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Let $K$ be a number field and let $A$ be an abelian variety of dimension $g$ over $K$. Let $L$ be a CM field and suppose that $[L:{\bf Q}]=2g$. Suppose that there exists an embedding $\iota:L\hookrightarrow{\rm End}_K(A)\otimes{\bf Q}$, where ${\rm End}_K(A)$ is the ring of $K$-endomorphisms of $A$. I shall say that the couple $(A\to{\rm Spec}\ K,\iota)$ is an abelian variety with complex multiplication by $L$ over $K$.

My question is: what is the simplest proof of the fact that there is no abelian variety with complex multiplication by a CM field over $\bf Q$ ?

The fact that there is no such abelian variety is for instance a consequence of a theorem of Ribet, which says that if $B$ is an abelian variety over a number field $K$, then $B(K(\mu_{\infty}))_{\rm tors}$ is a finite group (see 'Torsion points on abelian varieties in cyclotomic extensions', appendix to: N. M. Katz and S. Lang, Finiteness theorems in geometric class field theory, Enseign. Math. 27 (1981), 285--319) but the proof of Ribet's theorem uses p-adic Hodge theory and I have the impression that using it to answer the above question is an overkill.

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The action of $L$ on global 1-forms would give an embedding of $L$ into the algebra $M$ of $g$-by-$g$ matrices over $\mathbb{Q}$ (since we're in characteristic 0). But any maximal commutative $\mathbb{Q}$-subalgebra of $M$ is $g$-dimensional.

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  • $\begingroup$ Or dually, End(A) acts linearly on the tangent space at 0, which has dimension g=dim(A) over $\mathbb{Q}$. But a field of degree 2g over $\mathbb{Q}$ can't act on a vector space of dimension g. $\endgroup$
    – anon
    Commented Jun 24, 2015 at 21:00
  • $\begingroup$ Nice answer... thank you for that. I wonder why this argument does not appear in any of the usual references for the theory of complex multiplication (unless I missed something ?). $\endgroup$
    – Damian Rössler
    Commented Jun 25, 2015 at 10:41

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