to 1.
As a consequence, we deduce that, for any positive integer 𝑔, there exists an absolutely simple abelian variety over 𝐾 with dimension equal to 𝑔 and rank equal to 1.