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Suppose $1\to X\to \mathcal{G}\to \Gamma \to 1$ is an exact sequence of pro-p groups, with $X$ abelian and $\Gamma \cong {\mathbb{Z}}_p$. Conjugation gives $X$ a compact Galois module structure. The closed commutator subgroup is $(\gamma -1)X$: its image is closed by compactness, and after quotienting by it, a lift of $\gamma$ centralizes $X$ and topologically generates the remaining quotient. Hence $0\to X_{\Gamma }\to {\mathcal{G}}^{\mathrm{ab}}\to \Gamma \to 0$ is exact.

A compact Galois module is a compact profinite abelian group equipped with a continuous Galois group action. An abelian pro-p group is naturally a compact ${\mathbb{Z}}_p$-module. A continuous action of the Galois group of a Zp-extension extends to its Iwasawa algebra of a Zp-extension. This compact topology differs from the discrete topology usually used in Galois cohomology.

Let $G=G_K$ and choose a profinite Sylow subgroup $P$ at the prime $p$. Its image in every finite quotient of $G$ is a Sylow $p$-subgroup. Put $F=K_s^P$. Every finite intermediate field $K\subseteq L\subseteq F$ has degree prime to $p$, because its corresponding open subgroup contains $P$. Condition (iii), together with the preceding Kummer reduction, gives $H^{n+1}(L,{\mu }_p)=0$ for each such field. The cohomology continuity at closed subgroups property saysfor a discrete $G$-module $M$. Applying it to $M={\mu }_p$ gives $H^{n+1}(P,{\mu }_p)=0$.

The action of the pro-p group $P$ on ${\mu }_p$ is trivial. Indeed it maps continuously to $\mathrm{Aut}({\mu }_p)\cong {\mathbb{F}}_p^{\times }$, whose order $p-1$ is prime to $p$; every finite image of $P$ is a $p$-group, so that image is trivial. Choosing a primitive root identifies this module with the trivial module ${\mathbb{F}}_p$. Thus $H^{n+1}(P,{\mathbb{F}}_p)=0$.

We spell out why this controls arbitrary $p$-primary coefficients. A finite discrete $p$-primary $P$-module has a composition series with trivial ${\mathbb{F}}_p$ factors: its action factors through a finite $p$-group, whose only simple module in characteristic $p$ is the trivial one. Induction through the long exact cohomology sequence therefore gives vanishing of $H^{n+1}$ for every such finite module. Any discrete $p$-primary $P$-module is the filtered union of finite $P$-stable submodules. The orbit of an element is finite by continuity, and its orbit generates a finite abelian $p$-group. Continuous cohomology commutes with these filtered unions, sofor every discrete $p$-primary module $M$. This is the mechanism behind trivial coefficients detect the cohomological dimension of a pro-p group.

Now let $M$ be any discrete $p$-primary $G$-module and $a\in H^{n+1}(G,M)$. Its restriction to $P$ is zero. Continuity at the closed subgroup $P$ makes its restriction zero in some open $U\supseteq P$. The index $m=[G:U]$ is prime to $p$, and the restriction-corestriction identity in group cohomology for the corestriction map in group cohomology givesBut $a$ has $p$-power order, so multiplication by $m$ is invertible on the cyclic group it generates. Hence $a=0$. This proves $H^{n+1}(G,M)=0$ for all discrete $p$-primary $M$. Dimension shifting through an acyclic coinduced module gives vanishing in every higher degree, which is exactly ${\mathrm{cd}}_p(K)\le n$.

Therefore (iii) implies (i), completing the equivalence of all three conditions. The key reason prime-to-$p$ extensions suffice is that they approximate a pro-$p$ Sylow subgroup, where the cyclotomic module becomes trivial; prime-to-$p$ transfer then detects every $p$-primary cohomology class.

Write $\Gamma =\mathrm{Gal}(K_{\infty }/K)\cong {\mathbb{Z}}_p$. Its only finite subgroup is trivial. In particular, a real place cannot acquire complex inertia in this Zp-extension: the possible nontrivial inertia at a real place has order two. Thus every infinite place splits in the tower.

Suppose that no finite prime ramified. Every finite layer would then be an abelian everywhere unramified extension of number fields, including splitting at real places. Every layer would lie in the ordinary Hilbert class field of $K$, a finite extension. Their degrees are unbounded, a contradiction. Therefore some finite prime must ramify.

Let $v$ have residue characteristic $\ell \ne p$. Apply local class field theory to the corresponding decomposition group, a closed subgroup of $\Gamma$. The image of the unit group ${\mathcal{O}}_{K_v}^{\times }$ is the inertia group. Its first principal-unit subgroup is a pro-l group, so its continuous image in the pro-p group $\Gamma$ is trivial. The residue-unit quotient is the finite group ${\mathbb{F}}_{q_v}^{\times }$. Its image is finite, and is therefore also trivial in the torsion-free group ${\mathbb{Z}}_p$. The entire inertia group is trivial. HenceEquivalently, the tame ramification relation with a Frobenius element would force a tame inertia generator to satisfy ${\tau }^{q_v-1}=1$, which is impossible nontrivially in ${\mathbb{Z}}_p$.

For the final assertion, the cyclotomic Zp-extension of $K$ is $K\mathbb{Q}[p^{\infty }]$. Fix $v\mid p$ and put $E=K_v$, a finite extension of ${\mathbb{Q}}_p$. Let $C_n$ be the completion of $\mathbb{Q}[p^n]$ at its unique prime over $p$. This is a totally ramified extension of ${\mathbb{Q}}_p$ of degree $p^n$. In a compatible local algebraic closure,Here the inequality follows from the tower through $C_n$. The right side is unbounded. These composita occur among the completions of the cyclotomic tower over $K$, so $v$ has unbounded ramification index, in particular nontrivial inertia group. ThusThis argument allows an arbitrary finite intersection between $K$ and the rational cyclotomic tower; it does not assume disjointness.

Let $\mathcal{G}=\mathrm{Gal}(M_{\infty }/F)$. The Galois group sequence isFor $g\in \Gamma$ and $x\in X$, choose a lift $\tilde{g}\in \mathcal{G}$ and defineThis is independent of the lift because $X$ is abelian. Write $X$ additively. As an abelian pro-p group, it is a compact ${\mathbb{Z}}_p$-module, with $ax$ defined by the limit of integer powers, and the conjugation action is continuous and ${\mathbb{Z}}_p$-linear. It extends to the Iwasawa algebraIndeed, use the finite quotients of $X$ obtained by intersecting it with open normal subgroups of $\mathcal{G}$; on each quotient the action factors through one of these finite group rings. This defines the natural compact Galois module structure, even when $X$ is not finitely generated. Choosing $\gamma$ identifies $\Lambda$ with ${\mathbb{Z}}_p[[T]]$ through $T=\gamma -1$.

The subgroup $(\gamma -1)X$ is closed: it is the image of the compact group $X$ under a continuous map. Every element $(\gamma -1)x$ is a commutator of a lift of $\gamma$ with $x$. ConsequentlyConversely, in the quotient by $(\gamma -1)X$, the lift of $\gamma$ centralizes the image of $X$. Its powers are dense in the procyclic subgroup they generate. Since these powers map densely onto $\Gamma$, compactness shows that $\mathcal{G}$ is generated by $X$ and that procyclic subgroup. The quotient is therefore abelian. HenceOne can also see this by choosing a lift of $\gamma$. Since $\mathcal{G}$ is a pro-p group, the closed subgroup generated by this lift is ${\mathbb{Z}}_p$, and its map to $\Gamma$ is an isomorphism. This gives a semidirect product splitting.

The maximal abelian subextension $M/F$ is the fixed field of the closed commutator subgroup. Because $F_{\infty }/F$ is abelian, $F_{\infty }\subseteq M$. Restriction of the Galois correspondence to $X$ now gives the coinvariant formulaThe closed image above is important: the formula is a quotient of compact groups, not a quotient by an unclosed abstract subgroup. This is abelianization over a Zp-extension.

Let $F_n$ be the degree-$p^n$ layer, ${\Gamma }_n={\Gamma }^{p^n}$, andConjugation makes $Y$ a compact $\Lambda$-module. The goal is to show that it is finitely generated and has Iwasawa-module rank zero. This proof applies to every Zp-extension, not only a cyclotomic one, and makes no Leopoldt conjecture assumption.

By the ramification argument above, only primes over $p$ ramify in $F_{\infty }/F$, and at least one does. Every nonzero closed inertia group in $\Gamma \cong {\mathbb{Z}}_p$ is open. After replacing $F$ by a finite layer, every prime which ramifies is totally ramified in the remaining tower. The number $s\ge 1$ of such primes then stays constant. This replacement does not affect whether $Y$ is a torsion module: ${\mathbb{Z}}_p[[\Gamma ]]$ is finite free over the Iwasawa algebra of an open subgroup, and the two module ranks vanish together.

Let ${\mathcal{G}}_n=\mathrm{Gal}(L_{\infty }/F_n)$, and let $D_n$ be its maximal abelian quotient. The abelianization over a Zp-extension formula gives an exact sequenceChoose a lift of a generator of ${\Gamma }_n$ if a splitting is desired. We will bound the ${\mathbb{Z}}_p$-rank of the middle term independently of $n$.

At each of the $s$ ramified primes, inertia in $L_{\infty }/F_n$ maps isomorphically onto ${\Gamma }_n$: its kernel is inertia in the unramified extension $L_{\infty }/F_{\infty }$, hence trivial, and the map onto the totally ramified base inertia is surjective. Thus each image in $D_n$ is procyclic and has ${\mathbb{Z}}_p$-rank at most one. No other finite prime contributes inertia. Quotienting $D_n$ by the closed subgroup generated by these $s$ images gives an abelian extension of $F_n$ unramified at all finite primes. Its Galois group is finite by class field theory, using the ordinary or narrow ideal class group according to the treatment of real places. Any infinite-place inertia is finite and does not affect the rank bound. Therefore $D_n$ is finitely generated over ${\mathbb{Z}}_p$ andThe preceding exact sequence now yields the uniform boundIt does not assert that these coinvariant modules are always finite when several primes ramify.

At $n=0$, $Y/TY$ is finitely generated over ${\mathbb{Z}}_p$, so $Y/(p,T)Y$ is finite. Lift a finite basis of this quotient to $Y$. The Compact Nakayama lemma shows that these lifts generate $Y$ over $\Lambda$. Briefly, the quotient by their compact generated image satisfies $Q=(p,T)Q$. Every finite continuous $\Lambda$-quotient of $Q$ has nilpotent action by $(p,T)$: $p$ is nilpotent, and modulo $p$ a finite pro-$p$ action makes $\gamma -1$ nilpotent. Such a quotient must be zero. Finite quotients separate points of a compact pro-p group, so $Q=0$.

Suppose now that $Y$ had positive $\Lambda$-rank $r$. The Iwasawa module structure theorem provides a pseudo-isomorphism towith $E$ a finitely generated torsion module. Passing to ${\Gamma }_n$-coinvariant modules leaves a finite cokernel. Butis free of rank $p^n$ over ${\mathbb{Z}}_p$. Thus the coinvariant modules would have rank at least $r{p}^n$, contradicting the uniform bound $s-1$. Equivalently their ranks have asymptotic form $r{p}^n+O(1)$, with the torsion elementary divisors contributing only a bounded rank.

It follows that $r=0$. Since $\Lambda$ is an integral domain, rank zero means every element is killed by a nonzero scalar. ThereforeThis is the unramified Iwasawa torsion theorem. The crucial arithmetic inputs are finite class fields and the bounded number of ramified primes; neither vanishing of a $\mu$-invariant nor a cyclotomic main conjecture is required.

A profinite abelian group is an inverse limit of finite abelian groups. Its topology is compact and totally disconnected. An abelian pro-p group acquires scalar multiplication by p-adic integers by taking limits of integer multiples.

For a pro-p group $P$, vanishing of $H^{n+1}(P,{\mathbb{F}}_p)$ implies vanishing in that degree for every discrete $p$-primary module. Finite such modules have composition factors equal to the trivial module ${\mathbb{F}}_p$, so the long exact sequence gives the result by induction. General modules are unions of finite stable submodules, and continuous cohomology commutes with filtered unions. Dimension shifting gives ${\mathrm{cd}}_p(P)\le n$. The converse is immediate from the definition.