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For a principal unit $u$, apply the Hensel lemma to $F(T)=T^{p-1}-u$. The residue class one is a root, and $F^{\prime }(1)=p-1$ is a unit. Thus there is a unique $v\in 1+{\pi }_K{\mathcal{O}}_K$ withIn particular it is a unit of ${\mathcal{O}}_K$. Choose this $v$ for the $u$ just obtained and put $\alpha ={\pi }_K/v$. Then ${\alpha }^{p-1}=-p$, so ${\mathbb{Q}}_p(\alpha )\subseteq K$. The polynomial $T^{p-1}+p$ is Eisenstein, giving degree $p-1$ for the left-hand field. The p-adic cyclotomic extension $K$ also has degree $p-1$, so

The geometric sum givesSince ${\Phi }_p(1)=p$, its factorization at the nonidentity $p$th roots of unity yieldsEach factor is a unit, and Wilson theorem gives $w\equiv (p-1)!\equiv -1(\mathrm{mod}{\pi }_K)$. Set $u=-w^{-1}$. Then $u$ is a principal unit andThe minus sign comes from the product of the nonzero elements of the residue field, not from an arbitrary choice of uniformizer.

Logarithms on sufficiently deep units. Suppose $\mathrm{char}K=0$, let $\ell$ be the residue characteristic, and put $e=v(\ell )$. For every integer $r>e/(\ell -1)$ the p-adic logarithm and p-adic exponential function, evaluated in $K$, converge on the required domains:The bounds $v(m!)=e{v}_{\ell }(m!)\le e(m-1)/(\ell -1)$ and $v(m)\le e(m-1)/(\ell -1)$ show that, for $x,y\in {\pi }^r{\mathcal{O}}_K$, every term beyond the linear one has valuation strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp (y)\in 1+{\pi }^r{\mathcal{O}}_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphismsThus the logarithm isomorphism on deep principal units isThe right-hand group is torsion-free.

Roots of unity in ${\mathbb{Q}}_p$. For odd $p$, take $\ell =p,e=1,r=1$. The principal unit group $1+p{\mathbb{Z}}_p$ is torsion-free by the logarithm isomorphism on deep principal units. The Teichmuller representative splitting leaves precisely the $p-1$ roots in ${\mu }_{p-1}$.

For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in 1+4{\mathbb{Z}}_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. ConsequentlyThis computes the roots of unity in the p-adic numbers.

The units of ${\mathbb{Q}}_2(i)$. Set $\pi =1-i$. Its polynomial $T^2-2T+2$ is Eisenstein, soThe Eisenstein polynomial facts justifying this are stated in the next solution. Write $U_s=1+{\pi }^s{\mathcal{O}}_K$. Reduction gives ${\mathcal{O}}_K^{\times }=U_1$, and the successive quotients of principal-unit groupshave order two for $s\ge 1$. Hence $[U_1:U_3]=4$. The four roots in ${\mu }_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, sinceThey exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the logarithm isomorphism on deep principal units gives $U_3\cong ({\mathcal{O}}_K,+)$. Therefore the unit decomposition of the 2-adic Gaussian field isIn particular these four elements are all its roots of unity.

Quadratic extensions. A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^{\times }\cong {\pi }^{\mathbb{Z}}\times {\mathcal{O}}_K^{\times }$. Squaring on the decomposition above acts as squaring on ${\mu }_4$ and multiplication by two on the additive ring ${\mathcal{O}}_K={\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2}i$. Thus the square-class group of the 2-adic Gaussian field isBy quadratic extensions from square classes, in characteristic different from two the nontrivial square classes classify quadratic extensions $K(\sqrt{a})$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ quadratic extensions up to $K$-isomorphism.

Use the simple-root form of Hensel lemma: if $F\in {\mathbb{Z}}_p[T]$ and $t_0\in {\mathbb{Z}}_p$ satisfy $F(t_0)\equiv 0(\mathrm{mod}p)$ and $F^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, there is a unique $t\in {\mathbb{Z}}_p$ with $F(t)=0$ and $t\equiv t_0(\mathrm{mod}p)$. More generally the same assertion holds over a complete discrete valuation ring, with its maximal ideal in place of $(p)$.

Every nonzero 3-adic number has a unique expressionIndeed its normalized P-adic valuation fixes $k$, and its unit residue modulo three fixes $ϵ$. Cubing is an automorphism on $\{1,-1\}$, so it remains to compute the cube map on the principal units.

For $t\in {\mathbb{Z}}_3$,Thus $U_1^3\subseteq U_2:=1+9{\mathbb{Z}}_3$. Conversely, given $1+9a\in U_2$, apply Hensel lemma toModulo three it is $T-a$, and $F^{\prime }(T)=1+6T+9T^2$ is always a unit. The residue root $a\mathrm{mod}3$ therefore lifts, giving $t$ with $(1+3t{)}^3=1+9a$. HenceThis is the cube map on 3-adic principal units. Introducing $T$ is essential: applying simple-root lifting directly to $X^3-u$ would fail because its derivative is divisible by three at every unit.

The map $1+3t\mapsto t\mathrm{mod}3$ induces an isomorphism $U_1/U_2\cong (\mathbb{Z}/3\mathbb{Z},+)$, since multiplication adds the first principal-unit digits modulo three. The valuation factor contributes $\mathbb{Z}/3\mathbb{Z}$ independently, so the cube-class group of the 3-adic numbers isExplicit generators are the classes of $3$ and $4=1+3$. Representatives are $3^a{4}^b$ with $0\le a,b<3$. The class of $3$ is detected by valuation modulo three, and that of $4$ by its nonzero first principal-unit digit, so the two generators are independent.

For odd $p$ there is a decompositioninto the valuation factor, the Teichmuller representative factor, and the group of principal units. If $x$ is a $((p-1)m)$th power for every $m$ coprime to $p$, its valuation is divisible by every such $(p-1)m$, and is therefore zero. Its residue in ${\mathbb{F}}_p^{\times }$ is a $(p-1)$st power, hence is one. Thus $x\in 1+p{\mathbb{Z}}_p$.

Conversely, exponentiation by any integer $n$ coprime to $p$ is an automorphism of $1+p{\mathbb{Z}}_p$. This follows either from the principal-unit logarithm, under which it becomes multiplication by $n$, or by applying the Hensel lemma to $Y^n-x$. Since $(p-1)m$ is coprime to $p$, every $x\in 1+p{\mathbb{Z}}_p$ has a $((p-1)m)$th root for every allowed $m$.

Write $x={\pi }^{r}u$ with $r\in \mathbb{Z}$ and $u\in {\mathcal{O}}_K^{\times }$. If $x$ is an $m$th power, then $m$ divides $r$. Divisibility by infinitely many $m$ forces $r=0$, so $x$ is a unit.

Conversely, if $m$ is coprime to both the residue characteristic $p$ and $q-1$, exponentiation by $m$ is an automorphism on the finite residue-unit group and on every principal unit quotient; equivalently, use Hensel's lemma on $Y^m-u$. Hence every unit is an $m$th power for infinitely many such $m$. Thereforeas recorded by elements that are powers of infinitely many degrees in a local field.

In a complete discrete valuation ring of residue characteristic $p$, the power map $u\mapsto u^m$ is a bijection on $1+\mathfrak{m}$ when $p\nmid m$. For a principal unit $a$, the Hensel lemma lifts the simple residue root one of $T^m-a$ uniquely. This can absorb principal-unit factors in tame radical descriptions of local fields.

The roots of unity in a finite extension $K/{\mathbb{Q}}_p$ form a finite group. Each element of their prime-to-$p$ part is a Teichmuller representative, while any $p$-power part lies among the principal units.