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Let $x\in \mathfrak{m}$ and choose any positive integer $n$ coprime to the residue characteristic. Forone has $f(1)\equiv 0(\mathrm{mod}\mathfrak{m})$ and $f^{\prime }(1)=n\not\equiv 0(\mathrm{mod}\mathfrak{m})$. The simple-root form of Hensel lemma produces $y\in K$ with $y^n=1+x$. Infinitely many integers are coprime to the residue characteristic, so $\mathfrak{m}\subseteq S$.

Solved by gpt-5.6-sol high.

One useful form of Hensel lemma is this: if $R$ is a complete discrete valuation ring, $f\in R[X]$, andthen there is a unique $a\in R$ such that $f(a)=0$ and $a\equiv a_1(\mathrm{mod}\pi )$. Indeed, after constructing $a_r$ with $f(a_r)\equiv 0(\mathrm{mod}{\pi }^r)$, choose the unique $t$ modulo $\pi$ for whichand put $a_{r+1}=a_r+{\pi }^{r}t$. The resulting Cauchy sequence converges by completeness, and the same first-order congruence proves uniqueness.

Apply this to $X^{p-1}-1$. Every nonzero class in ${\mathbb{F}}_p$ is a simple root, so it has a unique Teichmuller representative in ${\mathbb{Z}}_p$. These give all roots of unity of order prime to $p$. For odd $p$, the group $1+p{\mathbb{Z}}_p$ has no nontrivial torsion: if $u\ne 1$, then the binomial theorem gives $v_p(u^p-1)=v_p(u-1)+1$, which is incompatible with finite $p$-power order. HenceFor $p=2$, the subgroup $1+4{\mathbb{Z}}_2$ is torsion-free by the same argument, while $-1$ supplies the extra torsion element. Thus $\mu ({\mathbb{Q}}_2)=\{\pm 1\}\cong C_2$. This describes the roots of unity in a p-adic field for $K={\mathbb{Q}}_p$.

Solved by gpt-5.6-sol high.

By local factorization and extended absolute values, extensions of $|\cdot {|}_p$ to the number field $L=\mathbb{Q}(\alpha )$ correspond to the irreducible factors of $X^3-2$ over ${\mathbb{Q}}_p$.

For $p=2$, the polynomial is Eisenstein, hence irreducible, so there is one extension. For $p=3$, a root would be a unit $x$ with $x\equiv -1(\mathrm{mod}3)$, but then $x^3\equiv -1\equiv 8(\mathrm{mod}9)$, not $2(\mathrm{mod}9)$. A reducible cubic has a root, so the polynomial is again irreducible and there is one extension.

For $p=5$, reduction givesThe factors are coprime, and the quadratic has discriminant $3$, a nonsquare modulo $5$. Hensel lemma lifts this as one linear and one irreducible quadratic factor over ${\mathbb{Q}}_5$, giving two extensions. The requested numbers are thereforefor $p=2,3,5$, respectively.

Solved by gpt-5.6-sol high.

One form of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[X]$, and $f(a_1)\equiv 0(\mathrm{mod}\pi )$ while $f^{\prime }(a_1)\not\equiv 0(\mathrm{mod}\pi )$, then there is a unique $\alpha \in R$ with $f(\alpha )=0$ and $\alpha \equiv a_1(\mathrm{mod}\pi )$.

Inductively, if $f(a_n)\equiv 0(\mathrm{mod}{\pi }^n)$, choose $t$ modulo $\pi$ so thatand put $a_{n+1}=a_n+{\pi }^{n}t$. The unit $f^{\prime }(a_n)$ makes $t$ unique. The resulting sequence is Cauchy, so completeness gives a root $\alpha$. Applying the same first-order congruence to two roots proves uniqueness.

Solved by gpt-5.6-sol high.