6
$\begingroup$

For any imaginary quadratic field $F$, the Hilbert class field $H$ is generated by the $j$-invariant of any elliptic curve with complex multiplication (CM) by $\mathcal O$, the ring of algebraic integers of $F$.

What is the simplest generalization of this well known and useful fact? Since an imaginary quadratic field is the simplest example of a CM field, one is led to ask:

Is there an (albeit conjectural) invariant of abelian surfaces which generates the Hilbert class field of a CM field $K$ (of degree $4$ over $\mathbb Q$)?

Note that $K$ is a totally imaginary quadratic extension of a real quadratic field.

Improve this question
$\endgroup$
2

1 Answer 1

Reset to default
7
$\begingroup$

The simplest generalisation to abelian surfaces is, I believe, the statement (which is a theorem, not a conjecture) that the Igusa invariants of an abelian surface with CM by $K$ generate an abelian unramified extension of the reflex field of $K$. Note that the reflex field is not, in general, equal to $K$, and that there is no claim that the full Hilbert class field is obtained in this way. There is also a generalisation of this to higher dimensional principally polarised abelian varieties.

Improve this answer
$\endgroup$
1
  • $\begingroup$ thanks! could you please point to a reference? also, are there interesting cases where the reflex field is $K$? $\endgroup$
    – guest
    Commented May 19, 2017 at 1:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .