I read an article where it is said: $E_1(\mathbb{Q}_p)\cong \mathbb{Z}_p$ where $E$ is an elliptic curve over $\mathbb{Q}_p$ and $E_1(\mathbb{Q}_p)=\{P\in E(\mathbb{Q}_p):\tilde{P}=\tilde{O}\}$.

The author says that the proof is in "Arithmetic of elliptic curves" by J. Silverman, at page 191, but there it is said:

If $E$ is an elliptic curve over $\mathbb{Q}_p$ and $\hat{E}$ is the formal group, then:

$$E_1(\mathbb{Q}_p)\cong \hat{E}(p\mathbb{Z}_p)$$

So I don't know a good reference for the proof of $E_1(\mathbb{Q}_p)\cong \mathbb{Z}_p$.

  • 5
    $\begingroup$ The isomorphism between the formal group (the right-hand side in your displayed equation) and the p-adic integers must be in Chapter 4 of the same book, towards the end. It may be false for p = 2, by the way. $\endgroup$
    – RP_
    May 15 '20 at 15:52
  • $\begingroup$ Crossposted at MSE. When cross-posting, it is important to link all versions of the question to prevent needlessly duplicating work. $\endgroup$
    – KReiser
    May 22 '20 at 23:52

As RP says, there's a chapter in The Arithmetic of Elliptic Curves that discusses formal groups, and in particular the points of a formal group defined over a complete local ring. The specific result that you want is Chapter IV, Theorem 6.4(b), in the special case that $K=\mathbb Q_p$ and $R=\mathbb Z_p$ and $\mathcal M=p\mathbb Z_p$. That theorem says that there the formal logarithm gives an isomorphism $$ \log_{\mathcal F} : \mathcal F(\mathcal M^r) \longrightarrow \hat{\mathbb G}_a(\mathcal M^r), $$ provided that $r$ is an integer satisfying $r>v(p)/(p-1)$. For your case, $v(p)=1$, so the isomorphism is valid for all $r\ge1$ except, as noted by RP, when $p=2$, in which case you'll need $r\ge2$. And indeed, for $p=2$ you may need $r\ge2$, since there are formal groups over $\mathbb Z_2$ in which $\mathcal F(2\mathbb Z_2)$ has an element of order 2, hence it cannot possibly be isomorphic to the additive group.

  • $\begingroup$ thank you very much!! $\endgroup$ May 16 '20 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.