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

In the Deligne-Rapoport paper entitled "Les schemas de modules de courbes elliptiques" the following is written (I translated in english):

Let $E$ be an elliptic curve with $\Gamma(N)$-level structure defined over $\mathbb{C}((T))$. Let $E'$ be the minimal model of $E$ over $\mathbb{C}[[T]]$. It may happen that $E'$ has bad reduction (i.e. when one reduces modulo $T$). So let $A$ be the neron model of $E'$ over the d.v.r. $\mathbb{C}[[T]]$. Then it seems that the special fiber of $A$ (i.e. when $T=0$) is always isomorphic to $\mathbb{C}^{\times}\times\mathbb{Z}/kN$ for some suitable integer $k$.

Q: How come this $N$ shows up in the special fiber of $A$ ?

This is a little bit strange since in the definition of a Neron model no such $N$ appears.

share|improve this question
add comment

1 Answer

up vote 9 down vote accepted

Roughly speaking, the N-torsion defined over the base injects into the Neron model (in characteristic 0), so the special fiber of the Neron model needs to have a subgroup isomorphic to Z/NZ x Z/NZ, since by assumption there is full level N structure. The special fiber (since there's bad reduction) has the form C^* x F, where F is a finite cycle group. So F contains an N-torsion element, which means F = Z/kN.

share|improve this answer
    
So the fact that you need characteristic 0 means that it does not follow formally from the universal property of the Neron model. So how does the characteristic 0 assumption intervene? Is related somehow to the fact that in characteristic zero finite flat group schemes are etale? –  Hugo Chapdelaine Jan 6 '11 at 0:51
    
I wonder if you need to assume $N \ge 3$ to exclude $I_0^*$-reduction. In that case the component group is $\mathbb Z/2 \times \mathbb Z/2$. Or do you assume multiplicative Reduction right away? In that case characteristic zero is not needed. In charakteristic $p$, you have to be careful because additive reduction gives you a lot of $p$-Torsion. –  Holger Partsch Jan 6 '11 at 13:52
    
Hugo, your question was phrased over C, so I put in the characterstic 0 assumption to avoid saying something false, since I didn't really think it through in finite characteristic. If you assume a priori that you have multiplicative reduction, then I think you're fine in all characteristics, including p=2. If you simply assume bad reduction, then there are probably problems with p=2 and p=3 for various reduction types. But I haven't thought it through carefully. Note that if p divides N, then there is some p-torsion in the formal group, but none in the multiplicative group. –  Joe Silverman Jan 8 '11 at 15:41
add comment

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.