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Let $K$ be a number field and $\mathfrak{p}$ be a place of good reduction. It is easy to see that the reduction map on prime-to-$p$ torsion $A(K)[p'] \hookrightarrow A_{\mathfrak{p}}(\kappa(\mathfrak{p}))$ is injective.

But if $p > e(\mathfrak{p}/p) + 1$, the reduction map is even injective on $p$-torsion. This can be seen by showing that the formal group has no $p$-torsion.

Now I ask if one can show this without using the formal group. By a theorem of Raynaud in "Schémas en groupes de type $(p, \ldots, p)$", the inequality for $p$ implies that for a $p^n$-torsion commutative finite flat group scheme the generic fibre can be spread out uniquely over the special fibre and $\mathrm{Hom}(G,H) = \mathrm{Hom}(G(\bar{K}),H(\bar{K}))$. Can this be used somehow?

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up vote 4 down vote accepted

One has the finite flat group scheme $\mathbb Z/p$ over $\mathcal O_{K_{\mathfrak p}}$ (I write $K_{\mathfrak p}$ for the $\mathfrak p$-adic completion of $K$, and $\mathcal O_{K_{\mathfrak p}}$ for its integer ring), as well as the finite flat group scheme $A[p]$. Giving a $p$-torsion point over $K_{\mathfrak p}$ (and hence in particular over $K$) is the same as giving a closed embedding on generic fibres: $(\mathbb Z/p)\_{/ K_{\mathfrak p}} \hookrightarrow A[p]\_{/K_{\mathfrak p}}.$

Raynaud's results imply that this extends to a closed embedding over $\mathcal O_K$: $\mathbb Z/p \hookrightarrow A[p],$ which is another way of saying the that the non-zero $p$-torsion point has non-zero reduction.

Just to see concretely what can happen in the situation when $e \geq p-1$, suppose that $K = \mathbb Q$ and $p = 2$. Then we could have a map $(\mathbb Z/2)\_{/\mathbb Q_2} \hookrightarrow A[2]_{/\mathbb Q_2}$ which extends to a closed immersion $\mu_2 \hookrightarrow A[2].$ This would correspond to having a 2-torsion point in the kernel of the reduction map. (Note that $\mu_2$ has a non-trivial point in char. zero, which collapses down to the identity in char. two.)

[Added in response to unkwown's comment:]

The point is that one can form the scheme-theoretic closure in $A[p]$ of the image of $(\mathbb Z/p)\_K,$ which is some finite flat subgroup scheme over $\mathcal O_K$ which is embedding as a closed subgroup scheme of $A[p]$ (by construction: we formed it as a scheme-theoretic closure). And it has $(\mathbb Z/p)\_K$ as its generic fibre (again by construction).

Now when $e < p-1$, Raynaud's results show that this finite flat group scheme has no choice but to be $\mathbb Z/p$, and so we get a copy of $\mathbb Z/p$ embedding into $A[p]$, extending the original embedding of generic fibres. Thus the order $p$ point in $A[p](K)$ reduces mod $p$ to an order $p$ point.

But if e.g. $p = 2$, then this scheme-theoretic closure could be $\mu_2$. Now the non-trivial point ($-1$) of $\mu_2(K)$ specializes to the trivial point in char. 2, and so when we have a copy of $\mu_2$ inside $A[2]$, the non-trivial point of $\mu_2(K)$ lies in the kernel of the reduction mod 2 map.

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Matt's argument applies with $\mathbf{Z}/(p)$ replaced by a power, such as corresponding to basis of $A[p](K)$, so desired result follows. Sutble variant: is homomorphism $G' \rightarrow G$ between smooth affine gps over a dvr $R$ a closed immersion when it is on generic fibers? (Above is $G$ an abelian scheme and $G'$ a finite constant group.) In the important case that $G'$ has connected reductive fibers, it is OK away from exceptions for residue char. 2. This lies quite deep. See Cor. 1.3 in Prasad-Yu "Quasi-reductive groups" (not to be confused with pseudo-reductive groups...) – BCnrd Jun 21 '10 at 20:36
I added a more detailed explanation. – Emerton Jun 23 '10 at 21:07
As examples, consider an elliptic curve $E$ over $\mathbf{Q}$ with good reduction at 2 and full rational 2-torsion (impossible with odd primes in place of 2!): $E$ is defined by $y^2 = f(x)$ where $f$ is a split cubic over $\mathbf{Q}$. Then working over $\mathbf{Z}_2$, the $(\mathbf{Z}/(2))^2$ in the generic fiber over $\mathbf{Q}_2$ has schematic closure equal to the finite flat 2-torsion in the N\'eron model. Whether in the ordinary or supersingular reduction case, can then find a $\mathbf{Z}/(2)$ in the generic fiber whose closure has special fiber $\mu_2$ or $\alpha_2$ respectively. – BCnrd Jun 24 '10 at 5:06
To know that $A[p]$ is finite flat over the ring of integers. – Emerton Jun 24 '10 at 17:54
To add to Emerton's clarification, if we let $\mathcal{A}$ denote the N\'eron model of $A$ over the local ring $\mathcal{O}_ {K, \mathfrak{p}}$ then the Zariski closure of $A[p]$ in $\mathcal{A}$ is always flat, and consequently quasi-finite (since the generic fiber is finite) even in cases with non-semistable reduction (for which $\mathcal{A}[p]$ is non-flat and has positive-dimensional special fiber). So it is really the finiteness that is the key place where Emerton invokes good reduction (since quasi-finite + proper implies finite, and $\mathcal{A}$ is proper in the good reduction case). – BCnrd Jun 24 '10 at 18:51

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