$C(p^{\nu})$ is a subgroup in Nagell-Lutz theorem I am reading Rational points on elliptic curves by Silverman and Tate. On page 51, they are trying to prove Nagell-Lutz theorem. Especially, they are proving that $C(p^{\nu})$ is a subgroup of $C(\mathbb{Q})$. Let
$$C:y^{2}=x^{3}+ax^{2}+bx+c$$
be an elliptic curve, $a,b,c\in \mathbb{Z}$. Let $p$ be a prime. Define
$$C(p^{\nu})=\{(x,y)\in C(\mathbb{Q}):ord(x)\leq -2\nu \textrm{ and } ord(y)\leq -3\nu\}.$$
(If we write $x=(m/n)p^{\nu}$, where $p\nmid m,n$, the order of $x$ is defined to be $ord(x)=\nu$.) Define $t=x/y$ and $s=1/y$. Define $R=\{\alpha\in\mathbb{Q}:ord(\alpha)\geq 0\}$.
On page 51, it says:

Thus our point $(t,s)$ is in $C(p^{\nu})$ if and only if $t\in p^{\nu}R$ and $s\in p^{3\nu}R$.

Then on the same page,

Let $P_{1}=(t_{1},s_{1})$ and $P_{2}=(t_{2},s_{2})$ be distinct points in $C(p^{\nu})$. If $t_{1}=t_{2}$, then the vertical line $t=t_{1}$ intersects $C$ at $P_{1},P_{2}$, and a third point $P_{3}=(t_{1},s_{3})$, where $P_{3}$ may equal $P_{1}$ or $P_{2}$. Then $P_{1}+P_{2}=(-t_{1},-s_{3})$, so the $t$-coordinate of $P_{1}+P_{2}$ is in $p^{\nu}R$, which shows that $P_{1}+P_{2}\in C(p^{\nu})$.

My question is at the last step. Here we only have $-t_{1}\in p^{\nu}R$. How do we prove $-s_{3}\in p^{3\nu}R$?
I tried to combine the two equations $s=t^{3}+at^{2}s+bts^{2}+cs^{3}$ and $t=t_{1}$, and compute $s_{1}+s_{2}+s_{3}$. I got $s_{3}=-bt_{1}/c -s_{1}-s_{2}$. I am not sure what to do next because I don't know the order of $b$ and $c$.
Edit: the $a,b,c$ are assumed to be integers.
 A: First we fix some notation. For points $P\in C(\mathbb{Q})$ we write $x(P)$ and $y(P)$ to denote the coordinates in the $(x,y)$-plane, likewise $t(P)$ and $s(P)$ for coordinates in the $(t,s)$-plane.
We shall prove that for $P\in C(\mathbb{Q})$ if $\mathrm{ord}(t(P))=\nu$ and $\mathrm{ord}(x(P))<0$ then $P\in C(p^{\nu}).$

Proof: Since $\mathrm{ord}(x(P))<0$ then there exists $k\in\mathbb{N}$ such that $\mathrm{ord}(x(P))=-2k$ and $\mathrm{ord}(y(P))=-3k$ (see e.g. page 48 in the book for a proof of this). Therefore we have $$\nu=\mathrm{ord}(t(P))=\mathrm{ord}\left(\frac{x(P)}{y(P)}\right)=\mathrm{ord}(x(P))-\mathrm{ord}(y(P))=-2k-(-3k)=k.$$ This implies that $\mathrm{ord}(x(P))=-2\nu$ and $\mathrm{ord}(y(P))=-3\nu$, whence $P\in C(p^{\nu})$.

Now we use the result above in your case. We note that $x(P_{1}+P_{2})=-x(P_{1})$ so $$\mathrm{ord}(x(P_{1}+P_{2}))=\mathrm{ord}(-x(P_{1}))=\mathrm{ord}(x(P_{1}))<0$$ since $P_{1}\in C(p^{\nu})$. Likewise we see that $\mathrm{ord}(t(P_{1}+P_{2}))=\nu$, whence we conclude that $P_{1}+P_{2}\in C(p^{\nu})$.
