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Let $L/{\mathbb{Q}}_p$ be finite, with ring of integers ${\mathcal{O}}_L$ and maximal ideal ${\mathfrak{m}}_L$. The formal group of an elliptic curve on an integral Weierstrass model is a formal group law $F(X,Y)\in {\mathcal{O}}_L[[X,Y]]$, and $\hat{E}(L)$ denotes its points with parameter in ${\mathfrak{m}}_L$. Its multiplication series isSince $p\nmid m$, the linear coefficient $m$ is a unit. Construct a compositional inverse $g(T)=m^{-1}T+{\sum }_{j\ge 2}{b}_j{T}^j$ recursively: at degree $j$, the coefficient of $b_j$ in $[m{]}_F(g(T))$ is $m$, so the equation making that degree vanish uniquely determines $b_j\in {\mathcal{O}}_L$. The inverse on the other side is the same series, by uniqueness of formal composition inverses.

Both series converge on ${\mathfrak{m}}_L$, since their coefficients are integral and powers of an element of ${\mathfrak{m}}_L$ tend to zero. Their formal identities therefore hold as identities of convergent values. Thus for every $P$ there is a unique $Q$ with $[m]Q=P$, andThis proves unique divisibility for every finite extension, including ramified extensions and negative $m$. It uses prime-to-residue-characteristic multiplication on a formal group, not an unjustified logarithm isomorphism on the entire maximal ideal.

Now assume good reduction of an elliptic curve. For a finite local extension $M$, the reduction sequence isProperness defines reduction for all points, smoothness and the Hensel lemma make it surjective, and its kernel is precisely the formal group. Let $K$ be the maximal unramified extension and take $P\in E(K)$. Its coordinates lie in a finite unramified extension $L/{\mathbb{Q}}_p$. Over the algebraic closure of the residue field, the isogeny $[m]$ is surjective, so choose $\overline{Q}$ with $[m]\overline{Q}=\tilde{P}$. This residue point is defined over a finite residue extension. Choose the corresponding finite unramified extension $M/L$ inside $K$, and lift it to $Q_0\in E(M)$ by smooth reduction.

Then $P-[m]Q_0$ lies in the reduction kernel. The first argument gives a unique kernel point $R$ with $[m]R=P-[m]Q_0$. Hence $Q=Q_0+R\in E(M)\subseteq E(K)$ satisfies $[m]Q=P$. ConsequentlyEvery lifting and convergence argument was made in a finite complete extension; $K$ itself need not be complete. Unique divisibility is asserted only for the formal group. For $|m|>1$, good reduction also lifts the prime-to-$p$ torsion, so $E(K)$ has nonzero $m$-torsion and division there is not unique. The construction is unramified division points at good reduction.

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

An elliptic curve has good reduction of an elliptic curve at $p$ if it admits a Weierstrass equation of an elliptic curve over ${\mathbb{Z}}_p$ whose reduced projective cubic is nonsingular. Equivalently its minimal discriminant is a $p$-adic unit.

For the given integral short equation,It therefore directly supplies good reduction outside $2,3,5$. These three primes cannot be rescued by changing the model. An admissible change of Weierstrass model changes the discriminant by a twelfth power, so its valuation changes by a multiple of twelve. The displayed valuations $4,7,4$ are not congruent to zero modulo twelve. No integral model can have unit discriminant at any of those primes. The good primes are exactly $p\notin \{2,3,5\}$.

In fact, $2,3,5,7$ must all be primes of bad reduction.

At a prime of good reduction of an elliptic curve, the kernel $E_1({\mathbb{Q}}_p)$ of reduction is the formal group of an elliptic curve evaluated on $p{\mathbb{Z}}_p$, using its parameter at the identity. This kernel has no prime-to-$p$ torsion: for an integer $\ell$ prime to $p$, the multiplication series is $[\ell {]}_F(T)=\ell T+$ higher-degree integral terms, so for nonzero $T\in p{\mathbb{Z}}_p$ its first term has strictly smaller valuation than the others and $[\ell {]}_F(T)\ne 0$.

For each odd good prime, the entire given torsion subgroup, of order $16$, would therefore inject into $E({\mathbb{F}}_p)$. But the Hasse theorem for elliptic curves givesNone can contain a subgroup of order $16$. Hence $3,5,7$ are bad.

For the prime-two torsion bound for good reduction, use the deeper formal subgroup corresponding to $4{\mathbb{Z}}_2$. Part (i), with $m=2>1/(2-1)$, shows that $E_2({\mathbb{Q}}_2)$ is torsion-free and has index $2$ in $E_1({\mathbb{Q}}_2)$. Therefore any finite torsion subgroup of $E_1({\mathbb{Q}}_2)$ injects into $E_1/E_2$ and has order at most $2$.

If $E$ had good reduction at $2$, its rational torsion subgroup of order $16$ would consequently have reduction image of order at least $8$. The Hasse theorem for elliptic curves instead gives $\#E({\mathbb{F}}_2)\le 5$. This contradiction proves bad reduction at $2$ as well, and thereforeThe special formal-group argument at $2$ is necessary; prime-to-$p$ injectivity alone says nothing about the given $2$-primary torsion there.

The curve has good reduction when it admits a Minimal Weierstrass equation over ${\mathcal{O}}_K$ whose discriminant is a unit, so reduction modulo $\pi$ is a nonsingular elliptic curve $\tilde{E}/k$. Write a point of $E(K)$ in primitive projective coordinates $(X:Y:Z)$ over ${\mathcal{O}}_K$. Reducing all three coordinates definesPrimitivity ensures that the reduced triple is not zero. The addition morphism on the smooth Weierstrass model reduces to the addition morphism on $\tilde{E}$; equivalently, this follows from the valuative criterion for properness for the smooth proper group scheme. Therefore $\mathrm{red}(P+Q)=\mathrm{red}(P)+\mathrm{red}(Q)$.

The discriminant is $1909=23\cdot 83$, so $3,5,7$ are primes of good reduction of an elliptic curve. The reduction of torsion points on an elliptic curve injects the prime-to-$p$ torsion into $\tilde{E}({\mathbb{F}}_p)$.

If a prime $\ell$ divided the order of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, its primary subgroup would inject at every one of $3,5,7$ except possibly when $p=\ell$. For $\ell =3$, the counts at $5$ and $7$ have greatest common divisor one; for $\ell =5$ use the counts at $3$ and $7$; for $\ell =7$ use those at $3$ and $5$; every other $\ell$ would divide all three counts. Each possibility is excluded. Hence

An integral Weierstrass equation has good reduction outside the finitely many primes dividing its nonzero discriminant. This proves finiteness of the set of bad primes. To prove finiteness of rational torsion, choose two distinct good primes. The reduction of torsion points on an elliptic curve injects each primary component at a good prime of different residue characteristic, so the two finite reduced point groups bound every primary component of $E(\mathbb{Q}{)}_{\mathrm{tors}}$.

Forthe displayed equation is minimal andIts bad primes are therefore exactlyThe good reductions at $11$ and $13$ haveTheir coprime orders exclude every rational torsion primary component, including the residue-characteristic components by using the other prime. Hence

For good reduction of an elliptic curve over a complete local field of residue characteristic $p$, smoothness and the Hensel lemma make reduction surjective. Its kernel is a formal group on the maximal ideal, where multiplication by $q^m$, $q\ne p$, is bijective by prime-to-residue-characteristic multiplication on a formal group. Thus there is no nonzero $q$-power torsion in the kernel. Given a lift $P$ of a reduced point killed by $q^m$, choose the unique kernel point $R$ with $q^{m}R=q^{m}P$. Then $P-R$ is a torsion lift. This proves both surjectivity and uniqueness of torsion lifting. For finite residue field these groups are finite.

At a prime of good reduction of an elliptic curve, the reduction map is a group homomorphism on all local points. If $N$ annihilates the reduced finite group, every $NP$ reduces to the identity at infinity. A finite point with integral affine coordinates reduces to an affine point, so a nonidentity $NP$ cannot have integral affine coordinates. The identity has no affine coordinates and must be treated separately.

At a prime of good reduction of an elliptic curve with residue characteristic not dividing $m$, multiplication by $m$ extends to a finite étale map of the smooth proper elliptic group scheme over the valuation ring. A local rational point extends to a section by properness. Its division fibre is therefore a finite étale torsor, so the Kummer map of an elliptic curve gives an unramified cohomology class. This bounds ramification in the Weak Mordell-Weil theorem.

A direct proof describes the coordinates, not only the cohomology class. Reduce $P$ by properness and choose a point $\overline{Q}$ with $[m]\overline{Q}=\tilde{P}$ over a finite extension of the residue field. Over the corresponding unramified extension of the complete local field, lift $\overline{Q}$ by the Hensel lemma. The error $[m]Q_0-P$ is in the kernel of reduction of an elliptic curve; prime-to-residue-characteristic multiplication on a formal group corrects it uniquely. There is no $m$-torsion in that kernel, even after a ramified field extension, so every division point with that reduction equals this unramified lift. Consequently all division points of an elliptic curve above $P$ have unramified coordinate fields. The torsion case is discussed in J. S. Milne's Elliptic Curves, Chapter V, Proposition 8.1.