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Let $p$ be a prime. Let $K/\mathbb{Q}_p$ be a finite extension and $\mathcal{O}=\mathcal{O}_K$ its ring of integers.

Let $A_K$ be an abelian variety and $\mathcal{A}_\mathcal{O}$ denote its Neron model over $\mathcal{O}$. We don't assume that $\mathcal{A}_\mathcal{O}$ has good or semistable reduction.

Let $\ell$ be a prime different from $p$ and let $G=A[\ell]_\mathcal{O}$ be the scheme-theoretic kernel of multiplication by $\ell$ in $\mathcal{A}_\mathcal{O}$. We know that $G\times_{\text{Spec} ~\mathcal{O}} \text{Spec} ~K = A_K[\ell]$ is a finite subgroup scheme of order $\ell^{2g}$, where $g$ is the dimension of $A_K$.

I have a very stupid question here:

Is $A[\ell]_{\mathcal{O}}$ finite and flat? (or at least quasi-finite)? If not, are there some necessary conditions for $A[\ell]_{\mathcal{O}}$ to be finite and flat?

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  • $\begingroup$ See 20.7 of Milne, J. S. Abelian varieties. Arithmetic geometry (Storrs, Conn., 1984), 103--150, Springer, New York, 1986. MR0861974, if $A_\mathcal{O}$ has good reduction. $\endgroup$ – user1225 Nov 25 '17 at 3:18
  • $\begingroup$ If the generic fiber $A_K$ is an elliptic curve and the closed fiber is a multiplicative group, then the generic fiber of $A[\ell]_{\mathcal{O}}$ has length $\ell^2$, yet the closed fiber has length $\ell$. Thus, the affine, flat group scheme $A[\ell]_{\mathcal{O}}$ over $\text{Spec}\ \mathcal{O}$ is not finite (it is quasi-finite). $\endgroup$ – Jason Starr Nov 25 '17 at 3:34
  • $\begingroup$ See Lemma 2 of Section 7.3 of S. Bosch, W. Lutkebohmert, M. Raynaud, Neron Models (1990). If $A$ has semistable reduction, then $A[\ell]_{\mathcal{O}}$ is quasi-finite and flat. $\endgroup$ – user1225 Nov 25 '17 at 3:41
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In general the scheme $\mathcal A[l]_{\mathcal O}$ is not finite because of the following lemma.

Let $f:X\to Y$ be a separated quasi-finite flat morphism of noetherian schemes. Then it is finite iff the fibral rank is locally constant.

For a proof see the paper "Les schémas de modules des courbes elliptiques" by Deligne and Rappoport (II$.1.19$).

So, you actually need just to compute the fibral rank of $\mathcal A[l]_{\mathcal O}$. For example, if $A_K$ is an elliptic curve with additive or multiplicative reduction it is usually not finite. Note that there is a small mistake in Jason Starr's comment, the rank of $\mathcal A[l]_k$ might be bigger than $l$ in the case of multiplicative reduction b/c $\mathcal A[l]_k$ isn't necessary connected.

However, $\mathcal A[l]_{\mathcal O}$ is always quasi-finite provided that $l$ is invertible on $\mathcal O_K$. Moreover, it is always étale and you don't need any semi-stability assumptions for this claim. For a proof look at Lemma 7.3/2 in the book "Néron Models" by Bosch, Lutkebohmert, Raynaud.

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  • $\begingroup$ For the result that is highlighted, either assume $Y$ is connected or replace "constant" with "locally constant". $\endgroup$ – nfdc23 Nov 25 '17 at 4:20
  • $\begingroup$ (also, maybe change "sch´emas" to "schémas") $\endgroup$ – Noam D. Elkies Nov 25 '17 at 4:55
  • $\begingroup$ Oops, I forgot about the component group of the closed fiber. $\endgroup$ – Jason Starr Nov 25 '17 at 5:25
  • $\begingroup$ Thank you very much! For instance, if we take $\mathcal{A}=X_0(11)$, I think $\mathcal{A}[5]$ is finite even though $\mathcal{A}$ has multiplicative reduction at $11$, which is an example. $\endgroup$ – user1225 Nov 25 '17 at 5:52

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