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The Lutz–Nagell theorem says that ifhas nonzero discriminant and $P=(x,y)\in E(\mathbb{Q})$ is a torsion point, then $x,y\in \mathbb{Z}$ and either $y=0$ or

For integrality, fix a prime $p$. If a rational point has nonintegral coordinates, its primitive projective coordinates reduce to $O$, so it belongs to the kernel of reduction. The parameter $t=-x/y$ identifies this kernel with the formal group of an elliptic curve over $p{\mathbb{Z}}_p$. The formal logarithm, with the standard separate first-step argument at $p=2$, shows that this group has no nonzero rational torsion. A rational torsion point therefore has nonnegative $p$-adic valuations in both coordinates for every prime $p$, hence integral coordinates.

Suppose now that $y\ne 0$. The point $2P$ is again a nonzero torsion point and hence integral. Its $x$-coordinate is $m^2-2x$, whereThus $m^2$ is an integer. A rational number whose square is integral is integral, so $2y\mid 3x^2+a$ and in particular $y^2\mid (3x^2+a{)}^2$. The curve equation and the identitythen prove $y^2\mid 4a^3+27b^2$. This is the divisibility proof in the Nagell–Lutz theorem.

Let $K^{\mathrm{nr}}$ be the maximal unramified extension of $K$. Because $p\nmid n$, multiplication by $n$ on the special fibre is a separable isogeny and is surjective on $\tilde{E}(\overline{k})$. Choose a point $\tilde{Q}$ with $[n]\tilde{Q}=\tilde{P}$ and lift it, after a finite unramified extension, to $Q_0$. Then $R=P-[n]Q_0$ lies in the kernel of reduction.

On the formal group of an elliptic curve, multiplication by $n$ has the formSince $n$ is a unit in ${\mathcal{O}}_K$, the invertible morphism criterion for formal group laws makes $[n{]}_F$ an automorphism of $\pi {\mathcal{O}}_{K^{\mathrm{nr}}}$. Hence $R=[n]S$ for a unique point $S$ in the kernel of reduction, and $Q=Q_0+S$ satisfies $[n]Q=P$. Moreover, good reduction makes the finite group scheme $E[n]$ étale over ${\mathcal{O}}_K$, so all its points are defined over an unramified extension. Every point of $[n{]}^{-1}P=Q+E[n]$ is therefore unramified, proving that $K([n{]}^{-1}P)/K$ is unramified.

Multiplication by $n$ is already an automorphism of the formal kernel, so the exact reduction sequence shows that $E(K)/nE(K)$ is finite. Choose representatives $P_1,\dots ,P_s$. Each becomes $n$-divisible over a finite unramified extension; their compositum $L/K$ is still finite and unramified. Every class from $E(K)/nE(K)$ then maps to zero in $E(L)/nE(L)$.

Fix a prime $p$. If a rational point $T=(x,y)$ has $v_p(x)<0$, the integral projective Weierstrass equation forcesfor some $m>0$. Hence its formal parameter $t=-x/y$ lies in $p{\mathbb{Z}}_p$, so $T$ belongs to the formal group of an elliptic curve at the identity. This formal kernel has no nonzero torsion for the present equation. Multiplication by an integer prime to $p$ is a formal-group automorphism, while the formal logarithm rules out $p$-power torsion for odd $p$. For $p=2$, inversion sends $t$ to $-t$ because the equation has no $xy$ or $y$ term, and thereforeFor $v_2(t)\ge 1$, its leading term has strictly smaller valuation than every higher term, so $[2{]}_F(t)$ cannot vanish; iterating excludes all two-power torsion as well.

Thus a nonzero torsion point cannot have $v_p(x)<0$. Since this holds for every prime, $x\in \mathbb{Z}$. The integral equation then makes $y^2$ an integer; a rational number whose square is integral is itself integral, so $y\in \mathbb{Z}$. This is the integrality assertion in the Lutz–Nagell theorem.

For a minimal integral Weierstrass equation, let $\tilde{E}$ be the reduced cubic and ${\tilde{E}}_{\mathrm{ns}}(k)$ its nonsingular points, with their induced group law. Define the filtration of elliptic-curve points over a local field byandThe parameter $z=-x/y$ identifies $E_1(K)$ with the formal group of an elliptic curve on $\pi {\mathcal{O}}_K$. Part (a) therefore gives $E_1(K)[n]=0$. Reduction restricts to the exact sequenceIts restriction to $n$-torsion has trivial kernel, yielding the injection

For $E:y^2=x^3+x+1$, the duplication formula isAt $P=(0,1)$, the tangent slope is $1/2$, so $x(2P)=1/4$ and $v_2(x(2P))=-2$. If $v_2(x(Q))=-2m<0$, the unique lowest-valuation terms in the numerator and denominator are respectively $x(Q{)}^4$ and $4x(Q{)}^3$, givingInduction yields $v_2(x(2^{m}P))=-2m$.

For a minimal integral equation, let $t=-x/y$ be the parameter of the formal group of an elliptic curve. Definewhere $E_1$ is the kernel of reduction to the identity. The parameter identifies $E_r$ with the formal group on $p^r{\mathbb{Z}}_p$. For odd $p$, the formal logarithm converges on $p{\mathbb{Z}}_p$ and is an analytic group isomorphism

For $p=2$, the logarithm gives $E_2({\mathbb{Q}}_2)\cong 4{\mathbb{Z}}_2$. The formal duplication series satisfies $[2](T)=2T+O(T^3)$, so $[2](2u)/4\equiv u(\mathrm{mod}2)$; successive lifting makes $[2]:E_1\to E_2$ surjective. Its kernel is rational 2-torsion, but $x^3+x+1$ has no root in ${\mathbb{Q}}_2$ because it has no root modulo $2$. Thus $[2]$ is an isomorphism and

For the filtration of elliptic-curve points over a local field, defineandFor $r\ge 1$, use the formal group of an elliptic curve with parameter $z=-x/y$ and putReduction induceswhile the coefficient of $p^r$ in the formal parameter gives

For $d\in {\mathbb{Q}}^{\times }$ the rational torsion subgroup of $X^3+Y^3+d{Z}^3=0$ is cyclic of order $3$ if $d$ is a cube, cyclic of order $2$ if $d/2$ is a cube, and trivial otherwise.

To bound torsion, use the Weierstrass equation of an elliptic curve $y^2=x^3-432d^2$. At every good prime $\ell \equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_{\ell }$, so the number of points is $\ell +1$. Reduction injects the prime-to-$\ell$ torsion, since multiplication by an integer prime to $\ell$ is invertible in the formal group of an elliptic curve. For any prime $q>3$, the Dirichlet theorem on primes in arithmetic progressions supplies a good $\ell \equiv 2(\mathrm{mod}3)$, $\ell \equiv 1(\mathrm{mod}q)$, excluding $q$-torsion. Good primes $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good primes $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point is killed by $6$.

The flexes of a diagonal cubic give rational three-torsion exactly when $d$ is a cube, and then it has order $3$. Rational two-torsion requires $x^3=432d^2$, whose sole possible rational root exists exactly when $d/2$ is a cube. This equivalence follows by comparing prime valuations modulo $3$, including the valuation at $2$. The two conditions cannot hold together because $2$ is not a rational cube. The asserted classification follows.

A good reduction in characteristic $p$ is supersingular when its geometric $p$-torsion is trivial, equivalently its formal group of an elliptic curve has height two. Good inert primes of a CM elliptic curve are supersingular. The ordinary bounded-measure and torsion-Selmer formulation must then be modified, for instance using signed local conditions in the cyclotomic tower.

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.