In Silverman's AEC, Proposition VII.2.2, the author has a point $(x, y) \in E(K)$, where $K$ is a local field with uniformizer $\pi$ for the maximal ideal of its ring of integers $R$, and the author says, "Since $(x,y)$ reduces modulo $\pi$ to the point at infinity on $\tilde{E}(k)$, we see that $v(x) < 0$ and $v(y) < 0$."
I think this is a basic question but I fail to see why $v(x) < 0$ and $v(y) < 0$. In particular, if the points $(\pi, 1 / \pi)$ is on $E$, then $v(\pi) = 1 > 0$, but $(\pi, 1/\pi) = [\pi : 1 / \pi : 1] = [\pi^2 : 1 : \pi]$ in projective coordinates, and $[\pi^2 : 1 : \pi]$ reduces to $[0 : 1 : 0]$, the point at infinity on $\tilde{E}(k)$. So somewhere I have gone wrong, can anyone point out where?
If either $x$ or $y$ has non-negative valuation, then the other one does, thanks to the normalized Weierstrass equation $y^2 = x^3 + ax + b$. For example, your counterexample can't be a point on an elliptic curve with normalized Weierstrass equation since $y^2$ has negative valuation but the right hand side has non-negative valuation.
But then if both $x$ and $y$ have non-negative valuation, then the point $(x, y)$, which really denotes the point $[x: y: 1]$ in $\mathbb{P}^2$, cannot possibly reduce to $[0: 1: 0]$.