 Solution

The needed Hensel lemma says that if $f\in {\mathcal{O}}_K[X]$ and $a\in {\mathcal{O}}_K$ satisfy $f(a)\equiv 0(\mathrm{mod}\pi )$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\pi )$, then $a$ lifts uniquely to a root of $f$ in ${\mathcal{O}}_K$ with the prescribed residue. At every affine point of the smooth curve $\tilde{E}$, one partial derivative of its Weierstrass equation is nonzero. Fixing the other coordinate and applying Hensel's lemma lifts that point to $E(K)$; $O_E$ lifts itself. Thus reduction is surjective.

Use the local parametersso $x=t/z$ and $y=-1/z$. Substitution in a general integral Weierstrass equation givesFor fixed $0\ne t\in \pi {\mathcal{O}}_K$, the difference between the two sides, viewed as a polynomial in $z$, is congruent to $z$ modulo $\pi$ and has derivative congruent to one. Hensel's lemma gives a unique $z\in \pi {\mathcal{O}}_K$, and therefore a unique point $\theta (t)=(t/z,-1/z)$ in the kernel of reduction of an elliptic curve. Together with $\theta (0)=O_E$, this identifies that kernel with the parameters $t\in \pi {\mathcal{O}}_K$.