TITLE: $p$-adic values of rational points on elliptic curves
QUESTION [6 upvotes]: The following question came up naturally whilst studying diophantine equations: given an elliptic curve $E$ of the form $Y^2 + aY = X^3 +bX^2 + cX + d$ defined over $\mathbb{Q}$, consider the subset $C \subseteq \mathbb{Q}$ of numbers which appear as either the first or the second coordinate of a rational point on $E$. 
If we assume that $E$ has infinitely many points then $C$ is infinite. I would like to understand how 'large' $C$ can get, in particular: can we choose $E$ such that $C$ has unbounded $p$-adic value for all prime numbers $p$? Maybe we can at least choose $E$ such that $C$ has unbounded $p$-adic value for all primes in a given finite set of prime numbers? I know almost nothing about the topic, so any pointers you might have to articles or books studying the set $C$ would be very helpful.

REPLY [4 votes]: I believe most of your questions can be answered using the material from Silverman's 'The Arithmetic of Elliptic curves'. The relevant section for your questions on the $p$-adic valuation of the rational solutions would be the section on elliptic curves over local fields.
Here are some highlights of the theory: if $E$ is an elliptic curve over $\mathbb{Q}_p$ (or any finite extension of such field, but if you're only interested in $\mathbb{Q}$ these suffice) defined by a Weierstrass equation, there is a filtration of subgroups of $E(\mathbb{Q}_p)$:
$$E(\mathbb{Q}_p)\supset E_0(\mathbb{Q}_p) \supset E_1(\mathbb{Q}_p) \supset E_2(\mathbb{Q}_p) \supset \cdots$$
with the following properties:

each of the successive quotients is finite: more precisely $E_r(\mathbb{Q}_p)/E_{r+1}(\mathbb{Q}_p) \simeq (\mathbb{F}_p,+)$ if $r\geq 1$, $E_0(\mathbb{Q}_p)/E_1(\mathbb{Q}_p) \simeq \tilde{E}_{ns}(\mathbb{F}_p)$ where $\tilde{E}_{ns}$ are the nonsingular points on the curve $E$ 'reduced modulo p' (i.e. by looking at the Weierstrass equation modulo $p$) and $E(\mathbb{Q}_p)/E_0(\mathbb{Q}_p)$ is trivial when $E$ has good reduction (i.e. when $E$ modulo $p$ has no singular points) but can be a nontrivial (finite) group when $E$ has bad reduction.
if $p\geq 3$ then there's an isomorphism $E_1(\mathbb{Q}_p) \simeq (\mathbb{Z}_p,+)$ of topological groups. If $p = 2$ then likewise $E_2(\mathbb{Q}_2) \simeq (\mathbb{Z}_2,+)$. 
For $r\geq 1$ we can explicitly describe $E_r(\mathbb{Q}_p)$ as $$E_r(\mathbb{Q}_p) = \left\{(x,y) \in E(\mathbb{Q}_p) \mid  v(x) \leq -2r , v(y) \leq -3r \right\} \cup \{O_E\} $$
(where $O_E$ is the identity of $E$ i.e. the unique point at infinity)
and $$E(\mathbb{Q}_p) \setminus E_1(\mathbb{Q}_p) = \{(x,y) \in E\mid v(x),v(y)\geq 0 \}$$

Now for your questions: if $P \in E(\mathbb{Q}_p)$ is of infinite order and $r\geq 1$ then $n.[E(\mathbb{Q}_p):E_r(\mathbb{Q}_p)].P$ is in $E_r(\mathbb{Q}_p)$ for $n\geq 1$ and it doesn't equal $O_E$. It follows that for every prime $p$ we have that in your notation $S _p =\{v_p(x) \mid x\in C\}$ is never bounded below if the rank of $E$ is positive. 
It seems that the set $S_p$ should be bounded above but I don't see an immediate argument why this is the case. Hope this helps.