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Let $E$ be an elliptic curve over $\mathbb{Q}_{p}$. Do we know anything about the order of the group $E(\mathbb{Q}_{p})/pE(\mathbb{Q}_{p})$? I know that it's finite, but do we know anything else?

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Are you assuming $E$ has good reduction or anything? –  Matt Jan 26 '13 at 23:57
    
Ah right, I forgot to add that assume that $E$ has good reduction at $p$. –  ljk128 Jan 27 '13 at 0:40
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1 Answer 1

Let $E$ be an elliptic curve over $\mathbb{Z}_p.$ Then attached to $E$ there is a formal group $\hat{E}$ such that the sequence

$$0 \rightarrow \hat{E}(p\mathbb{Z}_p) \rightarrow E(\mathbb{Q}_p) \rightarrow E(\mathbb{F}_p) \rightarrow 0$$

is exact. This induces, by the snake lemma (and the fact that $\hat{E}(p\mathbb{Z}_p)[p] = 0$), an exact sequence

$$0 \rightarrow E(\mathbb{Q}_p)[p] \rightarrow E(\mathbb{F_p})[p] \rightarrow \hat{E}(p\mathbb{Z}_p)/p\hat{E}(p\mathbb{Z}_p) \rightarrow E(\mathbb{Q}_p)/pE(\mathbb{Q}_p) \rightarrow E(\mathbb{F}_p)/pE(\mathbb{F}_p) \rightarrow 0.$$

We'll try to get a handle on the order of $ E(\mathbb{Q}_p)/pE(\mathbb{Q}_p)$ by estimating the other groups which appear in this sequence and then appeal to the fact that the euler charactertic (the alternating sum of $\mathbb{F}_p$-dimensions) of any finite exact sequence of finite dimensional vector spaces is trivial. First, as $E(\mathbb{F}_p)$ is a finite abelian group

$$|E(\mathbb{F}_p)[p]| = |E(\mathbb{F}_p)/pE(\mathbb{F}_p)|$$

so these groups kill each other in our Euler characteristic calculation.

Now let's consider $\hat{E}(p\mathbb{Z}_p)/p\hat{E}(p\mathbb{Z}_p).$ Let $[p]_{\hat{E}}$ be the multiplication by $p$ endomorphism of $\hat{E}.$ The endomorphism $[p]_{\hat{E}}$ is a power series over $\mathbb{Z}_p$ without constant term and with linear term equal to $p.$ Moreover, the coefficientwise reduction of $[p]_{\hat{E}}$ to $\mathbb{F}_p[[X]]$ yields a series which is equal to $X^{p^h} \mod X^{p^h+1}$ where $h$ is either $1$ or $2.$ It follows therefore, if $a \in p\mathbb{Z}_p,$ the Newton polygon of $[p]_{\hat{E}} - a$ has a single segment of negative slope if $v_p(a) = 1$ and two segments of negative slope otherwise. In the former case, the slope of the single segment is $-1/p^h$ and therefore $[p]_{\hat{E}} - a$ has no roots in $p\mathbb{Z}_p$ and in particular $a$ defines a nontrivial element of $\hat{E}(p\mathbb{Z}_p)/p\hat{E}(p\mathbb{Z}_p).$ In the latter case, the first negative slope appearing in the newton polygon of $[p]_{\hat{E}} - a$ has length $1$ and therefore $[p]_{\hat{E}} - a$ has a root in $p\mathbb{Z}_p$ and in particular $a \in p\hat{E}(p\mathbb{Z}_p).$

It follows

$$\hat{E}(p\mathbb{Z}_p)/p\hat{E}(p\mathbb{Z}_p) \cong \hat{E}(p\mathbb{Z}_p/p^2\mathbb{Z}_p) \cong p\mathbb{Z}_p/p^2\mathbb{Z}_p,$$

and hence

$$|\hat{E}(p\mathbb{Z}_p)/p\hat{E}(p\mathbb{Z}_p)| = p.$$

Now, as promised, we apply the theorem about the euler characteristic to the exact sequence above and obtain,

$$dim_{\mathbb{F}_p}(E(\mathbb{Q}_p)/pE(\mathbb{Q}_p)) = 1 - dim_{\mathbb{F}_p}(E(\mathbb{Q}_p)[p]).$$

(N.B. $E(\mathbb{Q}_p)[p] = 0$ unless $E$ is ordinary and $E(\mathbb{Q}_p)[p] = \mu_p \oplus 1)$.

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