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Let $E$ be an elliptic curve on $\Bbb Q_p$. $E_0(\Bbb Q_p)$ is points of $E(\Bbb Q_p)$ reduced to nonsingular points. How to prove #$E(\Bbb Q_p)$＝$∞$ directly ?

According to Silverman's book 'the arithmetic of elliptic curves', using formal group theory and finiteness of $[E(\Bbb Q_p):E_0(\Bbb Q_p)]＜∞$, he proves $E(\Bbb Q_p)$ contains subgroup which is isomorphic to $\Bbb Z_p^＋$, so in particular, it has continuous cardinality.

But is it difficult to prove #$E(\Bbb Q_p)$＝$∞$(continuous cardinarity) directly from definitions ?

In my intuition, $\Bbb Q_p$ is close to $\Bbb R$, so rational points seems continuouslike, so it's rational points has continuous cardainarity.

I posted this question to mathstack, but I couldn't get reaction. That is why I'm asking this here. If this is not suitable for this place, please answer in mathstack form. https://math.stackexchange.com/questions/4272525/how-to-prove-e-bbb-q-p＝∞-directly

Thank you in advance.

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    $\begingroup$ The proof using the formal group, is nothing but the implicit function theorem. The point $O$ is non-singular, so locally the points on the elliptic curve close to $O$ are in bijection with an open of $0\in\mathbb{Q}_p$, which is $\mathbb{Z}_p$ up to scaling. With the compactness of $E$, one can show that a finite number of such opens cover $E(\mathbb{Q}_p)$. $\endgroup$
    – Chris Wuthrich
    Commented Oct 10, 2021 at 19:50
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    $\begingroup$ Isn't Hensel's Lemma alone enough to show that there are infinitely (even uncountably) many Q_p-points that reduce to any given F_p-point? $\endgroup$
    – R.P.
    Commented Oct 10, 2021 at 19:53
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    $\begingroup$ When considering cross posting from Math SE you should really wait at least a couple days, not just half a day. In just 12 hours there is a lot of people who simply wouldn't get a chance to see the question on the original site. $\endgroup$
    – Wojowu
    Commented Oct 10, 2021 at 23:58
  • $\begingroup$ $y^2 = x^3 + Ax + B$, for $x\in \mathbb{Z}_p^\times$ the condition that $x^3 + p^{4000}\cdot Ax + p^{6000}\cdot B$ be a square (thus correspond to a $\mathbb{Q}_p$-point with $X$-coordinate $x/p^{2000}$) is a congruence mod $p^{1000}$ (mod $p$ if $p > 2$, mod $8$ if $p = 2$), there are such $x$’s in $\mathbb{Z}/p^{1000}$ (e.g. $1$), so everybody in $\mathbb{Z}_p^\times$ reducing to such an $x$ gives such a point. (Exactly as @RP_ said! @ ChrisWuthrich ‘s answer is a much more conceptual viewpoint of course.) $\endgroup$
    – alpoge
    Commented Oct 11, 2021 at 0:21

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