4
$\begingroup$

Let $F$ be a totally real number field, and let $K$ be a quadratic extension of $F$ which cannot be embedded into $\mathbb{R}$. Then $K$ is a so called CM field. For instance, take $F = \mathbb{Q}(\sqrt2)$, and $K = \mathbb{Q}(\zeta_8) = F(i)$ where $\zeta_8$ is a primitive complex $8$-th root of unity. Denote by $\mathcal{O}_K$ the ring of integers of $K$. My question is:

How to construct a number field $L$ and an abelian variety $A/L$ such that $\mathrm{End}(A) \cong \mathcal{O}_K ?$

For an example with elliptic curves, take $F = \mathbb{Q}$, and $K = \mathbb{Q}(\sqrt{-14})$ so that $\mathcal{O}_K = \mathbb{Z}[\sqrt{-14}]$. Evidently, $\mathcal{O}_K$ is a lattice in $\mathbb{C}$ so $E_0 = \mathbb{C}/\mathcal{O}_K$ is an elliptic curve with $\mathcal{O}_K \subseteq \mathrm{End}(E_0)$. Since $\mathrm{End}(E_0)$ is an order in an imaginary quadratic number field, we have $\mathrm{End}(E_0) \cong \mathcal{O}_K$ as $\mathcal{O}_K$ is the maximal order. Furthermore, $j(E_0)$ is an algebraic number (in fact, an algebraic integer) which generates a field isomorphic to $L = \mathbb{Q}(\sqrt{2\sqrt{2}-1})$ over $\mathbb{Q}$. It follows that we can define an elliptic curve $E_1/L$ such that $j(E_1) = j(E_0)$ so that $E_1 \cong_{\mathbb{C}} E_0$ and in particular $\mathrm{End}(E_1) \cong \mathrm{End}(E_0) \cong \mathcal{O}_K$ as required.

Can I tell a similar story (with abelian vaieties instead of elliptic curves) for any CM field?

It is only clear to me that the answer is positive for imaginary quadratic number fields.

$\endgroup$
  • 3
    $\begingroup$ Yes. Your notation "End($A$)" must mean endomorphisms after scalar extension to $\mathbf{C}$. The endomorphisms over $\mathbf{C}$ all descend to a common finite extension of $L$; this has arithmetic significance for elliptic curves in terms of a ring class field (for the CM order), and such descent for endomorphisms (really from $\mathbf{C}$ to $\overline{\mathbf{Q}}$) fails badly for projective spaces. See Theorem 1.7.2.1 and Prop. 1.7.4.5 in the book "Complex Multiplication and Lifting Problems" for an affirmative answer (with no explicit $L$) using modern algebraic geometry. $\endgroup$ – nfdc23 Jul 10 '16 at 14:34
  • 1
    $\begingroup$ @nfdc23 could you please expand your comment to an answer? I have hard time finding this book. $\endgroup$ – Pablo Jul 10 '16 at 14:54
  • 1
    $\begingroup$ If your university doesn't have a copy of that book then ask the library to get it for you via inter-library loan. Rather than explaining much that is already documented in detail there, I should point out that section A.4.6 in that book (see especially Corollary A.4.6.5) gives an improvement on Borovoi's comment below by avoiding any need to split the 3-torsion (building on a complete treatment of the Main Theorem of CM and its "converse" using modern algebraic geometry in A.2--A.3). $\endgroup$ – nfdc23 Jul 10 '16 at 15:31
  • $\begingroup$ @nfdc23 Thanks for your help, I will do the reading. $\endgroup$ – Pablo Jul 10 '16 at 15:39
2
$\begingroup$

A reasonably good answer to your question of finding a number field $L$ is given in the comments to Is there an excplicit number field of definition for an Abelian Variety $A/\mathbb{C}$ with CM? . However, the answer is roughly "no, there's no easy way to find $L$". But first you need to understand the difference between the field of moduli and fields of definition. For elliptic curves, the field of moduli is generated by the $j$-invariant (and if you also want the CM maps defined, you need to adjoin $j$ to the CM field, at least if $\text{End}(A)$ is the full ring of integers). For abelian varieties, the field of moduli is the field generated by the point in moduli space, or alternatively the fixed field of $$ \{\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q) : \sigma(A)\cong A\}.$$ But there need not be a model for $A$ that is defined over this field. There's a brief discussion of CM abelian varieties in Shimura's Introduction to the Arithmetic Theory of Automorphic Forms (Section 5.5). More detailed information is in books on CM abelian varieties such as those by Shimura-Taniyama and Lang.

$\endgroup$
  • $\begingroup$ So do you know what is the answer in the case of $K = \mathbb{Q}(\zeta_8)$ mentioned by me in the beginning of my question? $\endgroup$ – Pablo Jul 10 '16 at 14:27
  • 3
    $\begingroup$ @Pablo: Yes, such $L$ does exist. One can compute a field over which $A$ together with a polarization, the action of $L$ with given CM-type, and the points of order 3 is defined. In particular, $A$ is defined over this field. This field is given by the theory of Shimura-Taniyama. See e.g. Deligne, Travaux de Shimura. $\endgroup$ – Mikhail Borovoi Jul 10 '16 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.