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For a prime $p$ and $n\ge 1$, the unique residue $y$ modulo $p^n$ satisfying $y\equiv x(\mathrm{mod}p)$ and $y^p\equiv y(\mathrm{mod}{p}^n)$ is the displayed residue. Existence follows from Fermat's little theorem and prime-power contraction under pth powers. Uniqueness follows by raising any other such residue repeatedly to the $p$th power and applying the same contraction to its congruence with $x$. This is the finite-modulus version of a Teichmuller representative, including the zero class.

If a non-Archimedean local field $K$ has uniformizer $\pi$, valuation ring ${\mathcal{O}}_K$, maximal ideal ${\mathfrak{m}}_K$ and residue field ${\mathbb{F}}_q$, thenThe first factor records the discrete valuation, the Teichmuller representatives split reduction ${\mathcal{O}}_K^{\times }\to {\mathbb{F}}_q^{\times }$, and $1+{\mathfrak{m}}_K$ is the group of principal units.

The preceding constructions lead to the Iwasawa theorem on local cyclotomic units. Its significance is that an explicitly calculated local unit quotient is governed by a p-adic L-function; class field theory then suggests that the same analytic object should govern global arithmetic. The latter prediction is the Iwasawa main conjecture, rather than a consequence of the local calculation alone.

Fix compatible roots of unity ${\zeta }_{p^n}$ and put $K_n={\mathbb{Q}}_p({\zeta }_{p^n})$, ${\pi }_n={\zeta }_{p^n}-1$. Let $V_{\infty }$ be the norm limit of the local principal units. Take its even part under complex conjugation to obtain $U_{\infty }^1$, the local principal-unit limit in the real tower. Let $C_{\infty }^1$ be the closed subgroup generated by norm-compatible real cyclotomic units, with their finite-order residue factors removed. WriteHere the last map is augmentation. Normalize the p-adic zeta pseudomeasure ${\zeta }_p$ byThe expression $x^k$ is well defined on $G$ for even $k$. A pseudomeasure lies in the total fraction ring of the Iwasawa algebra; the pole on its trivial finite-character component is removed by the augmentation ideal, so $I(G){\zeta }_p\subset {\Lambda }_G$. In this normalization the local theorem isThe following steps outline its proof and explain where the earlier questions enter.

First encode local units by Coleman power series. Evaluation at ${\pi }_n$ is surjective from $R$ to ${\mathcal{O}}_{K_n}$, and the product defining the Coleman norm operator givesThus norm-fixed series give norm-compatible sequences of local units. Conversely, lift $u_m$ from such a sequence to $f^{(m)}\in R^{\times }$ and take $g_m=N^{m-r_m}{f}^{(m)}$, where $r_m=\lfloor m/2\rfloor$. This has $g_m({\pi }_{r_m})=u_{r_m}$. The estimates in Question 2 give $N{g}_m/g_m\in 1+p^{m-r_m+1}R$. Consequently for fixed $n\le r_m$, the series $N^{r_m-n}{g}_m$ differs from $g_m$ modulo $p^{m-r_m+1}$ and its value at ${\pi }_n$ is $u_n$. The compactness of $R$ in the coarser of the topologies on integral formal power series supplies a convergent subsequence. Continuity of the Coleman norm operator and of evaluation shows that its limit is norm-fixed and takes every value $u_n$. It is a unit because the original principal units have residue $1$. Uniqueness follows from the Weierstrass preparation theorem: a nonzero series in $R$ has only finitely many zeros in the open unit disk, whereas the distinct ${\pi }_n$ give infinitely many interpolation points. This constructs the Coleman power series $f_u$.

Next turn its multiplicative information into an integral p-adic measure. For a norm-fixed unit $f$, use the Frobenius-corrected logarithmThe ratio belongs to $1+pR$ by reduction modulo $p$, and the logarithm divided by $p$ is integral because $p$ is odd. Equivalently $\mathcal{L}(f)=(1-\varphi /p)\log f$, interpreting the constant coefficient with the p-adic logarithm on units. The averaging identity for the logarithm is understood on the open $p$-adic unit disk, where the translated logarithmic series converge; the corrected result itself lies in $R$. The logarithmic product formula and Question 1 giveUnder the Amice transform, the root-of-unity average projects a p-adic measure onto its restriction to $p{\mathbb{Z}}_p$: the averaged factor $(1+T{)}^x$ survives exactly for $x\in p{\mathbb{Z}}_p$. Consequently $S\mathcal{L}(f)=0$ says that $\lambda (\mathcal{L}(f))$ is supported on ${\mathbb{Z}}_p^{\times }$. This gives a Galois group-equivariant homomorphism from local principal units to the Iwasawa algebra of ${\mathbb{Z}}_p^{\times }$.

The integral lifting step in this proof is the Coleman unit-measure exact sequenceHere $\mathrm{Col}(u)=\lambda (\mathcal{L}(f_u))$. The last map is surjective, since $m_1({\delta }_1)=1$, and it has the cyclotomic action on its target. Its composition with $\mathrm{Col}$ is zero: $D\varphi =p\varphi D$ implies $D\mathcal{L}(f)(0)=0$. The kernel on the unit side consists of norm-compatible $p$-power roots of unity. To see the shape of the kernel, write $z=\log (1+T)$; the equation $\mathcal{L}(f)=0$ becomes $h(pz)=ph(z)$ for $h(z)=\log f(e^z-1)$, so only its linear term can survive. Integral solutions are $f=(1+T{)}^c$, $c\in {\mathbb{Z}}_p$, after excluding finite residue factors by the principal unit condition. For the image, use the following explicit integral lifting of the corrected cyclotomic logarithm.

Take a p-adic measure $\mu$ on the units with first moment zero, and write $g$ for its Amice transform. Then $Sg=0$ and $g^{\prime }(0)=0$. Put $H=\exp (pg)\in 1+pR$ and solve $f^p=H\varphi (f)$ with $f(0)\in 1+p{\mathbb{Z}}_p$. The constant equation is solved by $f_0=\exp (p{g}_0/(p-1))$. In degree one the terms involving $f_1$ cancel, and the remaining condition is $H_1=0$, exactly $g_1=0$; choose any $f_1\in {\mathbb{Z}}_p$. For $n\ge 2$, after fixing the lower coefficients, the coefficient multiplying the unknown $f_n$ isThe other terms are divisible by $p$: for the already integral truncated series, $f^p\equiv \varphi (f)(\mathrm{mod}p)$, while $H\equiv 1(\mathrm{mod}p)$. Division therefore gives an integral $f_n$, since $f_0$ and $1-p^{n-1}$ are units. Recursion constructs $f\in R^{\times }$ with $\mathcal{L}(f)=g$. Now $pSg=\log (Nf/f)=0$, and $Nf/f\in 1+pR$ by Question 2. The p-adic logarithm is injective there, so $Nf=f$. This supplies the desired Coleman power series of a norm-compatible sequence of local units. The freely chosen linear coefficient accounts precisely for the kernel $(1+T{)}^c$. It proves the integral surjectivity step, including the integrality that rational inversion of $1-\varphi /p$ would leave unproved.

Complex conjugation acts by $-1$ on both copies of ${\mathbb{Z}}_p(1)$. As $p$ is odd, taking the even part is exact and removes both end terms. It follows that $\mathrm{Col}:U_{\infty }^1\simeq {\Lambda }_G$. The integral exact sequence is essential here: the support calculation alone would not prove this isomorphism.

It remains to calculate the image of the cyclotomic units. For $a,b\in {\mathbb{Z}}_p^{\times }$, consider the symmetric cyclotomic unit seriesThe quotient has an invertible constant term $a/b$, because both numerator and denominator have a simple zero at $T=0$. The factor $X^{(b-a)/2}$ makes the expression unchanged under $X\mapsto X^{-1}$. The product over the roots of unity gives $N{f}_{a,b}=f_{a,b}$; one may verify this first for integer $a,b$ prime to $p$ and then extend by continuity. Dividing by the norm-compatible Teichmuller representative makes the series correspond to principal units and does not change its logarithm. The generating function for the Bernoulli numbers, with $z=\log X$, givesThe symmetric factor cancels the linear term. Applying $1-\varphi /p$, and then the moment formula of Question 3, yieldsfor even $k\ge 2$. These moments determine an even p-adic measure on the units: polynomials are dense by the Mahler theorem, even projection preserves density, and multiplication by $x^2$ is invertible on the units, so the positive even-degree polynomials are dense there as well. ThusThe closed ideal generated by all differences $[b]-[a]$ is $I(G)$. The closed subgroup generated by the corresponding cyclotomic units, including its Galois group translates, therefore has image exactly $I(G){\zeta }_p$. Taking the quotient of the isomorphism $U_{\infty }^1\simeq {\Lambda }_G$ proves the boxed local theorem.

Finally introduce global arithmetic. Let ${\overline{E}}_{\infty }^1$ be the closure of norm-compatible global units in the real tower, let $X_{\infty }^{+}$ be its p-ramified Iwasawa module, and let $Y_{\infty }^{+}$ be its unramified Iwasawa module. With compatible unit and splitting conventions, class field theory gives the class-field unit sequence; quotienting its unit terms by $C_{\infty }^1$ givesIn the cyclotomic tower the prime above $p$ is principal, so the unramified class-field convention also makes it split. On the torsion character components, multiplicativity of characteristic ideals impliesThe local theorem has already calculated the first factor analytically. Equality of the global unit and class-group factors would therefore identify the characteristic ideal of $X_{\infty }^{+}$ with the same ideal of the p-adic L-function. That is the content predicted by the Iwasawa main conjecture. More explicitly, for a nontrivial even finite character $\chi$, the character component of $I(G)$ is the whole component and the prediction isHere ${\Lambda }_{\chi }$ is the power-series algebra of the pro-$p$ part with coefficients in the character-value ring. The trivial component requires the augmentation correction for the zeta pole. Equivalently the characteristic ideals of ${\overline{E}}_{\infty }^1/C_{\infty }^1$ and $Y_{\infty }^{+}$ agree componentwise. A formulation using odd class-group characters must include the appropriate cyclotomic twist and inversion, as in Kummer reflection in Iwasawa theory. The motivation is an analytic formula for local units together with a global class-field exact sequence; the global equality is the additional arithmetic assertion.

The related class-number growth theorem illustrates why passage to an Iwasawa module is powerful. For the cyclotomic ${\mathbb{Z}}_p$-extension, let $p^{e_n}$ be the $p$-part of the class number at level $n$. The unramified Iwasawa torsion theorem, the Iwasawa module structure theorem and the class-group control relations give, for sufficiently large $n$,Here $\mu ,\lambda$ are the Iwasawa invariants. To see the two growth terms, use a topological generator $\gamma$ and, after a fixed sufficiently high level $n_0$, the norm polynomial ${\nu }_{n,n_0}=({\gamma }^{p^n}-1)/({\gamma }^{p^{n_0}}-1)$. An elementary summand $\Lambda /(p^a)$ contributes $a(p^n-p^{n_0})$ to the length of its quotient by this polynomial. A summand associated with a distinguished polynomial of degree $d$ contributes $dn$ plus a constant: at each root $\alpha$ the valuation of the norm polynomial is $n$ plus a fixed constant for large $n$. If $1+\alpha$ is a root of unity already killed at level $n_0$, its value is instead exactly $p^{n-n_0}$ and gives the same linear contribution. The finite errors in the structure and control maps stabilize and contribute to $\nu$. Thus module structure explains growth, while the Iwasawa main conjecture goes further by identifying the characteristic series of an Iwasawa module themselves with analytic ones.

We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.

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.

The valuation ring and local compactness. The opening, unlettered requests useThe ultrametric inequality makes ${\mathcal{O}}_K$ a valuation ring; its elements outside $\mathfrak{m}$ are exactly the units, so $\mathfrak{m}$ is its unique maximal ideal, and $k$ is the residue field.

Here the customary nontrivial-valuation hypothesis for a local field is necessary. With the trivial absolute value, any infinite field is discrete and locally compact, but its residue field is infinite and there is no nonzero uniformizer. We use the nontrivial valuation in the conclusions below.

Local compactness first gives a compact neighborhood of zero. Choose a sufficiently small nonzero $t$ so that the closed set $t{\mathcal{O}}_K$ is contained in that compact neighborhood. Then $t{\mathcal{O}}_K$, and hence ${\mathcal{O}}_K$, is compact. Since $\mathfrak{m}$ is open, its cosets give a discrete compact quotient $k$, which must be finite. Write $|k|=q$. The subgroup $\mathfrak{m}$ now has finite index and is closed in ${\mathcal{O}}_K$, hence compact. The continuous function $|\cdot |$ attains on $\mathfrak{m}$ a maximum $c$ with $0<c<1$. Choose $\pi$ with $|\pi |=c$. Every $x\in \mathfrak{m}$ has $|x/\pi |\le 1$, soSuccessive division by $\pi$ shows that every nonzero element of ${\mathcal{O}}_K$ is a unit times a power of $\pi$: division must terminate because $|\pi {|}^j\to 0$. Thus the value group is discrete. A Cauchy sequence eventually lies in a translate of the compact ring ${\mathcal{O}}_K$, has a convergent subsequence, and consequently converges itself. These arguments prove the discrete valuation from nontrivial local compactness and the completeness of locally compact nontrivially valued fields.

The iterated-power limit. Normalize the discrete valuation by $v(\pi )=1$, and write $q={\ell }^f$, where $\ell$ is the residue characteristic. For $u\in 1+{\pi }^s{\mathcal{O}}_K$ with $s\ge 1$, the binomial theorem givesThe linear term gains a factor $q\in \mathfrak{m}$ (or is zero in positive characteristic), while all terms of degree at least two have valuation at least $2s\ge s+1$.

For $a\in {\mathcal{O}}_K^{\times }$, reduction gives $a^{q-1}\in 1+\mathfrak{m}$, because $k^{\times }$ has order $q-1$. Iterating the inequality yieldsThe sequence is Cauchy and converges to a unit $\omega (a)$. Since the reduction of $a^{q^j}$ is always $\bar{a}$, its limit has the same residue. Taking limits after shifting the sequence gives $\omega (a{)}^q=\omega (a)$, whenceThis is the Teichmuller representative. The Hensel lemma applied to $T^{q-1}-1$, whose derivative is a unit at every nonzero residue, shows uniqueness for each residue. Consequently the prime-to-$\ell$ roots give a canonical splitting

Write $q=\#k_K=p^f$. For every $a\in k_K$, the polynomial $F(T)=T^q-T$ has $a$ as a residue root and $\overline{F^{\prime }}=-1$. By the Hensel lemma there is a unique element $[a]\in {\mathcal{O}}_K$ such thatThis defines the Teichmuller lift, including $[0]=0$. The two displayed properties determine each value uniquely by Hensel's uniqueness statement.

They also imply the standard Teichmuller lift properties. The elements $0,1$ already satisfy their defining equations, so $[0]=0$ and $[1]=1$. The product $[a][b]$ reduces to $ab$ and satisfies $([a][b]{)}^q=[a][b]$, so uniqueness gives $[ab]=[a][b]$. Likewise $[a{]}^p$ reduces to $a^p$ and is fixed by its $q$th power, givingFor $a\ne 0$, division of $[a{]}^q=[a]$ by $[a]$ gives $[a{]}^{q-1}=1$. Thus the nonzero Teichmuller representatives are precisely the prime-to-$p$ roots of unity lifting the nonzero residue classes. This is a multiplicative section of reduction; additivity is not asserted in mixed characteristic.

For the convergence, use the normalized discrete valuation $v_K$ with $v_K(\pi )=1$, and put $e=v_K(p)\ge 1$. A useful binomial estimate isIndeed, writing $u=v+\delta$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, so all terms of $(v+\delta {)}^p-v^p$ have valuation at least $\min \{e+r,pr\}$. The ultrametric inequality gives the bound for their sum.

Finite fields are perfect, so the successive $p$th roots $x_n$ exist uniquely and satisfy $x_n^{p^n}=x$. Let $a_n=[x_n]$. Since $y_n-a_n$ lies in the maximal ideal, iterating the estimate $n$ times givesMultiplicativity, or the Frobenius identity above, gives $a_n^{p^n}=[x_n^{p^n}]=[x]$. ThereforeThis proves the approximation of Teichmuller lifts by iterated pth powers for completely arbitrary choices of the lifts $y_n$, including $x=0$. The original PDF confirms $x\in k_K$, correcting the capital $K_K$ in the TeX transcription.

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 $m$.

The cyclotomic extension of a p-adic fieldhas degree $p-1$, is Galois with group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, and is totally ramified. Indeed, ${\zeta }_p-1$ is a root of the Eisenstein polynomial

Put $\alpha =\sqrt[p-1]{-p}$. Its polynomial $X^{p-1}+p$ is Eisenstein, so $L_2={\mathbb{Q}}_p(\alpha )$ is totally ramified of degree $p-1$. Every $(p-1)$st root of unity lies in ${\mathbb{Q}}_p$ by the Teichmuller lifts, so every root $\omega \alpha$ of this polynomial lies in $L_2$. Hence $L_2/{\mathbb{Q}}_p$ is also Galois.

For either $L_i/{\mathbb{Q}}_p$, the norm has every possible valuation because the residue-field degree is one. The norm units in a tamely totally ramified extension lie in the principal units $1+p{\mathbb{Z}}_p$: reduction of a unit norm is the $(p-1)$st power of its residue, hence is $1$. Part (b) says that this unit norm subgroup has index $p-1$, exactly the index of $1+p{\mathbb{Z}}_p$ in ${\mathbb{Z}}_p^{\times }$. ConsequentlyThe uniqueness clause in the existence theorem of local class field theory now gives

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 $K={\mathbb{Q}}_p$.

Since ${\bar{a}}^{q-1}=1$ in the residue field, $u=a^{q-1}\in 1+\pi {\mathcal{O}}_K$. Powers $u^{q^n}$ tend to one, by the binomial theorem initially and the $p$-adic logarithm once they enter its convergence domain. Therefore $a^{q^n}$ is Cauchy; let its limit be $\omega$. Reduction modulo $\pi$ gives $\bar{\omega }=\bar{a}$, whileThis is the Teichmuller representative of $\bar{a}$.

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.

For odd $p$, Teichmuller representatives provide the $p-1$ prime-to-$p$ roots, while the logarithm isomorphism on deep principal units makes $1+p{\mathbb{Z}}_p$ torsion-free. At two, the decomposition ${\mathbb{Z}}_2^{\times }=\{1,-1\}\times (1+4{\mathbb{Z}}_2)$ leaves exactly two roots.