# Isomorphism on p-torsion of Neron models

Let $A$, $B$ be abelian varieties over $\mathbb{Q}$, with corresponding Neron models $\mathcal{A}$, $\mathcal{B}$ over $X=Spec{\mathbb{Z}}$. Let $p$ be an odd prime of good reduction for both $A$ and $B$ such that we have an isomorphism $A[p]\cong B[p]$ over $\mathbb{Q}$. Keerthi Sampath has explained to me how this implies that we get an isomorphism $\mathcal{A}[p]\cong\mathcal{B}[p]$ of etale group schemes over $U=Spec{\mathbb{Z}[\frac{1}{p}]}=X-Spec{\mathbb{F}_p}$.

(a) I want to know if the same hypotheses as above also implies that we get an isomorphism $\mathcal{A}^0[p]\cong\mathcal{B}^0[p]$ on the identity components over $U$.

(b) Is it true that $\mathcal{A}^0[p]$ (the $p$-torsion on $\mathcal{A}^0$) is the same as $\mathcal{A}[p]^0$ (the identity component of $\mathcal{A}[p]$)?

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How do you define $\mathcal{A}[p]^0$ in (b)? If $G$ is a group scheme of finite type over a field, $G^0$ is a well-defined open subgroup scheme of $G$. Over a more general base scheme $S$, all you can do is define $G^0$ as a subfunctor of $G$: a point of, say, $G(S)$ is in $G^0$ iff its restriction over every point $s$ is in $G_s^0$. This functor is representable if $G$ is smooth (e.g. Néron models) but not in general (think of $\mu_p$ over $\mathbb{Z}_p$). –  Laurent Moret-Bailly Jun 5 '11 at 6:38
I suppose that I defined $\mathcal{A}[p]^0$ as an intersection $\mathcal{A}^0\cap\mathcal{A}[p]$ of group schemes. –  Saikat Biswas Jun 5 '11 at 20:41

(a) is not going to be true in general (and as Moret-Bailly says, (b) needs some more explanation).

Here's the "moral" reason (a) isn't true. Let $E$ be an elliptic curve over the rationals. If $E[p^n]$ is unramified at $\ell$ for all $n\geq1$, then $E$ will have good reduction. But if $E[p]$ is unramified then this is not enough -- this is "level-lowering" a la Mazur/Ribet, which was crucial in the proof that STW -> FLT. So if you take, for example, a counterexample to FLT and let $E$ be the associated Frey curve, then $E[p]$ will be unramified at all odd $\ell\not=p$ but $E$ could still have bad reduction at many such $\ell$.

But one does not need a counterexample to FLT to build such a phenomenon. One just needs two elliptic curves $A$ and $B$ with $A$ having good reduction and $B$ having, say, multiplicative reduction at $\ell$, such that $A$ and $B$ are congruent mod $p$. Then the $j$-invariant of $B$ will have valuation at $\ell$ a multiple of $p$, so the number of components will be a multiple of $p$, and this will give you your counterexample (if I've understood the question correctly) because over $\mathbf{Z}_\ell$ the identity component of the special fibre of $B$ is missing some $p$-torsion.

My memory is that Ribet-Stein is full of explicit examples of these phenomena.

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Thank you for the counterexample. Having read Section 1 of Mazur's Rational Isogenies paper, I tried to work out some of the details in your counterexample and (although I'm not entirely sure I have it right) it seems that over $\mathbb{Z}_{\ell}$, there might be a closed immersion of the finite flat multiplicative type subgroup-scheme of $\mathcal{B}^0[p]$ into $\mathcal{A}^0[p]$. Is this true? –  Saikat Biswas Jun 8 '11 at 6:34
I don't think it can be true because the generic fibres are isomorphic and dense, so how can anything be a closed subgroup of anything? –  Kevin Buzzard Jun 8 '11 at 20:02
Oh -- maybe I'm misunderstanding. What is "the finite flat multiplicative type subgroup-scheme of $B^0[p]$"? Note that $B^0[p]$ is not flat. I guess it will contain as a subscheme some flat lifting of the special fibre though, which will I guess then be a closed subscheme of $A^0[p]$. Maybe that's what you're asking, in which case it sounds true to me. –  Kevin Buzzard Jun 8 '11 at 23:21
Indeed. We know that, over $U$, $\mathcal{B}^0[p]$ is quasi-finite flat and separated, and it has a finite flat subgroup scheme, say $F\mathcal{B}^0[p]$. As I mentioned, I suppose that there is a closed immersion of $F\mathcal{B}^0[p]$ into $\mathcal{A}^0[p]$ but not sure exactly why –  Saikat Biswas Jun 9 '11 at 3:04