Unramified halving fields for a split cubic ($S=\{v:v(2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma ))>0\}$)

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

For distinct integral roots $\alpha ,\beta ,\gamma$ with $m=\max (|\alpha |,|\beta |,|\gamma |)$, the finite set in unramified halving fields for a split cubic contains only primes at most $2m$. A quadratic Galois character unramified away from $s$ primes is represented by a signed square-free integer supported on those primes, giving at most $2^{s+1}$ characters. The halving cocycle with rational two-torsion has two independent coordinates in $\mathbb{Z}/2\mathbb{Z}$, so the quotient has at most $2^{2s+2}$ elements. Since $m\ge 1$ and there are at most $m$ primes at most $2m$, this gives $2^{2m+2}$. The bound is deliberately crude but depends only on the root size.