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Complex conjugation sends a complex number $z=a+ib$ to its complex conjugate $\overline{z}=a-ib$. It is an involution and a field automorphism of $\mathbb{C}$, with fixed field $\mathbb{R}$. Applied componentwise to a complex vector, it produces the conjugated vector appearing in a Choi state for a rank-one Kraus operator.

The field automorphism group fixing $K$ consists of all field automorphisms of $L$ fixing $K$ pointwise, with composition. It is defined without assuming that $L/K$ is Galois. For $L=K(\alpha )$ with algebraic degree $d$, the action on the roots of the minimal polynomial inside $L$ is faithful, so this group embeds in $S_d$ and is finite.

Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.

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.

The polynomial $X^m-1$ is separable, since its derivative $m{X}^{m-1}$ has no common root with it when the characteristic does not divide $m$. Its roots form a cyclic multiplicative group of order $m$. For completeness, any finite subgroup of a field's multiplicative group is cyclic: if its exponent is $e$, all its elements are roots of $X^e-1$, so its size is at most $e$, while the elementary decomposition of a finite abelian group gives an element of order $e$.

Choose a primitive root $\zeta$. Then $L=K(\zeta )$, and every field automorphism sends $\zeta$ to ${\zeta }^a$ for a unique unit $a$ modulo $m$. Composition multiplies the exponents, and fixing $\zeta$ fixes all of $L$. This is the required injective homomorphismFor $K={\mathbb{F}}_q$, the least extension ${\mathbb{F}}_{q^d}$ containing a primitive $m$th root is characterized by $m\mid q^d-1$, since its multiplicative group is cyclic of order $q^d-1$. Thus $[L:K]$ is the multiplicative order of $q$ modulo $m$. The successive powers of $4$ modulo $11$ are $4,5,9,3,1$, so

Galois cohomology and finite quotients. Let $k$ be a number field and $G_k=\mathrm{Gal}(\overline{k}/k)$ its absolute Galois group. For a discrete abelian $G_k$-module $M$, $H^0(k,M)=M^{G_k}$. A continuous 1-cocycle is a map $c:G_k\to M$ with $c_{\sigma \tau }=c_{\sigma }+\sigma c_{\tau }$; a coboundary has the form $c_{\sigma }=\sigma m-m$. Their quotient is $H^1(k,M)$. Short exact sequences of modules give the long exact sequence in group cohomology; the connecting map measures the obstruction to lifting a fixed element to a fixed preimage. This interpretation gives a concrete descent map for an elliptic curve.

For an integer $m\ge 2$, multiplication on the geometric points gives the exact sequenceHere multiplication is a separable isogeny in characteristic zero, with $m^2$ geometric kernel points. For $P\in E(k)$ choose $Q$ with $mQ=P$. Then $c_{\sigma }=\sigma Q-Q$ lies in $E[m]$, satisfies the cocycle identity, and changing $Q$ changes it by a coboundary. Its class $\delta (P)$ vanishes exactly when $P\in mE(k)$: if $c_{\sigma }=\sigma T-T$ for $T\in E[m]$, the point $Q-T$ is rational and divides $P$. Thus the Kummer exact sequence of an elliptic curve includesThe construction also proves that $\delta$ is a homomorphism. The final group can be interpreted as classes of elliptic-curve torsors killed by $m$; imposing soluble completions leads to the Selmer group of an elliptic curve. For weak Mordell-Weil, the important task is to bound the arithmetic image of $\delta$, not to assume that the entire first cohomology group is finite.

Choose a finite set $S$ containing the bad primes, primes dividing $m$, and archimedean places. At a good finite place $v\notin S$, the class $\delta (P)$ is unramified. Here is the local reason. Over a finite extension of the residue field choose an $m$-division point of the reduction of $P$; multiplication is surjective on the points over its algebraic closure. Pass to the corresponding unramified extension of $k_v$ and lift this nonsingular point by the Hensel lemma. The discrepancy between $m$ times its lift and $P$ lies in the kernel of reduction. The formal group of an elliptic curve describes that kernel, and multiplication by $m$ is bijective there because its leading coefficient is a unit. Correcting the lift therefore gives an actual division point in the unramified closure. Inertia fixes that point, so the Kummer cocycle vanishes on inertia. The same good-reduction argument shows that $E[m]$ itself is unramified outside $S$.

We now prove the required finiteness of unramified cohomology classes. Choose a finite Galois extension $L/k$ over which $E[m]$ and the $m$th roots of unity are all rational, and enlarge $S$ to include its ramified primes. The inflation-restriction exact sequence shows that restriction to $G_L$ has finite kernel: that kernel comes from $H^1(\mathrm{Gal}(L/k),E[m])$, a finite group since both its group and coefficient module are finite. Over $L$, a choice of basis identifies $E[m]$ with ${\mu }_m^2$. The multiplicative Kummer sequence, together with Hilbert 90, givesFor clarity, Hilbert 90 says that a multiplicative cocycle for a finite Galois extension is a coboundary. If its values are $c_{\sigma }$, choose $a$ so that $b={\sum }_{\sigma }{c}_{\sigma }\sigma (a)\ne 0$. Then $\tau (b)=c_{\tau }^{-1}b$, so $c_{\tau }=\tau (b^{-1})/b^{-1}$. Such an $a$ exists because distinct field automorphisms are linearly independent: a relation with the fewest nonzero terms, evaluated on $at$ and compared with its value on $t$ multiplied by one automorphism's value on $a$, would produce a shorter nonzero relation. Continuity reduces the absolute-Galois version to a finite extension. This proves the cohomological vanishing used in the multiplicative Kummer sequence.

An unramified Kummer class represented by $a\in L^{\times }$ has $v_w(a)\equiv 0(\mathrm{mod}m)$ at every finite $w\notin S_L$. Indeed an $m$th root of $a$ fixed by inertia belongs to an unramified local extension, whose value group is still the integers. DefineFactoring the fractional ideal away from $S_L$ as an $m$th power gives the exact sequenceTo check it, send $[a]$ to the ideal class of its $m$th root ideal. Changing $a$ by an $m$th power changes that ideal by a principal ideal. The kernel consists precisely of S-units modulo $m$th powers, and every $m$-torsion ideal class has a representative whose $m$th power is principal, proving surjectivity. The S-unit group is finitely generated by the Dirichlet unit theorem and the finite collection of allowed prime valuations. The localized ideal class group is a quotient of the finite ordinary ideal class group. Both end groups, and hence $L(S_L,m)$, are finite. This is the finiteness of S-unramified Kummer classes argument.

The restrictions of the elliptic Kummer image lie in two copies of this finite group, and the restriction kernel is finite. The injection $\delta$ therefore provesThis is the Weak Mordell-Weil theorem. The unramified restriction is essential: even $H^1(k,{\mu }_2)=k^{\times }/k^{\times 2}$ is generally infinite. Cohomology turns divisibility into a cocycle; good reduction restricts its ramification; units and ideal classes supply the arithmetic finiteness.

Height functions and finite generation. Write $d=[k:\mathbb{Q}]$ and use absolute values extending the usual rational ones, with local degrees $n_v=[k_v:{\mathbb{Q}}_v]$. The logarithmic height of a projective pair isThe product formula makes it independent of scaling. For $t\in k$, $h(t)=h(t:1)$ is nonnegative. On a short Weierstrass equation $E:y^2=x^3+Ax+B$, put $h_x(P)=h(x(P))$ and $h_x(O)=0$.

Bounded height gives only finitely many rational points. To see the scalar finiteness behind the Northcott theorem, an algebraic number $t$ of degree $e$ has primitive minimal polynomial $a_0{\prod }_{i=1}^e(T-t_i)$, and the product formula gives $e^{eh(t)}=|a_0|{\prod }_i\max (1,|t_i|)$. If $e\le d$ and $h(t)\le H$, every coefficient has absolute value at most $2^d{e}^{dH}$. There are only finitely many such integer polynomials and hence finitely many roots. Each x-coordinate lifts to at most two points of $E$, with $O$ handled separately. Thus $\{P\in E(k):h_x(P)\le H\}$ is finite.

The second ingredient is height expansion under doubling:Indeed the duplication formula is the degree-four morphism of the projective x-line represented by the coprime quartic formsThey are coprime because the elliptic curve is nonsingular. At each place their coefficient bounds give $\max (|F{|}_v,|G{|}_v)\le C_v\max (|X{|}_v,|Z{|}_v{)}^4$. Bezout identities expressing powers of each coordinate as combinations of $F,G$ give the reverse inequality up to another local constant. One obtains these identities by the Euclidean algorithm on the two affine charts and clearing denominators. Only finitely many non-Archimedean places have a constant different from one. Taking weighted logarithms and summing proves the uniform bounded error. This is height growth under a morphism of the projective line, applied with degree four.

For a fixed $R=(r,s)\in E(k)$, the addition formula also givesFor $x(P)\ne r$, its x-coordinate isThe curve equation bounds $|y{|}_v$ by a local constant times $\max (1,|x{|}_v{)}^{3/2}$. Both numerator and denominator are consequently bounded by a local constant times $\max (1,|x{|}_v{)}^2$. The projective-height definition gives the claimed inequality; the finitely many exceptional points with $x=r$ are absorbed into the constant. For $R=O$ the estimate is immediate.

By weak Mordell-Weil, choose finitely many representatives $R_1,\dots ,R_t$ for $E(k)/2E(k)$. Every $P$ can be written $P=2Q+R_i$. Combining the last two inequalities giveswith one constant $C_0\ge 0$ for all the representatives. This is naive height contraction in elliptic halving descent. If $h_x(P)>2C_0+1$, the new point $Q$ has height at least one half less than that of $P$. Repetition must therefore reach the finite set of points of height at most $2C_0+1$. Reversing the identities $P_j=2P_{j+1}+R_{i_j}$ shows that every point is generated by this finite terminal set together with the finite list of representatives. ThusThe structure theorem then gives $E(k)\cong E(k{)}_{\mathrm{tors}}\times {\mathbb{Z}}^r$ with finite torsion subgroup: this is the Mordell-Weil theorem. Finiteness modulo two alone does not imply finite generation; the contracting height argument is what bridges that gap.

The bounded duplication error also constructs the canonical height of an elliptic curve:The successive differences are bounded by a summable geometric sequence, so the limit exists, $\hat{h}(P)=\frac{1}{2}{h}_x(P)+O_E(1)$, and $\hat{h}(2P)=4\hat{h}(P)$. It is nonnegative. Torsion points have canonical height zero because their multiples form a finite set. Conversely, if the canonical height is zero, all $2^{j}P$ have bounded naive height; Northcott finiteness makes two of them equal and hence makes $P$ torsion. The canonical height removes the bounded error from the arithmetic growth law and identifies exactly the torsion directions, while the elementary naive-height descent already proves finite generation.

A finite two-transitive group action is transitive on ordered pairs of distinct points. Equivalently, the action is transitive and a point stabilizer subgroup is transitive on the remaining points. Its rank of a transitive permutation group is two, and the orbit-stabilizer theorem gives $n(n-1)\mid |G|$ for degree $n$. Such an action is primitive: a block containing a point and another point must, by the transitivity of the stabilizer subgroup, contain every point.

These groups illustrate two quite different sources of symmetry. The structural theorem for finite two-transitive groups says that they have either a regular elementary-abelian normal subgroup, giving the affine case, or a nonabelian simple normal subgroup, giving the almost simple case. The full classification of the second case depends on the classification of finite simple groups; the useful point here is to describe the constructions, actions, and representative families rather than treat two-transitivity as an abstract property of a group without a specified action.

In the affine case let $V$ be a finite vector space over ${\mathbb{F}}_p$ and let $H\le GL(V)$ act transitively on $V\setminus \{0\}$. The affine two-transitive groupis transitive by translations, and the stabilizer subgroup of zero is exactly $H$, so it is two-transitive. Conversely, if a two-transitive group has an abelian minimal normal subgroup $V$, normal-subgroup orbits and primitivity make $V$ transitive. An abelian transitive faithful permutation subgroup is regular: an element fixing one point commutes with every element moving that point and hence fixes them all. Minimal normality makes $V$ elementary abelian, since its prime-primary subgroups and its subgroup of pth powers are characteristic. Conjugation by the stabilizer subgroup acts faithfully as linear automorphisms of $V$, and regularity gives $G=V⋊G_0$. This recovers precisely the construction above.

Many examples follow at once. The group $AG{L}_d(q)$ acts on the $q^d$ vectors of ${\mathbb{F}}_q^d$; $G{L}_d(q)$ sends any nonzero vector to any other by a basis map. For $d\ge 2$, the subgroup $AS{L}_d(q)$ also gives a two-transitive action, since the determinant of a basis map can be adjusted on a second basis vector while retaining the prescribed first vector. The affine group with linear part $S{p}_{2m}(q)$ gives another family: the nonzero-vector transitivity was proved using symplectic transvections in part 4(ii). Here the additive group of ${\mathbb{F}}_q^d$, for $q=p^f$, is an elementary abelian p-group of rank $fd$.

In particular $AG{L}_1(q)$ consists of $x\mapsto ax+b$ with $a\ne 0$. Given distinct $x,y$ and distinct $u,v$, the equations $ax+b=u$, $ay+b=v$ have the unique solution $a=(v-u)/(y-x)$, $b=u-ax$. Thus this action is sharply two-transitive, of degree $q$ and order $q(q-1)$. Adding field automorphisms gives $A\Gamma L_1(q)$, still two-transitive, with elements $x\mapsto a{x}^{p^i}+b$.

An almost simple group lies between a nonabelian simple group $T$ and its automorphism group. One must still specify an action: a group can have several permutation representations with different degrees or different transitivity properties. The natural actions of $S_n$ for $n\ge 2$ and $A_n$ for $n\ge 4$ are the most familiar examples of a two-transitive group action. For $n\ge 5$, the simple group $A_n$, a normal subgroup, places these groups in the almost simple case; the smaller two-transitive examples belong to the affine case. The symmetric group sends any ordered distinct pair to any other; for the alternating group one can adjust an odd map by swapping two letters outside the target pair. Thus both are two-transitive in the indicated ranges.

The projective linear examples are just as concrete. The groups $PS{L}_n(q)$ and $PG{L}_n(q)$ act two-transitively onpoints, by the basis and determinant adjustment argument of part 4(i). Their field-automorphism extensions are also two-transitive. On the projective line, $PG{L}_2(q)$ is sharply three-transitive: three distinct points determine a unique Möbius transformation. Its degree is $q+1$ and its order is $q(q^2-1)$. For even $q$, $PS{L}_2(q)=PG{L}_2(q)$; for odd $q$ the special projective group has half that order and remains two-transitive, but cannot be three-transitive. In projective dimension at least two, triples of distinct points may be collinear or noncollinear, so the natural action is not three-transitive.

The projective special unitary group over a finite field $PS{U}_3(q)$, for $q\ge 3$, acts on the isotropic points of a space with a nondegenerate Hermitian form of dimension three over ${\mathbb{F}}_{q^2}$. Its degree and order areFor two distinct isotropic lines the Hermitian form pairing is nonzero, since a three-dimensional nondegenerate Hermitian space has maximal totally isotropic dimension one. Scale their representatives to form a hyperbolic pair and extend to a basis adapted to the Hermitian form. The corresponding isometry sends one ordered pair to another, and a norm-one scalar on the orthogonal one-dimensional complement adjusts its determinant without moving the two lines. Thus the special projective unitary action is two-transitive.

Two further families of simple groups have analogous rank-two actions:The Suzuki group of Lie type and the Ree group of type G2 have stabilizer subgroups containing subgroups acting regularly on all the other points in these actions; this gives two-transitivity. For example $Sz(8)$ has degree 65 and order 29120. These are exceptional Lie-type families, distinct from the preceding linear and unitary constructions.

The symplectic group over a finite field in characteristic two has a particularly illuminating pair of two-transitive actions which are not their projective-vector actions. For $m\ge 3$, $S{p}_{2m}(2)$ acts on the two classes of quadratic forms whose polar form is the given alternating bilinear form, distinguished by the Arf invariant of a quadratic form. Their degrees areTransitivity on each class follows by choosing symplectic bases giving the standard quadratic-form normal forms. Fix a form $Q$. Every other form with the same polar form is $Q_a(x)=Q(x)+B(a,x)$, and has the same Arf invariant of a quadratic form exactly when $Q(a)=0$. The stabilizer subgroup of $Q$, an orthogonal group over a finite field, is transitive on its nonzero isotropic vectors: extend such a vector to a hyperbolic pair and map adapted quadratic-form bases. It is therefore transitive on the other forms of the same type. This proves two-transitivity of both actions. In particular $S{p}_6(2)$ has actions of degrees 28 and 36. The degree-six action of $S{p}_4(2)\cong S_6$ in part 4(ii) is the smaller-dimensional precursor.

Finally the Mathieu groups supply exceptional examples with even higher transitivity:In particular all five natural actions are two-transitive. Together these examples exhibit affine spaces, projective and Hermitian geometries, symplectic quadratic forms, and exceptional finite simple groups as different sources of the same pair symmetry.

The organizing distinction is affine versus almost simple; the action, its degree, and its stabilizer subgroup are essential data. The explicitly verified affine, symmetric, alternating, projective, unitary, and symplectic actions already give many infinite families, while the exceptional families show why no single matrix construction covers every example.

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.

A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so

LetSuppose $F_n=x^k$ is a perfect power with $x\ge 2$. The finitely many small $n$ can be absorbed into the final effective constant. The Binet formula gives $x^k\asymp {\varphi }^n$ andThus, forthe local Lipschitz equivalence of $u$ and $e^u-1$ at zero yieldsThe form cannot vanish: applying the nontrivial field automorphism of $\mathbb{Q}(\sqrt{5})$ to ${\varphi }^n=\sqrt{5}{x}^k$ would give ${\psi }^n=-\sqrt{5}{x}^k$, whose absolute values are incompatible.

Apply the Baker lower bound for a homogeneous linear form in logarithms with the variable-height number $x$ placed last. The parameters belonging to $\varphi$ and $\sqrt{5}$ are absolute constants, while $\log A_x\asymp \log x$. Moreover,so $n/\log A_x\ll k$ and therefore $B^{\ast }\ll k$. The refined lower bound becomesComparison with the exponential upper bound gives $k\le C_3\log k$. Since $k/\log k$ tends to infinity, this bounds $k$ by an effective absolute constant. Enlarging it to cover the discarded small indices proves the claim.