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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.

For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv 1(\mathrm{mod}8)$. The same conclusion holds for every 2-adic unit by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.

Conversely let $u\in {\mathbb{Z}}_2^{\times }$ with $u\equiv 1(\mathrm{mod}8)$. For digit lifting for square roots of 2-adic units, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u(\mathrm{mod}{2}^n)$, considerFor $n\ge 3$,since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of ${\epsilon }_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u(\mathrm{mod}{2}^{n+1})$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in {\mathbb{Z}}_2^{\times }$ with $x^2=u$.

The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore

Every nonzero P-adic number in ${\mathbb{Q}}_2$ is uniquely $2^{m}u$ with $m\in \mathbb{Z}$ and $u$ a 2-adic unit. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the square-class group of the 2-adic numbers isIts eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.

In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt{d})$ for a nonsquare $d\in K$. If $K(\sqrt{d})=K(\sqrt{e})$, write $\sqrt{e}=a+b\sqrt{d}$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne 0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven quadratic extensions of the 2-adic numbers are exactly

The unique unramified extension in this list is ${\mathbb{Q}}_2(\sqrt{5})$. To verify this, $\alpha =(1+\sqrt{5})/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over ${\mathbb{F}}_2$, so its residue field contains ${\mathbb{F}}_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the residue degree is two and the ramification index is one.

Conversely an unramified quadratic extension has residue field ${\mathbb{F}}_4$. A root there of $X^2+X+1$ lifts by Hensel lemma to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta -1{)}^2=5$, so the extension contains ${\mathbb{Q}}_2(\sqrt{5})$ and must equal it. Every other field in the list is therefore ramified.

For the requested ramification groups, choose $L={\mathbb{Q}}_2(\sqrt{2})$ and $\pi =\sqrt{2}$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi )=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi )=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in {\mathbb{Q}}_2$, and the two summands have valuations of opposite parity:Thus $a+b\pi$ is integral exactly when $a,b\in {\mathbb{Z}}_2$, proving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$ directly.

The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma :v_L(\sigma x-x)\ge i+1\text{ for every }x\in {\mathcal{O}}_L\}$, the ramification in the quadratic extension generated by the square root of two is thereforeThe lower ramification break is two; the valuation used here is the integral normalization on $L$.