Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Does anybody have a reference for the following fact?

All abelian varieties with complex multiplication and same CM type are isogenous over $\overline{\mathbb{Q}}$?

Here abelian variety with complex multiplication means an abelian variety A over $\overline{\mathbb{Q}}$ such that there exists a CM number field $K$ of degree twice the dimension of A and an embedding of $K$ into $End(A) \otimes \mathbb{Q}$. The CM type is obtained by looking at the action of $K$ into $H^0(A, \Omega^1_A)$.

share|improve this question
    
I think these notes should do it: math.stanford.edu/~conrad/vigregroup/vigre04/cm.pdf –  Keerthi Madapusi Pera Feb 23 at 12:46
1  
The main point is that the CM type completely determines the $\mathbb{Q}$-Hodge structure attached to $A$ as a $K$-module. –  Keerthi Madapusi Pera Feb 23 at 12:48
1  
@KeerthiMadapusiPera: One needs a bit more: if a pair of CM abelian varieties (for a common CM field) over an algebraically closed field $k$ become isogenous (linearly over the CM field) over an extension $K/k$ then they're isogenous (linearly over the CM field) over $k$. As you know, this is a standard "specialization" argument, by descending from $K$ to a finitely generated $k$-subalgebra $R$ (so now working with abelian schemes over $R$) and then passing to fibers over a $k$-point of $R$. (The much stronger results on descent of homomorphism for abelian varieties are not needed.) –  user76758 Feb 23 at 14:28
    
@user76758: the descent of homomorphisms needed here can be obtained by noticing that the graph of a homomorphism has a dense subset of torsion points and thus must descend to $k$ (because Zariski closure commutes with field extensions). –  Damian Rössler Feb 23 at 21:30
add comment

2 Answers 2

This is true over $\mathbb{C}$, and for any algebraically closed subfield $k$ of $\mathbb{C}$ the functor from CM abelian varieties over $k$ to CM abelian varieties over $\mathbb{C}$ is an equivalence of categories. All of this can be found in the notes on Complex Multiplication on Milne's webpage.

share|improve this answer
add comment

I do not know a reference, but it is easy to prove. Let $K/F$ be a totally imaginary quadratic extension $K$ of a totally real number field $F$ of degree $n$ over $\mathbb Q$. Let ${\mathfrak a}$ be a nonzero fractional ideal in $K$. Now, $K\otimes _{\mathbb Q} {\mathbb R}$ is isomorphic as an $\mathbb R$-algebra to the $n$ fold product of $\mathbb C$ with itself. Then $A={\mathbb C}^d/{\mathfrak a}$ is an abelian variety and its ring of complex multiplications tensored with ${\mathbb Q}$ is precisely $K$. \vskip 5mm

Moreover all the abelian varieties whose ring of endomorphisms tensored with $\mathbb Q$ is $K$ is obtained in this way, up to isogeny. This is proved by showing that a lattice in ${\mathbb C}^d$ which is stable under an order $R$ in $K$ is isogenous to ${\mathbb C}^d/R$.

share|improve this answer
    
This is proved in Mumford's book on abelian varieties in section 22 (see "First Example", and the Theorem there). As in Mumford's discussion, the notation should keep track of the choice of CM type $\Phi$ giving the identification of $K \otimes_{\mathbf{Q}} \mathbf{R}$ with $\mathbf{C}^d$ as $\mathbf{R}$-algebras; this is implicit to make sense of the quotient of $\mathbf{C}^d$ modulo orders in $\mathscr{O}_K$ or fractions ideals of $K$ (though since the CM order is not preserved by isogeny, it might be "better" to avoid formulating things in terms of orders). –  user76758 Feb 23 at 15:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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