Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 3 / Solution

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.