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This question is related to this one. Tate module of CM elliptic curves

There seem to be several versions of "complex multiplication". Fact 1: We say $E/\mathbb{C}$ has CM if $End_C(E) \supsetneq Z$. It is equivalent that $End_C(E)$ is an order in a quadratic field.

Fact 2: Now suppose $E$ defined over $\it{ANY}$ field $K$, we say that $E$ has CM $\it over$ $\it K$ if $End_K(E) \supsetneq Z$. By the method Pete pointed out in the comment of the above MO question, if can be shown that in this case, the Galois group $Gal(\bar{K}/K)$ action on the Tate module $T_{\ell}(E)$ for $\ell \neq char(K)$ is abelian.

Question 1: Suppose $E$ defined over $\it{ANY}$ field $K$ with $ch K = 0$, if $E$ has CM $\it over$ $\it K$, is it true that $End_K(E)$ is an order in a quadratic field? And what about $End_{\bar{K}}(E)$, is it an order in a quadratic field? We require $ch K =0$ because when $K$ is a finite field, $End_{\bar{K}}(E)$ is always bigger than $Z$ and could be an order in a quaternion algebra.

QUestion 2: Now suppose $E$ defined over a $\it number$ $\it field$ $K$, thus it could be seen as an elliptic curve over $C$ as well.Is it true that $End_K(E) \supsetneq Z$ if and only if $End_C(E) \supsetneq Z$?

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5 Answers 5

Allow me to say something which is not so much an answer to this question as to a (very natural) question that I sense is coming in the future.

There are two possible pitfalls in the definition of "has complex multiplication" for abelian varieties over an arbitrary ground field $F$.

1) Let's first talk only about complex abelian varieties. There is a serious discrepancy between the terminology "has complex multiplication" as used in the classical literature (up to and including, say, some of Shimura's papers in the 1960s) and the way it is almost invariably used today.

Classically, an abelian variety over the complex numbers was said to "have complex multiplications" whenever its endomorphism ring was strictly larger than $\mathbb{Z}$. This is never the generic case, but starting in dimension $2$ it allows abelian surfaces with multiplication by an order in a real quadratic field (parameterized by Hilbert modular surfaces), by an order in an indefinite rational quaternion algebra (parameterized by Shimura curves), and so forth.

Nowadays one means something much more restrictive by "complex multiplication": one wants the endomorphism ring to be an order in a $\mathbb{Q}$-algebra whose dimension is large compared to the dimension, say $g$, of $A$. In the case of an abelian variety without nontrivial abelian subvarieties ("simple"), CM means precisely that the endomorphism ring is an order in a number field of degree $2g$: it then follows from this (not so obviously) that the number field is totally complex and has an index $2$ totally real subfield. For nonsimple abelian varieties, the generally agreed upon definition is that $\operatorname{End}(A)$ should contain an order in a commutative semisimple $\mathbb{Q}$-algebra of dimension $2g$. Such things do not vary in moduli, even in higher dimensions.

2) If you are interested in abelian varieties over an arbitrary ground field $F$ (let me concentrate on the case of characteristic 0), then you need to distinguish between the ring of endomorphisms which are rationally defined over $F$ and the ring of endomorphisms which are defined over an algebraic closure of $F$ (hence, it turns out, over any algebraically closed field containing $F$, i.e., one cannot acquire endomorphisms by passing from $\overline{\mathbb{Q}}$ to $\mathbb{C}$). If one does not specify, "having CM" means having CM over the algebraic closure.

As in the responses above, it is very often the case that you have to extend the ground field slightly in order to get all the endomorphisms rational over the ground field. E.g. if $E_{/\mathbb{C}}$ is any elliptic curve, it can be minimally defined over $\mathbb{Q}(j(E))$. If $E$ has complex multiplication (over $\mathbb{C}$), then $j(E)$ is a real [not necessarily totally real] algebraic number, from which it follows that $\operatorname{End}_{\mathbb{Q}(j(E))}(E) = \mathbb{Z}$.

In particular, the $\ell$-adic Galois representation of a "CM abelian variety" [using standard conventions as in 1) and 2) above] need not have abelian image, quite. It has "almost abelian image", meaning that once you make a finite extension of the ground field to make all the endomorphisms rational, then the image will be abelian. (Both M. Emerton's and my arguments in the one-dimensional case extend easily to this case, although I think his is more insightful.)

What I have said carries over to fields of positive characteristic, but the general picture of endomorphism algebras looks different, especially over finite fields: every abelian variety over a finite field has complex multiplication in the above sense.

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Regarding question 2: No, the elliptic curve defined by $E: y^2=x^3-x$ has CM over $\mathbb{C}$ and $End_\mathbb{Q}(E)=\mathbb{Z}$.

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Thank you Ben! Can you tell me how to verify that End_Q(E)=Z? –  natura Feb 4 '10 at 20:44
    
The extra endomorphisms of $E$ over $\mathbb C$ are generated by $(x,y)\mapsto (-x,iy).$ This is evidently not defined over ${\mathbb Q}$. To cement the matter, note that it acts on the differential $dx/y$ as multiplication by $i$, which is not in ${\mathbb Q}$. –  Emerton Feb 5 '10 at 3:53

Question 1. No. You answered it yourself. $End_{\bar K}(E)$ could be an order in a quaternion algebra and the same will happen over some subfield of $\bar K$, which can be taken to be a finite field. As to whether the endomorphisms can only be seen in an extension field, Ben's answer to question 2 also works for finite fields ${\mathbb F}_q$ if $q \equiv 3 \mod 4$.

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ah I made a mistake, I meant for char K =0! –  natura Feb 4 '10 at 19:25

A more specific answer to questions 2: $End_K(E)$ is the full endomorphism ring (i.e. $\mathcal{O} = End_{\overline{K}}(E)$) if $K$ contains the RCF (ring class field) of $\mathcal{O}$. I think that this may be if and only if. The field generated by the $j$-invariant of $E$ is a subfield of index 2 in the RCF. In the example given by Ben Linowitz the RCF of $\mathcal{O}$ is $\mathbb{Q}[i]$.

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1  
@VM: Yes, it's iff. –  Pete L. Clark Feb 4 '10 at 19:35

I am offering this answer in light of basic's comment following Ben's answer.

If $\phi$ is an endomorphism of $E$, then $\phi$ induces an endomorphism $d\phi$ of the space of holomorphic differentials on $E$. If the ground field $K$ is of char. 0, then this action is injective. On the other hand, the space of holomorphic differentials is 1-dim'l over $K$, and so this action induces an injective homomorphism of rings $End_K(E) \hookrightarrow K.$

Thus if $K$ is totally real (e.g. ${\mathbb Q}$), $End_K(E) = {\mathbb Z}$, since a totally real field can't contain an order in a quad. imag. field.

Somewhat more precisely we see that $K$ must at least contain the field of CM, for the CM to be defined over $K$. (This is not a sufficient condition however, just a necessary one. E.g. for $E$ over $K$ to admit CM by ${\mathbb Z}[\sqrt{-5}]$, it is both necessary and sufficient that $K$ contain $\sqrt{5}$, as well as $\sqrt{-5}$. This is related to the discussion in Victor's answer. (A little more precisely still: we need $\sqrt{5}$ to be able to write down an equation for the elliptic curve at all, since its $j$-invariant lies in ${\mathbb Q}(\sqrt{5})$, and then we need $\sqrt{-5}$ to be able to write down the extra endomorphisms.))

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Actually the condition that you listed as necessary is sufficient. Since, as I stated the field generated over $\mathbb{Q}$ by the $j$-invariant has index 2 in the RCF, if we throw in the CM field we get the RCF. –  Victor Miller Feb 5 '10 at 4:21
    
I think you're referring to my second use of necessary; I'm fixing the wording. Thanks. –  Emerton Feb 5 '10 at 4:27

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