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For a cyclic Galois extension $M/K$ of degree $m$, choose a generator $g$ and a primitive $m$th root of unity $\zeta \in K$. Reindexing the displayed sum gives $gR(w)=\zeta R(w)$. Linear independence of distinct field embeddings ensures this sum is not identically zero as a function of $w$. Choosing $\alpha =R(w)\ne 0$ therefore gives $m$ distinct conjugates ${\zeta }^j\alpha$, while ${\alpha }^m\in K$. Consequently $M=K(\alpha )$ is the splitting field of $X^m-{\alpha }^m$. Independence follows by taking a shortest supposed linear relation among distinct automorphisms, evaluating at $uv$ and subtracting a suitable multiple of the relation at $v$ to remove one term but not all terms.

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.

The hypotheses for Kummer theory are that $n\ge 1$ is prime to the characteristic of $K$ and that all $n$th roots of unity lie in $K$. Under these hypotheses, adjoining radicals produces abelian extensions of exponent dividing $n$, and conversely every such finite extension is obtained this way. The roots-of-unity hypothesis matters: without it, adjoining one radical need not give a Galois extension.

We first prove Hilbert's theorem 90, including its cocycle form. Distinct field automorphisms are linearly independent as functions into the field. Indeed, in a shortest nontrivial relation ${\sum }_i{c}_i{\tau }_i(y)=0$, choose $x$ with ${\tau }_1(x)\ne {\tau }_r(x)$. Subtract ${\tau }_r(x)$ times the relation at $y$ from the relation at $xy$. This eliminates the last term while leaving the first nonzero, contradicting minimality. This is linear independence of distinct field embeddings.

Let $L/K$ be finite Galois with group $G$, and suppose $a_{\sigma }\in L^{\ast }$ satisfy $a_{\sigma \tau }=a_{\sigma }\sigma (a_{\tau })$. By independence of the automorphisms, choose $c\in L$ for whichThenThus every multiplicative cocycle is a coboundary:For a cyclic group generated by $\sigma$ of order $m$, an element $a$ of norm one defines the cocycle $a_{{\sigma }^i}={\prod }_{j=0}^{i-1}{\sigma }^j(a)$; the condition at $i=m$ is exactly $N_{L/K}(a)=1$. HenceThe reverse implication follows by telescoping the product of conjugates.

Now let $\Delta$ be a finite subgroup of $K^{\ast }/K^{\ast n}$ and put $L=K(\sqrt[n]{\Delta })$. Finitely many representatives generate it; since ${\mu }_n\subset K$, the polynomials $X^n-a$ split in $L$ and are separable. Thus $L/K$ is finite Galois. For $\sigma \in G$ and $[a]\in \Delta$, choose ${\alpha }^n=a$ and define the Kummer pairingChanging the root or multiplying $a$ by an $n$th power does not change this value. It is multiplicative in both arguments. Since the radicals generate $L$, it embeds $G$ in $\mathrm{Hom}(\Delta ,{\mu }_n)$. Consequently $G$ is abelian of exponent dividing $n$.

For completeness, establish perfection rather than just this injection. Let ${\Delta }_L$ consist of all classes in $K^{\ast }/K^{\ast n}$ having an $n$th root in $L$. The same construction gives an injective map ${\Delta }_L\to \mathrm{Hom}(G,{\mu }_n)$: the character is trivial precisely when the root lies in $K$. It is surjective by Hilbert 90. Given $\chi :G\to {\mu }_n$, apply the cocycle theorem to $a_{\sigma }=\chi (\sigma {)}^{-1}$; it gives $\sigma (b)/b=\chi (\sigma )$, so $b^n\in K$ and its class has the required character. A finite abelian group of exponent dividing $n$ has as many ${\mu }_n$-characters as elements, by decomposition into cyclic factors. ThereforeIt follows that $\Delta ={\Delta }_L$ andConversely, for any finite abelian $L/K$ of exponent dividing $n$, Hilbert 90 supplies eigenvectors for every character. Their $n$th powers lie in $K$, and the intersection of their character kernels is trivial, so they generate $L$. This proves the Kummer classification and the inclusion-reversing duality between subgroups and intermediate extensions.

The simplest case is a cyclic extension of degree $m\mid n$: it has a generator $\alpha$ with ${\alpha }^m\in K$, and its automorphisms multiply $\alpha$ by $m$th roots of unity. Quadratic extensions in characteristic different from two are the case $n=2$. The cohomological summary is the exact sequence $1\to {\mu }_n\to K_s^{\ast }\stackrel{z\mapsto z^n}{\to }{K}_s^{\ast }\to 1$ together with Hilbert 90, giving $H^1(G_K,{\mu }_n)\cong K^{\ast }/K^{\ast n}$.

The automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the linear independence of distinct field embeddings. Therefore the operatoris not the zero map. Choose $\beta \in L$ for which $\alpha =T(\beta )\ne 0$. Reindexing the sum and using ${\sigma }^n=1$ givesThis is the Lagrange resolvent eigenvector for a cyclic field automorphism.

Let $F=L^{\sigma }$ be the fixed field. Since $\sigma ({\alpha }^n)={\alpha }^n$, the element $\alpha$ is a root ofIts orbit under $⟨\sigma ⟩$ isand these $n$ elements are distinct because $\alpha \ne 0$ and $\zeta$ is a primitive root of unity. Hence the minimal polynomial has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so

The fixed field isIt is a subfield because every field automorphism preserves sums, products and inverses.

Set $F=K^G$. If $[K:F]$ is infinite the desired finite lower bound is immediate. Otherwise let $b_1,\dots ,b_d$ be an $F$-basis of $K$. If $|G|>d$, the homogeneous system over $K$has a nonzero solution. For $a={\sum }_j{a}_j{b}_j$, with $a_j\in F$, every $\sigma$ fixes $a_j$, so the same coefficients give ${\sum }_{\sigma }{c}_{\sigma }\sigma (a)=0$. This contradicts linear independence of distinct field embeddings. Consequently