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

Let $A$ be an abelian variety defined over a number field, and let $MT(A)$ be its Mumford-Tate group. It is a conjecture of Morita that if $MT(A)$ is anisotropic-mod-center (that is, it has no $\mathbb{Q}$-rational unipotent elements), then $A$ has potentially good reduction. The basic example is when $MT(A)$ is a torus: in this case, this conjecture is one of the first results one proves in the theory of complex multiplication of abelian varieties.

One way that Morita's conjecture is formulated is as follows: The condition on $MT(A)$ ensures that the Shimura variety attached to it is compact, and the conjecture says that the Shimura variety in fact has proper reduction at any finite prime; a transcendental condition has been transported to the world of arithmetic. Strictly speaking, this is stronger than the formulation given above, since we are now talking about the reduction of a whole host of abelian varieties (ones that appear in the family over the Shimura variety) rather than just $A$. This leads me to:

Question: Is anything known about the converse to the original formulation? Suppose that $A$ has potentially good reduction everywhere; then is $MT(A)$ anisotropic-mod-center?

share|improve this question
6  
Doesn't an elliptic curve with integral $j$-invariant have potentially good reduction everywhere? The Mumford-Tate group is $GL_2$ which is not anisotropic modulo the centre. –  ulrich Dec 12 '11 at 4:51
    
Excellent. If you can leave this as an answer, I'll accept it. –  Keerthi Madapusi Pera Dec 12 '11 at 12:40

1 Answer 1

up vote 4 down vote accepted

An elliptic curve with integral $j$-invariant has potential good reduction everywhere. If it does not have CM then its Mumford-Tate group is $GL_{2,\mathbb{Q}}$ which is not anisotropic modulo its centre.

share|improve this answer

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.