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Choose an integral Minimal Weierstrass equation over ${\mathbb{Z}}_p$. Let $E_0({\mathbb{Q}}_p)$ be the points with nonsingular reduction, and let $E_1({\mathbb{Q}}_p)$ be those reducing to $O$. This distinction matters at bad reduction of an elliptic curve: the whole singular reduced cubic is not a group. The smooth locus is a group under the induced elliptic curve group law, and Hensel's lemma lifts each of its rational points, givingThe formal kernel of a minimal Weierstrass equation identifies $E_1({\mathbb{Q}}_p)$ with $p{\mathbb{Z}}_p$ equipped with $F$. Indeed, the integral series $w(t)$ converges there, and $x=t/w(t)$, $y=-1/w(t)$ give the unique point for each nonzero parameter. If $v_p(t)=r>0$, then $v_p(x)=-2r$ and $v_p(y)=-3r$; the parameter zero gives $O$. Conversely a point reducing to $O$ has these local coordinates of positive valuation, so is recovered by the series.

Define $E_m$ for $m\ge 1$ by $t\in p^m{\mathbb{Z}}_p$. The linear part of the formal group law is addition, and the higher terms have degree at least two. Hence the filtration of elliptic-curve points over a local field hasAt good reduction, $E_0=E$ and $E/E_1\cong \tilde{E}({\mathbb{F}}_p)$. At bad reduction, there is also the finite quotient $E/E_0$: $E({\mathbb{Q}}_p)$ is a compact group because the curve is projective, and $E_0$ is an open subgroup, so its index of a subgroup is finite. Together these quotients and the formal filtration describe the group, including the extra components at bad primes.

The formal logarithm makes the deepest part explicit. Write $L(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$. Since its derivative is the integral invariant differential of a formal group law, $n{c}_n\in {\mathbb{Z}}_p$. ConsequentlyThe inequality $v_p(n)\le (n-1)/(p-1)$ shows that for an integer $m>1/(p-1)$ every nonlinear term on $p^m{\mathbb{Z}}_p$ has strictly higher valuation than the linear term. After scaling, $L$ is the identity plus a strictly contracting series; solving $L(t)=u$ by successive correction gives its inverse formal group exponential. Thus the deep logarithm subgroup of a formal group givesOne may take $m=1$ for odd $p$ and $m=2$ for $p=2$. In particular, $E({\mathbb{Q}}_p)$ contains an open subgroup isomorphic to the additive p-adic integers, and the remaining information is in finite quotients and their group extensions.

Let $E_r({\mathbb{Q}}_2)$ be the $r$th term in the filtration of elliptic-curve points over a local field. Reduction givesof order four, whilehas order two. The formal logarithm is injective on $E_2({\mathbb{Q}}_2)$ and identifies it with an additive subgroup of ${\mathbb{Q}}_2$, so $E_2({\mathbb{Q}}_2)$ is torsion-free. A finite subgroup of $E({\mathbb{Q}}_2)$ therefore injects into $E({\mathbb{Q}}_2)/E_2({\mathbb{Q}}_2)$, whose order is $4\cdot 2=8$. Thus $|E({\mathbb{Q}}_2{)}_{\mathrm{tors}}|$ divides $8$.

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 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 a locally compact group $G$ on which multiplication by $n$ has finite kernel and cokernel, compare a Haar measure with its pushforward under $[n]$. On the one-dimensional $p$-adic Lie group $E({\mathbb{Q}}_p)$, the derivative of $[n]$ at the identity is $n$, henceThis can also be read directly from the successive quotients in the filtration of elliptic-curve points over a local field; factors prime to $p$ act invertibly on a sufficiently small formal-group neighbourhood.

The real Lie group $E(\mathbb{R})$ has one or two circle components. On its identity component, $[n]$ has degree $n$; the component-group kernel and cokernel have the same order. ThereforeOnly primes dividing $n$ contribute to the finite-place product, and unique factorization gives

Let $E/{\mathbb{Q}}_p$ have good reduction, and choose an integer $m>1/(p-1)$. Its filtration of elliptic-curve points over a local field has a torsion-free group $E_m$ identified by the formal logarithm with $p^m{\mathbb{Z}}_p$. The quotient group $G=E({\mathbb{Q}}_p)/E_m$ is finite. For $g\in G$ of order of a group element $N$, choose a lift $P$. There is a torsion point of an elliptic curve in this coset exactly when the displayed condition holds. Indeed, it is equivalent to the existence of $Q\in E_m$ with $[N]Q=[N]P$. Then $P-Q$ is killed by $N$; conversely every torsion representative has order $N$, since $E_m$ is torsion-free. The logarithmic division and the formal group exponential give the unique such $Q$. Changing the lift changes the logarithm by an element of $N{p}^m{\mathbb{Z}}_p$, so the criterion is well-defined. This finite-coset test includes residue-characteristic torsion; only the prime-to-p torsion lifts uniquely at good reduction are automatic.