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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.

The idele group is the multiplicative restricted productIts restricted product topology has basic open sets ${\prod }_{v\in S}{U}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains every infinite place, and $U_v$ is open in $K_v^{\times }$. In particular, the restriction to almost all local unit groups is part of the topology, not merely the unrestricted product topology.

The diagonal map $K^{\times }\to J_K$ is injective and well-defined by finite support of the principal ideal's valuations; its image consists of principal ideles. To prove discreteness of principal ideles, take the neighbourhood of $1$ which requires $x_v\in {\mathcal{O}}_v^{\times }$ at every finite place and $|x_v-1|<1/2$ in the ordinary real or complex modulus at every infinite place. A diagonal element in it is a global unit, so $x-1\in {\mathcal{O}}_K$. If $x\ne 1$, the nonzero integer $N_{K/\mathbb{Q}}(x-1)$ has absolute value at least one, but the archimedean bounds make its absolute value less than $2^{-[K:\mathbb{Q}]}$. This contradiction proves $K^{\times }$ is a discrete subgroup of $J_K$.

For a modulus of a number field, write $\mathfrak{m}$ for its finite ideal part together with any selected real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals coprime to the finite part. Let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$ locally at each finite place in the modulus, and $a>0$ at each selected real place. The ray class group isEquivalently it is the quotient $J_K/(K^{\times }{U}_{\mathfrak{m}})$, with the corresponding principal-unit factors at finite places and positive multiplicative factors at selected real places. The question now assumes that no infinite places occur.

Put $R_{\mathfrak{m}}={\prod }_{v\in S}{\mathcal{O}}_v^{\times }/(1+{\pi }_v^{m_v}{\mathcal{O}}_v)$. Global units map to $R_{\mathfrak{m}}$ by reduction. To map $R_{\mathfrak{m}}$ to the ray class group, choose, by the Chinese remainder theorem, an integral $a$ having the prescribed unit residues at all $v\in S$, and send the tuple to the ray class of $(a)$. A different choice changes the ratio by an element congruent to one at every such place, so the map is well-defined and multiplicative.

The forgetful map to the ordinary ideal class group is surjective: weak approximation for number fields multiplies an ideal by a principal ideal so that all its valuations in $S$ vanish. Its kernel consists exactly of classes represented by principal ideals coprime to $S$, and their generators supply the local unit residues. Finally, a residue tuple maps to the identity ray class precisely when its chosen $a$ differs from a congruence-one generator by a global unit. Thus the ray class exact sequence isThe leftmost arrow need not be injective, explaining the absence of an initial zero.

For $K=\mathbb{Q}(\sqrt{2})$, the ring of integers of Q of square root two is $\mathbb{Z}[\sqrt{2}]$. Its absolute field norm is a Euclidean function. Indeed, for $a+b\sqrt{2}\in K$, choose nearest integers $m,n$; writing $u=a-m$, $w=b-n$, both bounded by $1/2$, gives $|u^2-2w^2|\le 1/2<1$. Applying this to a quotient proves Euclidean division, hence the ring is a principal ideal domain and $\mathrm{Cl}(K)=0$. This is norm-Euclideanity of the integers adjoined square root two.

Write $\mathfrak{p}=(\sqrt{2})$ and $\mathfrak{q}=(1+2\sqrt{2})$. Their ideal norms are $2$ and $7$, respectively; ${\mathfrak{p}}^2=(2)$ and the residue field at $\mathfrak{q}$ is ${\mathbb{F}}_7$, with $\sqrt{2}\equiv 3$ because $1+2\sqrt{2}\equiv 0$. The finite local quotient product for the requested modulus isThe first factor consists of $1$ and $1+t$.

The given fundamental unit $ϵ=1+\sqrt{2}$ and $-1$ generate the global units. Their images areThe first image has order six; its cube is $(1+t,1)$ and its square is $(1,2)$, generating the first factor and the order-three subgroup of the second. The second image supplies the missing order-two element of ${\mathbb{F}}_7^{\times }$. Thus the global unit map is surjective onto all $12$ local unit residues. By the exact sequence, or the unit-surjectivity criterion for a trivial finite ray class group,There is no positivity condition here: the modulus has no infinite part, so using the unit $-1$ is legitimate and essential in the surjectivity calculation.

Classical Kummer theory relates extraction of $n$th roots to Galois cohomology. In characteristic zero the exact sequenceand Hilbert theorem 90 identify $H^1(K,{\mu }_n)$ with $K^{\times }/K^{\times n}$. For an elliptic curve, the corresponding Kummer exact sequence of an elliptic curve isMultiplication by $n$ is surjective over the algebraic closure. If $nQ=P\in E(K)$, the cocycle $\sigma \mapsto \sigma Q-Q$ takes values in $E[n]$. Changing the choice of $Q$ changes it by a coboundary, and changing $P$ by an element of $nE(K)$ does not change its class. Conversely, a trivial cocycle class allows $Q$ to be adjusted by an $n$-torsion point to become $K$-rational. Thus the Kummer map of an elliptic curve is an injectionThis is the elliptic form of Kummer theory. The full cohomology group need not be finite; the arithmetic restriction on these classes is essential.

Choose a finite set $S$ of places containing the archimedean places, the primes over $n$, and all primes of bad reduction of an elliptic curve. At a finite place outside $S$, $E$ has good reduction and $n$ is a unit. A division point of the reduction of $P$ exists over the algebraic closure of the residue field. Smooth lifting gives a point over a finite unramified extension whose multiple differs from $P$ by an element of the kernel of reduction of an elliptic curve. In the formal group of an elliptic curve, $[n](T)=nT+O(T^2)$ is an isomorphism by the invertible morphism criterion for formal group laws, and its integral inverse converges on the maximal ideal. Correcting that difference produces a division point in the maximal unramified extension. Hence the Kummer map of an elliptic curve class is unramified outside $S$.

To prove finiteness explicitly, choose a finite Galois extension $L/K$ containing all $E[n]$ and all $n$th roots of unity, and enlarge $S$ to include its ramified places. A basis of $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$ identifies it over $L$ with ${\mu }_n^2$. Classical Kummer theory then identifiesThe restriction of every class in the image of $\delta$ belongs to $L(S_L,n{)}^2$, where the S-unramified power class group isIndeed an unramified local Kummer extension at residue characteristic prime to $n$ has valuation divisible by $n$: in an unramified field containing a root, $nv(a^{1/n})=v(a)$ with integral valuations.

The finiteness of S-unramified Kummer classes follows from the exact sequenceTo see the final map, write the ideal of $a$ away from $S_L$ as ${\mathfrak{a}}^n$ and take the ideal class of $\mathfrak{a}$. Its kernel is represented by an S-unit, after division by an $n$th power; conversely an $n$-torsion ideal class yields such an $a$. The S-unit group is finitely generated by the Dirichlet unit theorem together with the finitely many inverted primes. The ideal class group of the localized ring is a quotient of the finite ordinary ideal class group. Both outer groups are therefore finite.

Finally restriction has finite kernel: inflation-restriction puts it in the finite group $H^1(\mathrm{Gal}(L/K),E[n])$. Thus the image of $\delta$ has finite restriction image and finite kernel, andThis proves the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the full Mordell-Weil theorem. Local restrictions at every place refine the finite group used here to the Selmer group of an elliptic curve, which is useful for explicit descent calculations.