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Let $C$ an elliptic curve over $\mathbb Q$. Assume that the rank of $C(ℚ)$ is equal to $r$. Then the cardinality of a maximal independent set in $C(ℚ)$ is $r$, thus there exists $r$ independent points ${P_1,P_2,\dots,P_r}$ of infinite order in $C(ℚ)$, i.e., $P_{k}=(x_{k},y_{k})\in\mathbb Q^2,\ k=1,..,r$ (we ignor the torsion part of $C(ℚ)$ to use the canonical height) such that if $∑_{k=1}^{r}α_{k}P_{k}=0$, then $α_{k}=0$ for all $k=1,..,r$.

My question is: Is it possible to say that every point $P$ in $C(ℚ)$ other than ${P_1,P_2,\dots,P_r}$ is of finite order?

Safwane
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  • But the multiples of those infinite order elements will also be of infinite order. – Tobias Kildetoft Apr 02 '13 at 14:02
  • So, how I can prove this result. – Safwane Apr 02 '13 at 14:03
  • What result? That any multiple of an element of infinite order has infinite order? – Tobias Kildetoft Apr 02 '13 at 14:05
  • Yes, this is the result . – Safwane Apr 02 '13 at 14:06
  • This is a general result about groups (so nothing to do with this coming from elliptic curves), and an easy exercise. – Tobias Kildetoft Apr 02 '13 at 14:11
  • Ok, and thank you very much. – Safwane Apr 02 '13 at 14:12
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    The group $C(\Bbb Q)$ is isomorphic to a group $T\times \Bbb Z^r$, where $T$ is a finite group. Then, almost tautologically, any element in the infinite set $T\times \Bbb Z^r$ which is not in the finite set $T\times 0$ has infinite order. (Else such an element would have finite order, thus being in $T\times 0$. If the second component is not $0$, then its projection on $\Bbb Z^r$ also has finite order, contradiction.) – dan_fulea Jul 11 '21 at 12:18

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