Why $E_1(\mathbb{Q}_p)\cong\mathbb{Z}_p$

Question

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 $E_1(\mathbb{Q}_p)\cong \mathbb{Z}_p$.

Answer 1

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.