= Unramified halving fields for a split cubic
{title2=$S=\{v:v(2(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma))>0\}$}
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>.
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