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For odd $p$, remove the finite torsion subgroup ${\mu }_{p-1}$ from the Galois group of $\mathbb{Q}({\mu }_{p^{\infty }})/\mathbb{Q}$ by taking its fixed field. For $p=2$, take the fixed field of $\{\pm 1\}$ instead. The remaining Galois group is ${\mathbb{Z}}_p$, using the p-adic logarithm on $1+p{\mathbb{Z}}_p$ or $1+4{\mathbb{Z}}_2$. For a number field $K$, the cyclotomic Zp-extension is the compositum with this rational tower.

For a cyclic extension of p-adic fields, sufficiently deep principal units are equivariantly isomorphic to an additive lattice by the p-adic logarithm. The normal basis theorem makes its Herbrand quotient one. Passing across finite unit quotients preserves it. The valuation exact sequence with quotient $\mathbb{Z}$ therefore gives the displayed formula.

Let $R={\mathbb{Z}}_p[[T]]$, $p$ odd, and $\varphi (T)=(1+T{)}^p-1$. An element $g\in R$ is the Frobenius-corrected logarithm of a unit $f$ with $f(0)\in 1+p{\mathbb{Z}}_p$ if and only if $g^{\prime }(0)=0$. Necessity follows by differentiating at zero, using ${\varphi }^{\prime }(0)=p$. For sufficiency put $H=\exp (pg)\in 1+pR$ and solve $f^p=H\varphi (f)$. Choose $f_0=\exp (p{g}_0/(p-1))$. The linear equation reduces to $H_1=0$, so $f_1$ is free. In degree $n\ge 2$ the unknown $f_n$ has coefficient $p{f}_0^{p-1}(1-p^{n-1})$. The remaining terms are divisible by $p$ because reduction of any integral truncated $f$ satisfies $f^p\equiv \varphi (f)(\mathrm{mod}p)$. Division gives a unique integral $f_n$. Two lifts differ by $(1+T{)}^c$, $c\in {\mathbb{Z}}_p$, as follows from the logarithmic equation in the coordinate $z=\log (1+T)$. If also $Sg=0$, the Coleman trace operator gives $\log (Nf/f)=pSg=0$. As $Nf/f\in 1+pR$, injectivity of the p-adic logarithm implies $Nf=f$.

On the set underlying a uniform pro-p group, define scalar multiplication by $\lambda x=x^{\lambda }$, and define addition and the Lie bracket byRoots are the unique roots in the appropriate power layers. The limits give a powerful Lie lattice. In a suitable complete normed group algebra, the p-adic logarithm identifies these operations with ordinary addition and the associative-algebra commutator.

For a mixed-characteristic local field with residue characteristic $\ell$, the p-adic logarithm and p-adic exponential function are mutually inverse continuous group homomorphisms on these domains. Bounds on the valuations of the denominators show that higher terms have greater valuation than the linear term. Dividing the logarithm by ${\pi }^r$ identifies the multiplicative higher principal-unit group with the additive integer ring; in particular it is torsion-free.

For a finite Galois extension $L/K$ with group $G$, there is $\theta \in L$ such that the elements $g(\theta )$ form a $K$-basis. Equivalently the additive $K[G]$-module $L$ is a regular representation. Over p-adic fields, its p-adic lattices are thus commensurable with regular lattices; this is useful in Herbrand quotient calculations after applying a p-adic logarithm to deep principal units.

We give the full local-field norm-index argument. Let $G=⟨\sigma ⟩$ be cyclic of order $n$. On an additive $G$-module set $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two cyclic Tate cohomology groups areWhen these are finite, the Herbrand quotient is $h_G(M)=|{\hat{H}}^0(G,M)|/|{\hat{H}}^{-1}(G,M)|$. The periodic complex alternating $D$ and $N$ yields a six-term periodic exact sequence from a short exact sequence of modules. Taking alternating products of orders in that exact sequence provesFor a finite module, $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$ show that the two Tate cohomology groups have equal order. Thus the Herbrand quotient of a finite module is one. For the trivial integral module $\mathbb{Z}$, the two groups are $\mathbb{Z}/n\mathbb{Z}$ and zero, giving $h_G(\mathbb{Z})=n$.

We also need cyclic cohomology of a regular lattice. In $R[G]$, fixed coefficient vectors are constant, and every such vector is a norm of a basis vector times its coefficient. The kernel of $N$ consists of vectors whose coefficient sum is zero. For such a vector $(a_i)$, solve $b_{i-1}-b_i=a_i$ successively around the cycle; the final equation holds precisely because ${\sum }_i{a}_i=0$. Thus $\ker N=DR[G]$. Both Tate cohomology groups vanish, so $h_G(R[G])=1$, for $R=\mathbb{Z}$ or ${\mathbb{Z}}_p$, and likewise for finite direct sums.

Now take a cyclic extension $L/K$ of p-adic fields of degree $n$. Write $U_L^r=1+{\mathfrak{p}}_L^r$. For $r>v_L(p)/(p-1)$, the convergent p-adic logarithm and p-adic exponential give mutually inverse $G$-equivariant group isomorphismswhere the right side is additive. The convergence assertion follows from the estimates $v_L(x^j/j)\ge j{v}_L(x)-v_L(p)v_p(j)$ and $v_L(x^j/j!)\ge j{v}_L(x)-v_L(p)j/(p-1)$; formal inverse identities therefore hold on this ball.

By the normal basis theorem, choose $\theta$ so that $\{{\sigma }^i\theta \}$ is a $K$-basis of $L$. The lattice $A={\sum }_i{\mathcal{O}}_K{\sigma }^i\theta$ is a regular ${\mathcal{O}}_K[G]$-module. It is a finite direct sum of regular ${\mathbb{Z}}_p[G]$-modules, hence its Tate cohomology vanishes. The lattices $A$ and ${\mathfrak{p}}_L^r$ are commensurable: their intersection has finite index in both. The exact sequences with these finite quotients establish finiteness of the Tate cohomology groups and show $h_G({\mathfrak{p}}_L^r)=1$. Thus $h_G(U_L^r)=1$. Since ${\mathcal{O}}_L^{\times }/U_L^r$ is finite, $h_G({\mathcal{O}}_L^{\times })=1$ as well.

The $G$-equivariant valuation exact sequence isIts last term has trivial $G$-action. The multiplication rule gives the Herbrand quotient of the local multiplicative group:All the Tate cohomology groups here are finite by the lattice argument and the exact sequence; finiteness of the norm index has not been assumed.

For completeness we prove the needed Hilbert theorem 90. If $a\in L^{\times }$ has field norm one, put $A_0=1$ and $A_i={\prod }_{j=0}^{i-1}{\sigma }^j(a)$. The linear map $c\mapsto b={\sum }_{i=0}^{n-1}{A}_i{\sigma }^i(c)$ is not zero by linear independence of distinct field embeddings. Choose $c$ with $b\ne 0$. The relation $A_n=1$ gives $a\sigma (b)=b$, so $a=b/\sigma (b)$. This proves ${\hat{H}}^{-1}(G,L^{\times })=0$. The zeroth Tate cohomology group is $K^{\times }/N_{L/K}{L}^{\times }$. Consequently the cyclic local norm index isWriting $n=ef$, the valuation of a field norm is $v_K(Nx)=f{v}_L(x)$. The valuation quotient has order $f$, and a norm with valuation zero comes from a unit. ThereforeIn particular every unit is a norm in an unramified extension. These results give the norm assertions used in the quadratic Hilbert symbol computation.

We finish with the Hasse norm theorem: for a cyclic extension of number fields, an element is a global norm if and only if it is a norm at every place. More precisely the local norm is the norm from the product $L{\otimes }_K{K}_v$; in a Galois extension its factors have the same norm subgroup. The forward implication is immediate. For the converse associate to $a\in K^{\times }$ the cyclic algebra $(L/K,\sigma ,a)$. The splitting criterion for a cyclic algebra, proved below, identifies its splitting with $a\in N_{L/K}{L}^{\times }$. At a place where the tensor product of fields has several factors, this algebra is a matrix algebra over the cyclic algebra of one decomposition-field factor, with the generator restricted to its cyclic decomposition subgroup. Thus the local norm assumption makes its Brauer class zero at every place. Injectivity in the Albert-Brauer-Hasse-Noether theorem makes the global Brauer class zero, giving the desired global norm. This is the required proof outline; cyclicity is essential to this reduction.

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

The proof of the linearity of a uniform pro-p group has three distinct steps: pass to a finite-dimensional Lie algebra, integrate a faithful representation on a sufficiently small open subgroup, and extend it to the entire group. Working first on a small subgroup is essential because a faithful Lie algebra representation need not send the original integral lattice into the convergence domain of the matrix exponential.

Let $G$ be a uniform pro-p group and let $L_G$ be its intrinsic powerful Lie lattice. Thenis a finite-dimensional Lie algebra over a field of characteristic zero. The Ado theorem, the classical Lie-theoretic ingredient in this proof, supplies a faithful finite-dimensional Lie algebra representationOne must use a faithful representation rather than just the Adjoint representation, which annihilates the center of a Lie algebra.

Choose a p-adic lattice $\Lambda \subset V$. Since $L_G$ is finitely generated as a module, the entries of the representing matrices of a module basis have bounded denominators. Hence for sufficiently large $k$,For odd $p$, the matrix exponential and p-adic logarithm converge and are inverse between $p\mathrm{End}(\Lambda )$ and $1+p\mathrm{End}(\Lambda )$. Put $H=G^{p^k}$; under the uniform pro-p group and powerful Lie lattice correspondence it is the BCH group on $p^k{L}_G$. DefineHere ${\exp }_G$ denotes the inverse of the intrinsic logarithm identification, not an assumed matrix representation of $G$. The Lie algebra homomorphism $\rho$ respects the convergent Baker--Campbell--Hausdorff formula, soThus $\tau$ is a continuous group homomorphism. If $\tau ({\exp }_{G}X)=1$, taking the matrix logarithm gives $\rho (X)=0$, whence $X=0$. Therefore $\tau$ is faithful and all its matrices preserve $\Lambda$. This proves linearity of the open normal subgroup $H$.

To finish, use the extension of a faithful representation from an open normal subgroup. Choose representatives $t_1=1,t_2,\dots ,t_r$ of the finite left cosets $G/H$, where $r=[G:H]$. On the finite free module $W={⨁}_{j=1}^r\Lambda$, define the action of $g\in G$ by writing uniquelyand mapping the $j$-th block to the $i$-th block by $\tau (h_{ij})$. If $g^{\prime }{t}_j=t_{\ell }{h}^{\prime }$ and $g{t}_{\ell }=t_{i}h$, then $g{g}^{\prime }{t}_j=t_{i}h{h}^{\prime }$; this proves the representation identity because $\tau (h{h}^{\prime })=\tau (h)\tau (h^{\prime })$. Each block matrix is invertible over ${\mathbb{Z}}_p$, and the map is continuous because the finite coset permutation is locally constant and the remaining matrix entries are obtained continuously from $\tau$.

If $g\notin H$, its permutation of $G/H$ moves the identity coset, so its block representation is not the identity. If $g\in H$ and its block representation is the identity, the block corresponding to $t_1=1$ is $\tau (g)=1$, and faithfulness of $\tau$ gives $g=1$. Thus this induced representation is faithful. Choosing a basis of $W$ provesThe continuous injection is a topological embedding, since $G$ is compact and the target is Hausdorff. The trivial group is immediate. For uniform groups at $p=2$, the same proof uses the usual stronger powerfulness condition and the convergence ball $4\mathrm{End}(\Lambda )$ instead; the preceding questions concern odd $p$.

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 give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.

The correspondence concerns uniform pro-p groups and powerful Lie lattices, with their natural morphisms. Put $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. A powerful pro-p group is a pro-p group whose closed commutator subgroup lies in its closed $p^{ϵ}$-power subgroup. It is uniform when it is topologically finitely generated and all successive indices in its lower p-series are equal. Equivalently, a finitely generated powerful pro-p group is uniform precisely when it is a torsion-free group. A powerful Lie algebra over p-adic integers means a Lie algebra over ${\mathbb{Z}}_p$ that is free of finite rank as a module and satisfiesHere a Lie algebra over a ring is a module with an alternating bilinear Lie bracket satisfying the Jacobi identity. Finite freeness is an essential part of the lattice convention. The zero-bracket module ${\mathbb{F}}_p$ satisfies the bracket inclusion but has torsion, and its additive group is not uniform. Thus allowing arbitrary torsion modules would make the categorical assertion false.

The basic structural facts behind the construction can be seen from the power layers. Write $P_n=G^{p^n}$ with $P_0=G$. Collection for a powerful pro-p group givesFor odd $p$, the power lifting is exactly the successive-correction argument of Question 3, followed by compactness; at $p=2$ the stronger fourth-power commutator condition makes the corresponding collection corrections work. The maps between successive layers are surjective linear maps. Uniformity makes their finite dimensions equal, so these maps are isomorphisms. Comparing the first layer in which two elements differ now shows that the power map $P_n\to P_{n+1}$ is injective, while the power-set formula makes it surjective. Thus all the roots used below exist uniquely in the indicated layer. If $g_1,\dots ,g_d$ lift a basis of $G/P_1$, the images of $g_i^{p^n}$ form a basis of $P_n/P_{n+1}$. Successively choosing their coefficients modulo $p$ gives the unique ordered expressionExistence follows by lifting one layer at a time and taking a limit; uniqueness follows by comparing the first nonzero coefficient layer. In particular $[G:P_n]=p^{nd}$ and these layers form a neighbourhood basis. These are the power-layer and coordinate facts needed in the proof, rather than an assumption that an arbitrary pro-p group has unique roots.

On the underlying set of $G$, set the additive zero to $1$, the additive negative of $x$ to $x^{-1}$, and defineThe p-adic integers act through limits of integer powers in the closed cyclic subgroup generated by $x$. In the second formula the product belongs to $P_n$; in the third the group commutator belongs to $P_{2n+ϵ}$. The symbols $p^{-n}$ and $p^{-2n}$ denote the unique group roots, not unrestricted division in a group. We next justify convergence and the Lie algebra laws together.

A useful concrete tool is a complete normed group algebra containing $G$. In its completed group algebra, put $b_i=g_i-1$. Ordered-coordinate uniqueness gives unique expansions ${\sum }_{\alpha \in {\mathbb{N}}^d}{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To see why, first work modulo $p$ and $P_n$: the ordered products of the $g_i$ with exponents below $p^n$ form the regular basis, and the change to the corresponding products of $g_i-1$ is triangular with diagonal entries one. Take inverse limits and then lift the coefficient digits successively in powers of $p$. Define the weightIt is a genuine separated, submultiplicative weight. Indeed reordering $b_j{b}_i$ produces a group-commutator correction in $P_{ϵ}$; the binomial expansion of a $p^{ϵ}$th power gives that correction weight at least $2ϵ$, so collection never decreases total weight. Also $v(pc)=1+v(c)$. Inverting $p$ and completing therefore gives a Banach algebra $B$ over ${\mathbb{Q}}_p$. For $g\in G$, ordered coordinates showwith value infinity at $g=1$. Thus the induced topology on $G$ is exactly its original topology. This construction supplies the ambient algebra without presupposing the uniform pro-p group and powerful Lie lattice correspondence. The weighted construction, including the case $p=2$, is described in Equivariant D-modules on rigid analytic spaces, Section 6.5.