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

Let $p$ be a prime. Suppose you have an Abelian scheme $A$ over $Spec\ \mathbb{Z}_p$. How do you prove that if $q$ is another prime, then the $q$-torsion of $A$ injects into the torsion of $A_p$, under the reduction map?

share|improve this question
add comment

4 Answers

up vote 7 down vote accepted

Let me try again at an alternate answer. If $A_{\mathbb{Z}_p}$ is an abelian scheme and $\ell \neq p$ is a prime, then for any positive integer $n$, the isogeny $[\ell^n]: A \rightarrow A$ is an etale map. [If I am not mistaken, the proof of this does not require formal groups!] Since the special fiber has $\ell^{2n}$ points over the algebraic closure, by Hensel's Lemma all of the $\ell^n$-torsion on $A$ is defined over the maximal unramified extension, and it follows that the reduction map over the maximal unramified extension is an isomorphism on the $\ell^n$-torsion, hence an injection over $\mathbb{Q}_p$.

share|improve this answer
    
MO is asking me to choose an answer. I myself do not know which is the best one; however I am leaning towards this one as it is not using formal groups. If someone more learned would give suggestions, my job would be easier. Or maybe I should wait for longer time? –  Anweshi Jan 5 '10 at 11:33
    
Both answers are correct, and two correct answers should be enough: I don't think that if you wait longer you'll get an obviously superior answer. Go ahead and choose whichever one seems best to you. (I upvoted Felipe's answer, if that influences your decision at all.) –  Pete L. Clark Jan 5 '10 at 11:56
    
I still prefer this answer, as it looks more "arithmetical" and I need to understand one thing less for figuring this out. Sorry if this erodes your opinion of me. –  Anweshi Jan 5 '10 at 15:12
    
You really don't need to apologize for accepting my answer! –  Pete L. Clark Jan 5 '10 at 15:25
2  
Personally, I find formal groups much easier to understand than anything with the word etale in it. Though, while I am sure I am not the only one with this opinion, I am also quite sure this is uncalled for and self destructive... –  Dror Speiser Jan 10 '10 at 21:38
show 1 more comment

If A is an abelian variety over a local field of char zero with good reduction, then any point in the kernel of reduction is in the so-called formal group. Now, to answer your question (properly understood), you want to show that the formal group does not have $q$-torsion for $q \ne p$. This follows because multiplication by $q$ is invertible as a (vector of) formal power series, so is a bijection in the formal group.

share|improve this answer
8  
@Anweshi: To get a feeling for the details of this argument, you could look at the elliptic curve case, which is discussed in Chapters IV and VII of Silverman, The arithmetic of elliptic curves. For a generalization of your statement, see the appendix to Katz, Nicholas M. Galois properties of torsion points on abelian varieties. Invent. Math. 62 (1981), no. 3, 481--502. –  Bjorn Poonen Jan 2 '10 at 3:54
1  
The simplest situation is that of the multiplicative group : 1+pZ_p has no l-torsion for l\neq p. That it has no p-torsion for p\neq2 is an accident : think of 1+(1-\zeta)Z_p[\zeta], where \zeta is a primitive p-th root of 1. –  Chandan Singh Dalawat Jan 2 '10 at 4:38
    
Oh so it must use formal groups. I tried to look into the book of Hazewinkel; but it is too big for me. The book of Serre on Lie Algebras and Lie Groups seem to explain this notion better. What is an easy introduction to formal groups? What is the difference between a formal group and a formal group law? Must such questions on abelian varities use formal groups for solution? And to what extend can you recover an abelian variety/algebraic group/lie group if you know its formal group? Perhaps is it better to ask this question separately? –  Anweshi Jan 2 '10 at 19:33
    
I heartily recommend Serre's LALG for a treatment of formal groups. –  Pete L. Clark Jan 2 '10 at 23:51
3  
Anweshi, first you should follow Bjorn's advice above. The original source is also very readable: Mattuck, Arthur Abelian varieties over $p$-adic ground fields. Ann. of Math. (2) 62, (1955). 92--119. jstor.org/stable/2007101. For your original question, $p$-divisible groups are irrelevant. If this is a new question, then open a new question. –  Felipe Voloch Jan 3 '10 at 17:12
show 1 more comment

To complete Pete and Milne's answers when A is not an abelian scheme (for example, when it is the Néron model of an abelian variety over ${\mathbb Q}_p$ with not necessary good reduction), then for any $n$ prime to $p$, the kernel $A[n]$ is still étale over ${\mathbb Z}_p$ (because the tangent map at $0$ of the multiplication by $n$ is just multiplication by $n$ for any commutative algebraic group), but not necessarily finite. There is a biggest closed subscheme $H$ of $A[n]$ which is étale and finite over ${\mathbb Z}_p$. The reduction map on $H$ is injective (see Pete's proof). The generic fiber of $H$ corresponds to the points of the generic fiber of $A[n]$ having specialization mod $p$. You may read Bosch, Lütkebohmert and Raynaud ''Néron Models'', § 7.3.

share|improve this answer
    
Thanks a lot! I suppose, the second part of your book is a good preparation for reading Bosch, et. al.? –  Anweshi Jan 24 '10 at 1:22
    
Yes I think so, but only to a part of their Chapter 9. The rest of ''Néron Models'' is far beyond. –  Qing Liu Jan 24 '10 at 1:39
add comment

This is Hartshorne, Exercise IV.4.19: Let $\mathcal{A}/\mathbb{Z} \setminus S =: T$ be an Abelian scheme. The multiplication by $n$ morphism is flat [I don't know how to show this, but I think it can be found in Katz-Mazur.], so the $n$-Torsion $\mathcal{A}[n] \to \mathcal{A}$ is also flat as it is a base change and $\mathcal{A} \to T$ also since it is flat. It is also proper and quasi-finite, and therefore finite. So we have a finite flat group scheme. Because of $(n,p) = 1$ $\mathcal{A}[n]$ is étale over $\mathbb{Z}_{(p)}$ (how to show this?). We have for $X/S$ finite étale

Consider the reduction map $\mathcal{A}[n]_\eta(\mathbf{Q}) = \mathcal{A}[n](T) \to \mathcal{A}[n]_p(\mathbf{F}_p)$ for $(n,p) = 1$, confer Liu, Chapter 10.1.3. Liu, Proposition 10.1.40(b) gives us one-point fibres of the reduction map.

share|improve this answer
4  
See 20.7 of Milne, J. S. Abelian varieties. Arithmetic geometry (Storrs, Conn., 1984), 103--150, Springer, New York, 1986. MR0861974, which is available at jmilne.org/math/articles/1986b.pdf. Once you know the kernel of multiplication by n is a finite etale group scheme, the rest is very easy. –  JS Milne Jan 10 '10 at 19:58
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.