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Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with Galois group $G$, normalized valuation $v_L$, integer ring ${\mathcal{O}}_L$, and residue characteristic $p$. The lower ramification numbering isThese higher ramification groups measure successively finer agreement with the identity. The group $G_0$ is the inertia group, the kernel of the action on the residue field. Each $G_i$ is normal in $G$: conjugation permutes integral elements and preserves their valuations. The groups eventually become trivial, since each nonidentity automorphism moves some integral element and its displacement has finite valuation.

Let $K_0=L^{G_0}$, the maximal unramified subextension of a local field extension. Then $G/G_0$ is the Galois group of the residue extension, cyclic and generated by a Frobenius automorphism. For a uniformizer $\pi$ of $L$, the uniformizer criterion for lower ramification groups givesTo see why one uniformizer suffices, the extension $L/K_0$ is totally ramified and its integer ring is ${\mathcal{O}}_{K_0}[\pi ]$. Elements of $G_0$ fix its coefficients. Differences of powers of $\pi$ are divisible by $\sigma \pi -\pi$, so the valuation bound for $\pi$ implies it for every integral element. The converse follows by testing $\pi$ itself.

The tame inertia quotient is seen explicitly fromIt is a homomorphism because inertia acts trivially on the residue field, and its kernel is $G_1$. Thus $G_0/G_1$ is cyclic of order not divisible by $p$. For $i\ge 1$, the homomorphismhas kernel $G_{i+1}$. Indeed composing two automorphisms adds the leading displacements, since $\sigma \pi /\pi \equiv 1$ and inertia fixes residue coefficients. Hence each $G_i/G_{i+1}$ is an elementary abelian p-group. Eventual triviality makes $G_1$ a p-group; the prime-to-p quotient makes it the unique Sylow subgroup of $G_0$. This is the wild inertia group. Tame ramification means $G_1=1$.

The filtration also measures the different exponent. For the totally ramified extension over $K_0$, the minimal polynomial of an algebraic element $g$ of $\pi$ has polynomial roots $\sigma \pi$, $\sigma \in G_0$. The derivative formula givesFor a fixed $\sigma$, its displacement valuation is the number of indices $i\ge 0$ for which it belongs to $G_i$. Interchanging the finite sums therefore gives different exponent from ramification groups:The unramified part contributes zero. More generally different in a tower follows by taking products of bases of a module dual for the trace pairing, and givesThus a tame extension has different exponent $e-1$, while nontrivial wild groups contribute additional positive terms.

The lower numbering restricts to subgroups: if $H=\mathrm{Gal}(L/E)$, its groups are $H_i=H\cap G_i$, directly from the definition using the same top-field valuation. Quotients need a different indexing. For real $s\ge 0$ put $G_s=G_{\lceil s\rceil }$ and define the Herbrand functionThis upper ramification numbering stretches or compresses intervals according to the inertia indices. The quotient theorem says that for a normal subgroup $H$, upper ramification groups commute with quotients:In particular the same upper index has a meaning compatible with passing to a smaller Galois extension, which lower indices generally lack.

For a tame totally ramified extension the picture is simple: $G_0=G$, all $G_i$ with $i\ge 1$ are trivial, and the different exponent is $|G|-1$. The extension ${\mathbb{Q}}_p({\zeta }_p)$ for odd $p$ illustrates this, giving exponent $p-2$.

A mixed-characteristic wild family is the lower ramification filtration of a prime-power cyclotomic extension. Take $L={\mathbb{Q}}_p({\zeta }_{p^r})$ with odd $p$ and $\pi ={\zeta }_{p^r}-1$. If ${\sigma }_a(\zeta )={\zeta }^a$ and $a\ne 1$ modulo $p^r$, thenThe last equality follows because ${\zeta }^{a-1}$ has order $p^{r-v_p(a-1)}$ and its uniformizer valuation scales by the ramification index in that cyclotomic tower. The lower breaks are $0,p-1,p^2-1,\dots ,p^{r-1}-1$. Integrating the successive indices $(p-1),p(p-1),\dots$ gives upper breaks $0,1,\dots ,r-1$.

In equal characteristic there is an equally explicit wild example. Start with $K={\mathbb{F}}_q((t))$ and an Artin–Schreier extensionThere is no polynomial root in $K$: a negative-valuation element has $v(x^p-x)=pv(x)$, while the right side has valuation $-m$. Adjoining a polynomial root splits the polynomial, whose polynomial roots are $y+c$ for $c\in {\mathbb{F}}_p$, so the extension is cyclic of degree $p$. The valuation equation forces total ramification, $v_L(t)=p$ and $v_L(y)=-m$. Choose $1\le b\le p-1$ with $-bm\equiv 1(\mathrm{mod}p)$ and set $a=(1+bm)/p$. Then $\pi =t^a{y}^b$ is a uniformizer. For a nonidentity automorphism $y\mapsto y+c$,the first binomial coefficient $b$ is nonzero modulo $p$, and all subsequent terms have strictly higher valuation. ThereforeThis is the ramification break of an Artin–Schreier pole. It shows concretely that wild breaks, unlike tame inertia, can occur arbitrarily far out in the filtration.

The polynomial is Eisenstein at every prime number $p$, so a polynomial root $\alpha$ generates a totally ramified cubic field $F/{\mathbb{Q}}_p$. The splitting-field Galois group is a transitive subgroup of $S_3$, hence $C_3$ or $S_3$. Its polynomial discriminant isThe product of the three polynomial root differences is a square root of $\Delta$, transformed by the sign of the polynomial root permutation. Thus the group is $C_3$ exactly when $\Delta$ is a square, and otherwise it is $S_3$. In the latter case the splitting field isIndeed adjoining one polynomial root leaves a quadratic factor, whose polynomial discriminant is $\Delta /f^{\prime }(\alpha {)}^2$, so the square root of $\Delta$ supplies its remaining polynomial roots.

Suppose first that $p>3$. The unit $-(4p+27)$ reduces to $-27$, and Hensel's lemma makes it a square exactly when $-3$ is a square modulo $p$. Quadratic reciprocity givesIf $p\equiv 1(\mathrm{mod}3)$, the field generated by $\alpha$ is already the cyclic splitting field, totally ramified of degree three. If $p\equiv 2(\mathrm{mod}3)$, the nonsquare unit gives an unramified quadratic field $E={\mathbb{Q}}_p(\sqrt{\Delta })$. Since $F$ and $E$ have coprime degrees, $[FE:{\mathbb{Q}}_p]=6$. Its residue degree is at least two and its ramification index at least three; their product is six, so they are exactly two and three. Thus its inertia group is the normal $C_3$ inside $S_3$.

In both cases the ramification index is three and is not divisible by $p$. The wild inertia group is trivial, so all positive lower groups are trivial. As a different check, $f^{\prime }(\alpha )=3{\alpha }^2+p$ has valuation two in $F$ when $p>3$, agreeing with tame different exponent $3-1$.

For $p=3$ the polynomial discriminant is $-351=-3^3\cdot 13$, which is nonsquare. Hence $G=S_3$. The quadratic subfield can be written $E={\mathbb{Q}}_3(\beta )$ with ${\beta }^2=-39$. It is ramified of degree two, by the Eisenstein polynomial $T^2+39$. Together with the totally ramified cubic field this forces $e(L/{\mathbb{Q}}_3)$ to be divisible by both two and three. Since $[L:{\mathbb{Q}}_3]=6$, the splitting field is totally ramified. In its normalized valuation,

All differences between distinct polynomial roots have the same valuation, since $S_3$ permutes the polynomial root pairs and preserves the absolute value on a field. Alsohas valuation six, so each polynomial root difference has valuation three. A nonidentity three-cycle fixes $\beta$, and thereforeThe transposition fixing $\alpha$ negates $\beta$, so sends $\pi$ to $-\pi$ and has displacement valuation $v_L(2\pi )=1$. The other transpositions have the same value by conjugacy. The uniformizer criterion now determines every group:For the wild case the different exponent is $5+2=7$. Independently, the cubic subfield has different exponent $v_F(3({\alpha }^2+1))=3$, and the quadratic extension $L/F$ is tame with exponent one. The tower formula gives $1+2\cdot 3=7$, confirming the filtration. Its positive upper break is $1/2$, because $[G_0:G_1]=2$. This completes ramification of the splitting field of X3 plus pX plus p for every odd prime number.

Put $a=\sqrt[3]{2}$ and $\zeta ={\zeta }_3$. The polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $2$, so $[\mathbb{Q}(a):\mathbb{Q}]=3$. This field is real and cannot contain the nonreal $\zeta$. Adjoining $\zeta$ gives a quadratic extension and hence $[F:\mathbb{Q}]=6$. Since $F$ is the splitting field of $X^3-2$, its Galois group acts faithfully on $a,\zeta a,{\zeta }^{2}a$ and has order six. Consequently

For a prime $\mathfrak{q}$ over a rational prime $p$, its absolute residue degree is the order of the cyclic quotient group $D_{\mathfrak{q}}/I_{\mathfrak{q}}$ from part (i). If that degree were six, then $D_{\mathfrak{q}}$, a subgroup of $S_3$, would have order at least six, so $D_{\mathfrak{q}}=S_3$ and $I_{\mathfrak{q}}=1$. This would make $S_3$ cyclic, a contradiction. The argument applies also at ramified primes. Thus

For the second request printed with the same subpart label, work over $E=\mathbb{Q}(\zeta )$. The Galois group of $F/E$ is cyclic of order three, with generator $\sigma (a)=\zeta a$ and $\sigma (\zeta )=\zeta$. The ideal norm of $(\zeta +3)$ is seven, sinceIn its residue field we have $\zeta \equiv -3\equiv 4(\mathrm{mod}{\mathfrak{p}}_1)$. The polynomial $X^3-2$ has no root in ${\mathbb{F}}_7$, because its cube residues are $0,1,6$. It is therefore irreducible there. Its discriminant $-108$ is a unit at ${\mathfrak{p}}_1$. To check the local index condition, the trace pairing matrix of $1,a,a^2$ has this unit determinant. For any integral element $b$ of $F$, its traces against these three basis elements are integral; solving with this invertible matrix shows that its basis coefficients are integral locally. Thus the relative ring of integers equals ${\mathcal{O}}_E[a]$ at ${\mathfrak{p}}_1$. The relative Kummer-Dedekind theorem gives one unramified prime above ${\mathfrak{p}}_1$, of residue degree three.

Use arithmetic Frobenius, characterized by $b\mapsto b^7$ on this residue field. As $a^3=2$,The three possible images $a,\zeta a,{\zeta }^{2}a$ remain distinct modulo $\mathfrak{q}$, since $a$ is nonzero and $1,4,2$ are distinct modulo seven. This congruence therefore identifies the actual automorphism uniquely:The geometric Frobenius convention would give its inverse. Since $F/E$ is abelian, the arithmetic element is independent of the choice of prime above ${\mathfrak{p}}_1$.

We use the Dirichlet density form of the Chebotarev density theorem, the version directly proved by the permitted information at $s=1$. If $L/K$ is a finite Galois extension of number fields with group $G$ and $C$ is a conjugacy class, let $S_C$ be the unramified prime ideals of $K$ whose Frobenius conjugacy class is $C$. ThenThe same conclusion holds for a union of conjugacy classes, replacing $|C|$ by the size of that union. The denominator may equivalently be the sum over all primes of $K$. Here and below $s>1$ is real. This specifies the density being proved; an unweighted prime-counting asymptotic is a stronger formulation and is not being inferred merely from a limit at one.

First establish the necessary pole for a general base field. The Dedekind zeta function as a permutation Artin L-function gives it using exactly the allowed assumptions. Take a finite normal closure $E$ of $K/\mathbb{Q}$, and let $W$ be the permutation representation on ${\mathrm{Hom}}_{\mathbb{Q}}(K,E)$. Its trivial constituent has multiplicity one: an invariant function on the transitive set of embeddings is constant. At a rational prime, a basis of $W^I$ consists of the sums over inertia orbits of embeddings. Frobenius automorphism permutes these orbit sums. Within the decomposition-group orbit belonging to a prime $\mathfrak{p}$ of $K$, the resulting cycle has length $f_{\mathfrak{p}}$, because the residue embeddings are permuted cyclically by finite-field Frobenius. A cycle of length $f$ has determinant $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Consequently the local factor of an Artin L-function givesincluding at ramified primes. Decomposing $W=1\oplus {⨁}_{\psi \ne 1}{m}_{\psi }\psi$ givesAll factors after $\zeta$ are holomorphic and nonzero at one by the allowed hypothesis. Thus ${\zeta }_K$ has a simple pole there andThis argument also explains why the trivial relative Artin L-function is ${\zeta }_K$, rather than always the Riemann zeta function.

Now let $\chi$ be an irreducible character of $G$, of degree $d_{\chi }$. At an unramified prime the Euler product has local factor $\det (1-{\rho }_{\chi }(g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}{)}^{-1}$, where $g_{\mathfrak{p}}$ is a Frobenius representative. The eigenvalues of ${\rho }_{\chi }(g)$ are roots of unity, since $g$ has finite order. Expanding the logarithm for $s>1$ therefore giveswhere the finitely many ramified local factors contribute a function bounded as $s\downarrow 1$. They are nonzero and finite near one because their eigenvalues have modulus one while $(N\mathfrak{p}{)}^{-1}<1$.

The terms $m\ge 2$ are uniformly bounded. Indeed $|\chi (g)|\le d_{\chi }$, and for $N\mathfrak{p}\ge 2$ and $s\ge 1$,There are at most $[K:\mathbb{Q}]$ primes of $K$ above each rational prime and each has norm at least that prime. Hence ${\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. Isolating $m=1$ proves the character prime sums in the Chebotarev proof identityFor nontrivial $\chi$, the allowed holomorphy and nonvanishing at one give a local holomorphic logarithm and a bounded right side. The logarithm specified by the Euler series differs from that local branch by a constant multiple of $2\pi i$ on the real interval near one, so it too is bounded. For the trivial character, the right side is $\log (1/(s-1))+O(1)$ by the pole of ${\zeta }_K$. Thus

Expand the conjugacy-class indicator by character orthogonality. For $c\in C$,To see the coefficient, take its inner product with $\chi$: $|G{|}^{-1}{\sum }_{g\in C}\overline{\chi (g)}=|C|\overline{\chi (c)}/|G|$. The irreducible characters form an orthonormal basis of class functions, so this is the full expansion. Substituting the prime-sum estimates leaves only the trivial character's logarithmic term:Division by the diverging logarithm proves the theorem. The same trivial-character calculation shows that all primes of $K$ have density one; adding or removing finitely many primes does not change a Dirichlet density.

There is a necessary qualification for the final polynomial application: the printed claim is false for degree one. For example, $X$ is monic and irreducible in $\mathbb{Z}[X]$, yet has the root zero modulo every prime. We prove the intended assertion for degree $n\ge 2$.

Let $L$ be the splitting field of the monic irreducible polynomial over $\mathbb{Q}$, and let $\Omega$ be its $n$ roots. The Galois group $G$ acts transitively on $\Omega$. There is a derangement in a transitive group action: count the pairs $(g,\alpha )$ satisfying $g\alpha =\alpha$. The orbit-stabilizer theorem givesIf every $g$ fixed at least one root, the identity, which fixes $n\ge 2$, would make this sum at least $|G|+n-1$, a contradiction. Therefore some $g$ fixes no root, and neither does any conjugate of $g$.

For completeness, relate this group action to reduction without assuming the desired splitting result. All roots are algebraic integers, since the polynomial is monic. Exclude the finitely many primes ramifying in $L$ or dividing the nonzero polynomial discriminant. At a prime $\mathfrak{q}$ over any remaining $p$, the roots reduce to distinct elements of $k_{\mathfrak{q}}$ and give all roots of the reduced polynomial. Arithmetic Frobenius permutes them by $b\mapsto b^p$. A reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed by this map. Because reduction is injective on the original root set, this happens exactly when the Frobenius permutation fixes an original root. This proves the Frobenius permutation and roots modulo a prime criterion.

The Chebotarev density theorem assigns the conjugacy class of our fixed-point-free $g$ the positive Dirichlet density $|C_g|/|G|$. All but finitely many of these primes have rootless reduction, so there are infinitely many. More precisely, applying the same argument to the union of all fixed-point-free classes gives

Let $L=F({\mu }_n)$. Its Galois group embeds into $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$, by sending an automorphism to the exponent in its action on ${\zeta }_n$. It is therefore finite abelian, of order dividing $\phi (n)$. The finite case of the Fundamental theorem of finitely generated abelian groups supplies a subgroup chain whose successive quotients are cyclic of prime order $\ell$. Every such $\ell <n$, since $\ell \le \phi (n)<n$ for $n>1$ (and for $n=2$ the possible prime is $2$, whose roots of unity already lie in every characteristic-zero field). Thus every intermediate base contains ${\mu }_{\ell }$, because $F$ contains all roots of unity of orders smaller than $n$.

Each cyclic degree-$\ell$ step $E^{\prime }/E$ is a Kummer extension. Here is the relevant criterion explicitly. For a generator $\sigma$ and a primitive $\ell$th root $\zeta \in E$, the operator ${\sum }_{j=0}^{\ell -1}{\zeta }^{-j}{\sigma }^j$ is not identically zero, by linear independence of distinct field automorphisms. Choose its nonzero value $a$; then $\sigma (a)=\zeta a$, so $a^{\ell }\in E$ and $a\notin E$. Since the degree is prime, $E^{\prime }=E(a)$ is obtained by adjoining an $\ell$th radical. Applying this to the subgroup chain proves $F({\mu }_n)/F$ is a succession of Kummer extensions.

Finally use strong induction on $n$, for arbitrary characteristic-zero base fields. Successively adjoining the roots ${\zeta }_1,\dots ,{\zeta }_{n-1}$ is contained in a Kummer tower by the induction hypothesis, applied over each current base. The just-proved assertion then adjoins ${\zeta }_n$ by further Kummer steps. Hence any cyclotomic extension is contained in a Kummer tower. In the given cyclotomic-and-Kummer tower for a soluble extension, replace every cyclotomic step by such a tower; enlarge subsequent steps by compositum with the new intermediate fields. Base change preserves radical adjunction and the requisite roots of unity, possibly making a step trivial. Thus every finite soluble extension is contained in a succession of Kummer extensions.

The previous permutation calculation shows that every automorphism either preserves $x^3,y^3$ separately or interchanges them. Their sum and product are therefore fixed by the Galois group and belong to $K$, proving the quadratic assertion.

For the depressed cubic, $\alpha +\beta +\gamma =0$, $\alpha \beta +\alpha \gamma +\beta \gamma =b$, and $\alpha \beta \gamma =-c$. Using $1+\zeta +{\zeta }^2=0$ givesand henceAlso $x^3{y}^3=-27b^3$. The desired polynomial isIts discriminant is $729c^2+108b^3=-27D$, where $D=-4b^3-27c^2$. Here $-27$ is a square in $K$, since $[3(\zeta -{\zeta }^2){]}^2=-27$.

One can justify the group criterion directly even when one resolvent vanishes. The nonzero root-difference producthas $d^2=D$ and changes sign exactly under odd root permutations. It is fixed by the entire group exactly when the group is contained in $A_3$. If $D$ is a square in $K$, then $d$ equals one of its two square roots in $K$; conversely, a fixed $d$ lies in $K$. ThereforeSeparability ensures $D\ne 0$, and the characteristic hypotheses ensure that all divisions and the sign argument are valid.

The irreducible cubic is separable, and its Galois group acts transitively on its three roots. The only transitive subgroups of $S_3$ areFor the cycle $\sigma :(\alpha ,\beta ,\gamma )\mapsto (\beta ,\gamma ,\alpha )$, the Lagrange resolvent for a cubic obeys $\sigma (x)={\zeta }^{2}x$. A transposition interchanging $\beta ,\gamma$ sends $x$ to $y=\alpha +\zeta \gamma +{\zeta }^2\beta$. Since $\zeta \in K$, these scalars are fixed by every automorphism. Thus the two possibilities for the image set areThese are descriptions of sets, so coincident values or a zero resolvent are not being asserted distinct.

The cyclotomic Galois group theorem statesfor $b$ coprime to $n$, with $[\mathbb{Q}({\mu }_n):\mathbb{Q}]=\phi (n)$, where $\phi$ is the Euler totient function.

Factorization over $K$ gives the orbits of the Galois group on the three resolvent roots. If $Q$ splits completely, the quotient is trivial and $G=V_4$. If $Q$ is a linear factor times an irreducible quadratic, the quotient is $C_2$ and $G=D_8$.

If $Q$ is irreducible, the quotient acts transitively on three roots, so it is $C_3$ or $S_3$. The product of their pairwise differences is a square root of the nonzero discriminant $\Delta$ and changes by the sign of the permutation. Thus $\Delta$ is a square in $K$ exactly when the quotient contains only even permutations. The criterion isThe assumed containment of $V_4$ is essential to reconstruct $G$ from the quotient by this table.

The three resolvent cubic roots correspond to the partitions $12|34$, $13|24$, $14|23$. Permuting the four roots of the quartic permutes these three quantities and defines a homomorphism $G\to S_3$. They are distinct: because the quartic has zero cubic coefficient, their differences includesince characteristic zero and irreducibility imply separability. Thus the kernel of the action is exactly $V_4$. The splitting field of the resolvent is the subfield generated by its roots; restriction from the quartic splitting field is surjective onto its Galois group, since automorphisms of a normal separable intermediate extension extend to the splitting field. HenceIn the four cases these groups are respectively the trivial group, $C_2$, $C_3$ and $S_3$. Computing the symmetric functions also gives $Q(X)=X^3+2b{X}^2+(b^2-4d)X-c^2$ with the source's root convention.

Over $K=\mathbb{Q}$, takeIts roots are $\pm \sqrt{2}\pm \sqrt{3}$. For $\alpha =\sqrt{2}+\sqrt{3}$, its inverse is $\sqrt{3}-\sqrt{2}$, so both square roots belong to $\mathbb{Q}(\alpha )$. Since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree four, $P$ is irreducible and this field is its splitting field. Independently changing the signs of the two square roots gives its Galois group $C_2\times C_2=V_4$, which is abelian and satisfies the hypothesis.

Write $d=\mathrm{deg}A$ for the degree of a central simple algebra, so ${\dim }_{K}A=d^2$. Choose a finite Galois splitting field of a central simple algebra $E/K$ for $A$ and an isomorphism $\varphi :A{\otimes }_{K}E\to M_d(E)$. The reduced norm isThe existence of a finite separable splitting field is a standard structural property of a central simple algebra; passing to its Galois closure supplies $E$.

Every $E$-algebra automorphism of $M_d(E)$ is inner. Here is the matrix-unit argument for the inner automorphisms of a matrix algebra result. For an automorphism $h$, choose $w\ne 0$ in the image of $h(E_{11})$, and set $w_i=h(E_{i1})w$. Then $h(E_{ij})w_k={\delta }_{jk}{w}_i$. The $w_i$ are nonzero and independent, and their number is $d$, so they form a basis. Relative to that basis $h$ acts on each matrix unit in the usual way. Thus $h(T)=PT{P}^{-1}$ for some $P\in {\mathrm{GL}}_d(E)$. Determinants are unchanged by conjugation, proving independence of $\varphi$.

For $\sigma \in \mathrm{Gal}(E/K)$, compare $\varphi$ with the isomorphism obtained by applying $\sigma$ to matrix entries and ${\sigma }^{-1}$ to the scalar factor of $A\otimes E$. Their difference is again inner. For $a\in A$, this shows $\sigma (\det \varphi (a))=\det \varphi (a)$. The determinant therefore belongs to $K$. In fact, for a $K$-basis $a_1,\dots ,a_{d^2}$, the same comparison shows that all coefficients ofare fixed by the Galois group, so the reduced norm is a homogeneous polynomial of degree $d$ over $K$. This coefficient argument also applies over finite fields, where equality merely as functions would not identify polynomials.

Finally, two finite splitting fields embed into a common finite splitting extension. Determinant commutes with scalar extension, and over that common extension the two matrix identifications differ by an inner automorphism. Hence the resulting polynomials agree over $K$. The reduced norm is independent of both the splitting field and the matrix identification. It is multiplicative, and $\mathrm{Nrd}(a)=0$ exactly when $a$ is not invertible: a matrix with nonzero determinant is invertible after scalar extension, and invertibility descends by the invertibility of the $K$-linear multiplication map. In a division algebra the only element of reduced norm zero is zero.

Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

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.

All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore