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For an elliptic curve over a local field with integral Minimal Weierstrass equation, let $E_0(K)$ consist of points with nonsingular reduction and let $E_1(K)$ be the kernel of $E_0(K)\to {\tilde{E}}_{\mathrm{ns}}(k)$. This definition also applies at a prime of bad reduction of an elliptic curve. The point at infinity is always smooth. With $t=-x/y$ and $w=-1/y$, the Weierstrass equation of an elliptic curve becomesThe unit linear coefficient in $w$ gives a unique integral formal power series $w(t)=t^3+O(t^4)$. Then $x=t/w$ and $y=-1/w$, so the nonidentity points of $E_1(K)$ have $v(x)=-2r$ and $v(y)=-3r$ for $r=v(t)>0$. The local parameter $t$ bijects $E_1(K)$ with ${\mathfrak{m}}_K$ and turns its addition into the formal group of an elliptic curve. At good reduction this agrees with the kernel of reduction of an elliptic curve.

The elliptic-curve discriminant is $37$, so the elliptic curve has good reduction at both $2$ and $3$. At a prime $\ell$ of good reduction, the reduction of torsion points on an elliptic curve is injective on torsion of order prime to $\ell$. Here is the formal reason: the kernel of reduction of an elliptic curve is described by an integral formal group law on $\ell {\mathbb{Z}}_{\ell }$, and its multiplication series is $[n](t)=nt+O(t^2)$. If $\ell \nmid n$, the first term has strictly smaller valuation than every higher term for $t\ne 0$. Thus this kernel has no nonidentity point killed by $n$.

Apply this to each prime-primary component of the rational torsion subgroup. A component at $2$ injects into ${\tilde{E}}_3({\mathbb{F}}_3)$, of order $7$, and is therefore trivial. A component at $3$ injects into ${\tilde{E}}_2({\mathbb{F}}_2)$, of order $5$, and is trivial. Every other prime-primary component injects into both groups, so its order divides $\gcd (5,7)=1$. This proves the two-prime bound on rational elliptic torsion in the present case:

Only the prime two can ramify. More precisely, every field $\mathbb{Q}(P_n)$ is unramified at every odd prime; actual ramification at two depends on $P$ and the chosen division point. Use $[2^n]P_n=P$, as printed in the original PDF.

Fix an odd prime $p$ and put $N=2^n$. The elliptic-curve discriminant $-2^8$ gives good reduction at $p$, and $N$ is a unit in ${\mathbb{Z}}_p$. The unramified division torsors at good primes imply that all division points of an elliptic curve above $P$ are defined over finite unramified extensions of ${\mathbb{Q}}_p$. Here is a direct lifting proof, which does not require $P$ to have integral affine coordinates.

Properness gives a reduction $\tilde{P}$. Choose a point $\overline{Q}$ with $[N]\overline{Q}=\tilde{P}$ over the algebraic closure of the residue field, and a finite residue extension containing it. Pass to the corresponding finite unramified extension $K/{\mathbb{Q}}_p$. The surjectivity of good reduction over a local field lifts $\overline{Q}$ to $Q_0\in E(K)$. Its error $[N]Q_0-P$ lies in the kernel of reduction of an elliptic curve. The prime-to-residue-characteristic multiplication on a formal group makes $[N]$ bijective on that kernel. Choose the unique kernel point $R$ with $[N]R=[N]Q_0-P$; then $Q=Q_0-R$ satisfies $[N]Q=P$ and has reduction $\overline{Q}$.

Every algebraic division point has some such reduction and is the unique lift in its division fibre: the difference of two lifts lies in the reduction kernel, which has no $N$-torsion. Thus every chosen $P_n$ lies in an unramified local extension. For every prime of its possibly non-Galois coordinate field above $p$, the ramification index is one. Globally,One cannot universally assert that two actually ramifies: $P=O$ and $P_n=O$ give the field $\mathbb{Q}$. It can ramify. For example, a half of $T=(-1,0)$ isThe elliptic curve group law verifies this by doubling. Its coordinate field contains $\mathbb{Q}(\sqrt{2})$, which has ramification in the quadratic extension generated by the square root of two, so the only possible bad prime does occur for this choice.

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.

Double $2Q=(12,52)$ using the tangent formula. The slope issoThis finite point is neither $P$ nor $-P$, since its abscissa is nonzero. Therefore $P\pm 4Q$ are both finite nonidentity points.

For a direct exact-coordinate proof, the line from $P$ to $4Q$ has slope $985/104$, while the line from $P$ to $-4Q=(264/169,-3944/2197)$ has slope $-493/429$. For a secant through $P=(0,0)$ and $R=(u,v)$, its slope is $m=v/u$, and the group law gives $x(P+R)=m^2-1-u$, $y(P+R)=13-mx(P+R)$. Simplification yieldsThe first abscissa has odd numerator and denominator $64$. The second has denominator $1089=3^2\cdot 11^2$ and numerator not divisible by either $3$ or $11$. Thus neither point has integral coordinates.

Reduction also explains the obstruction without requiring the full secant arithmetic. At the good prime $2$, $\overline{Q}=(0,1)=-\overline{P}$, and $\overline{P}$ has order three, so $\overline{P+4Q}=O$. At the good prime $11$, the displayed coordinates of $4Q$ reduce to $(0,0)=\overline{P}$, so $\overline{P-4Q}=O$. A finite point with integral affine coordinates reduces to an affine point, not to the identity at infinity. Membership in the kernel of reduction of an elliptic curve therefore certifies the same two nonintegrality conclusions, at primes $2$ and $11$ respectively.

Reduce $P$ to $\overline{P}\in \tilde{E}({\mathbb{F}}_p)$. Because $p\nmid n$, the multiplication-by-n morphism on the reduced elliptic curve is a separable, surjective morphism of degree $n^2$. Choose a point $\overline{Q}\in \tilde{E}({\overline{\mathbb{F}}}_p)$ with $n\overline{Q}=\overline{P}$. Its coordinates lie in some finite field ${\mathbb{F}}_{p^d}$.

Let $L/{\mathbb{Q}}_p$ be the finite unramified extension with residue field ${\mathbb{F}}_{p^d}$. The curve retains good reduction over $L$. Its smooth integral model and the Hensel lemma give surjectivity of good reduction over a local field, so lift $\overline{Q}$ to some $Q_0\in E(L)$. This lift need not satisfy $n{Q}_0=P$ exactly, but its errorhas zero reduction and belongs to $E_1(L)$, the kernel of reduction of an elliptic curve.

The parameter $t=-x/y$ at infinity identifies $E_1(L)$ with the maximal ideal ${\mathfrak{m}}_L$ equipped with the formal group of an elliptic curve. The integral power series for $[n]$ and its inverse from part (i) remain integral over ${\mathcal{O}}_L$. They converge on ${\mathfrak{m}}_L$: terms of degree $k$ tend to zero in valuation because their coefficients are integral and the argument has positive valuation. Hence multiplication by $n$ is bijective on $E_1(L)$. Choose its unique solution $T\in E_1(L)$ to $nT=R$, and put $Q=Q_0+T$. ThenThe residue-field extension supplies a division point modulo $p$, and the formal-group inverse corrects the error without introducing ramification. This is the local construction underlying the uniform unramified division field over a local field and unramified division torsors at good primes.

We use two precise facts about the formal group of an elliptic curve. At a prime $p$ of good reduction, its kernel of reduction of an elliptic curve is identified by the uniformizer $t=-x/y$ with the group $\hat{E}(p{\mathbb{Z}}_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral invariant differential of a formal group law has the form $(1+{\sum }_{j\ge 1}{c}_j{T}^j)dT$, with $c_j\in {\mathbb{Z}}_p$. Integrating constructs the formal logarithmIt is a group homomorphism to the additive group. For $0\ne t\in p{\mathbb{Z}}_p$ and $j\ge 2$,when $p$ is odd. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational torsion subgroup at an odd prime of good reduction, including its $p$-primary part.

For the present elliptic curve, $\Delta =64D^6$, so every odd $p\nmid D$ is a prime of good reduction. If $p\equiv 3(\mathrm{mod}4)$, the Legendre symbol of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the Legendre symbols of $x^3-D^{2}x$, yieldingThe rational torsion subgroup injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.

For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies infinitely many $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$. Discard the finitely many dividing $D$. Since $\ell \nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv 3(\mathrm{mod}8)$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid 4$.

There are already four rational 2-torsion points,which are distinct because a squarefree integer $D$ is nonzero. ThusIts order is four. In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

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 $t\in \pi {\mathcal{O}}_K$.

If an elliptic curve has good reduction over a local field $K$, its integral projective model is smooth. Every point over the residue field lifts by the Hensel lemma, applied in a chart with a unit partial derivative. Thus reduction gives an exact sequence $0\to E_1(K)\to E(K)\to \tilde{E}(k)\to 0$. The kernel of reduction of an elliptic curve is the maximal ideal of the valuation ring, with the formal group of an elliptic curve as its group operation.

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.