= Solution
The needed <Hensel lemma> says that if $f\in\mathcal O_K[X]$ and $a\in\mathcal O_K$ satisfy $f(a)\equiv0\pmod\pi$ and $f'(a)\not\equiv0\pmod\pi$, 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 $\widetilde E$, 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 parameters
$$
t=-x/y,
\qquad z=-1/y,
$$
so $x=t/z$ and $y=-1/z$. Substitution in a general integral Weierstrass equation gives
$$
z=t^3+a_1tz+a_2t^2z+a_3z^2+a_4tz^2+a_6z^3.
$$
For 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 $t\in\pi\mathcal O_K$.
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