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This is almost certainly known (and maybe written down somewhere?). Is there an example of two elliptic curves $C, E/k$ that are not isomorphic, yet there is an embedding $C\hookrightarrow E\times E$ as an abelian subvariety?

If this is known, is there a reference that talks about such things.

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2 Answers 2

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Giving a morphism $C \to E\times E$ is the same as giving a pair of morphisms $p,q: C \to E$, by the definition of the product. Translating $p$ and $q$ appropriately by some points of $E$, it is no loss of generality to assume that $p$ and $q$ both take the origin of $C$ to the origin of $E$, and hence are not only morphisms of curves, but homomorphisms of elliptic curves.

The map $C \to E \times E$ will be injective if and only if $ker(p) \cap ker(q) = 0$.

Now if $q^{\vee}$ denotes the dual of $q$, then $q^{\vee}\circ p$ is an endomorphism of $C$. If $C$ (equivalently, $E$) is not CM, then this will be multiplication by some integer $n$, in which case we find (composing with $q$) that $\deg(q) p = n q$, and a short argument shows that in fact there is an morphism $r: C \to E$ such that $p = m r$ and $q = n r$ for some integers $m$ and $n$. If $ker(p)\, $ and $ker(q)\, $ have trivial intersection, it follows that $r$ must be an isomorphism, so that $C \cong E$.

On the other hand, if $E$ has CM, then the situation you ask about can occur. The easiest case is to assume that $E$ and $C$ both have CM by the full ring of integers $\mathcal O$ in some imag. quad. field $K$, but are not isomorphic (which is possible iff $K$ has class number bigger than one). Then $Hom(C,E)$ is an $\mathcal O$-module, invertible (i.e. locally free of rank one) but not free. Such a module can always be generated by two elements. If we take $p$ and $q$ to be a pair of generators, then they will embed $C$ into $E\times E$.

If $E$ has CM by a field of class number one, then the same kind of construction should be possible, I think, by choosing $C$ to have CM by some proper order of $K$ with non-trivial class number.

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Fantastic answer, Matt! –  Georges Elencwajg Jul 24 '12 at 8:31
    
Thank you so much! How interesting. –  Matt Jul 24 '12 at 16:02
    
Maybe I'm confused about something here. For the "easy" case, if $E$ and $C$ have CM by the full ring of integers in $K$ doesn't this imply the class number is 1? Or is that only true when the curves are defined over $\mathbb{Q}$ or some other subtlety that I'm missing? –  Matt Jul 24 '12 at 17:05
    
@Matt: Dear Matt, A CM elliptic curve with CM by the full ring of integers of $K$ has $j$-invariant lying in the Hilbert class field of $K$, and the Galois conjugates of the $j$-invariant generate the Hilbert class field of $K$ (over $K$). Thus $K$ has class number one if and only if there exists an elliptic curve with CM by the full ring of integers whose $j$-invariant lies in $K$. (A more subtle consideration of Galois actions, taking into account the Galois action over $\mathbb Q$, and not just over $K$, shows that in fact the $j$-invariant then lies in $\mathbb Q$.) Regards, –  Matt E Jul 24 '12 at 17:23
    
P.S. Concretely, if $I$ is any ideal in $\mathcal O_K$, then $\mathbb C/I$ is an elliptic curve with CM by $\mathcal O$, so obviously there is no obstruction to making such curves for any $K$. (And indeed, this construction is the beginning of the theory of CM.) –  Matt E Jul 24 '12 at 17:24

By the Poincare Complete Reducibility Theorem, the isogeny category of abelian varieties over an arbitrary field is semisimple.

And now, with more words: if $A_{/k}$ is an abelian variety, then:

(i) There are simple abelian varieties (i.e., without proper, nontrivial abelian subvarieties) $B_1,\ldots,B_k$ such that $A \sim B_1 \times \ldots B_k$, where the relation $\sim$ denotes $k$-rational isogeny. (An isogeny $f: A \rightarrow B$ of abelian varieties is a surjective algebraic group homomorphism with finite kernel.)
(ii) The isogeny classes of the $B_i$'s are uniquely determined: if $C$ is any abelian subvariety of $A$, then $C$ is $k$-rationally isogenous to a direct sum of $B_i$'s.

Applying this result to your question, we find that the possible choices for an elliptic curve $C$ inside $E \times E$ are contained in the set of elliptic curves $k$-rationally isogenous to $E$. For instance if $k = \mathbb{C}$, this will be a countably infinite set. Note that we may well have a finite morphism $C \rightarrow E \times E$ but not an embedding. Studying abelian subvarieties in the strict sense -- i.e., not up to isogeny -- is significantly trickier.

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You know, it dawns on me that this is not really an answer to your question: it just limits the possible isomorphism classes of elliptic curves $C$ that could occur. I believe that the answer is that yes, it is possible for $C$ to be (isogenous but) not isomorphic to $E$...I'll have to think more about coming up with a proof / reference for that. –  Pete L. Clark Jul 24 '12 at 0:53
    
Thanks! Because of the context in which this came up, I actually already had that it had to be isogenous (for a completely different reason), so it is interesting that this stays consistent and could have been derived afterwards. As for the bounding the possibilities by countably infinitely many, it is interesting that a paper by Lenstra, Oort, and Zarhin called "Abelian Subvarieties" proves that there is actually only finitely many possibilities up to isomorphism (of embedded abelian subvarieties in any abelian variety). –  Matt Jul 24 '12 at 1:14

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