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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 a one-dimensional commutative integral formal group law, the formal logarithm identifies $F(4{\mathbb{Z}}_2)$ with $4{\mathbb{Z}}_2$. This is an index-two subgroup of $F(2{\mathbb{Z}}_2)$. Choose a topological generator $h$ of the subgroup and an element $v$ outside it, and write $2v=ch$. If $c$ is odd, $v$ generates the whole group. If $c$ is even, $v-(c/2)h$ has order two and splits off a cyclic factor. The formal additive group and formal multiplicative group realize the two cases.

For an integral short equation $y^2=x^3+ax+b$, negation acts on $t=-x/y$ by $t\mapsto -t$. Thus the integral formal multiplication series is odd: $[2](t)=2t+t^{3}g(t)$. If $v_2(t)=s\ge 1$, the leading term has valuation $s+1<3s$, so successive doubling never kills a nonzero point. Odd multiplication has unit linear coefficient. Combined with the formal logarithm at odd primes, this proves torsion-freeness of the formal identity neighbourhood at every prime for a short equation. The short-model hypothesis matters at two.

For odd $p$, a one-dimensional formal group law over ${\mathbb{Z}}_p$ has a formal logarithm ${\log }_F(T)=T+{\sum }_{j\ge 2}{c}_j{T}^j/j$ with $c_j\in {\mathbb{Z}}_p$. This follows by integrating its integral invariant differential. For $t\in p{\mathbb{Z}}_p\setminus \{0\}$, $v_p(c_j{t}^j/j)>v_p(t)$ for $j\ge 2$. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so its kernel is zero. As it is a homomorphism into a characteristic-zero additive group, there is no nonzero torsion. In particular, at an odd prime of good reduction over $\mathbb{Q}$, reduction injects the whole rational torsion subgroup, including its $p$-primary part.

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.