4
$\begingroup$

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.

2 Answers 2

11

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.

  • 0
    ... it is easily checked to be torsion free over $End(C)$, and is finitely generated over $\mathbb Z$ by general principles. Putting this altogether, and using that $End(C) = \mathcal O_K$ is a Dedekind domain, we deduce that $Hom(C,E)$ is an invertible $End(C)$-module, as claimed. The fact about two generators is proved exactly as you suggested. Regards,2012-07-25
6

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.

  • 0
    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).2012-07-24