Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 4

The discrete valuation gives a split exact sequencesplit by $n\mapsto {\pi }^n$. Reduction gives a second exact sequenceThe Teichmuller lift identifies the cyclic group ${\mathbb{F}}_q^{\times }$ with the group ${\mu }_{q-1}$ of prime-to-$p$ roots of unity in $K$, where $p$ is the residue characteristic. Consequently the multiplicative group of a non-Archimedean local field decomposes as

If $d=[K:{\mathbb{Q}}_p]$, the p-adic logarithm identifies a sufficiently deep higher principal-unit group with the additive group of a rank-$d$ ${\mathbb{Z}}_p$-lattice. The remaining kernel is precisely the finite group ${\mu }_{p^a}(K)$ of $p$-power roots of unity in $K$. Thus the principal-unit group of a mixed-characteristic local field has the topological group structure

The main theorem of local class field theory consists of Local Artin reciprocity and its existence theorem. There is a canonical continuous reciprocity mapwith dense image, normalized so that a uniformizer maps to a chosen Frobenius. For every finite abelian extension $L/K$, it induces an isomorphismMoreover, the existence theorem of local class field theory says that $L\mapsto N_{L/K}(L^{\times })$ is an inclusion-reversing bijection between finite abelian extensions of $K$ in $K^{\mathrm{ab}}$ and finite-index open subgroups of $K^{\times }$.

Let $L_n={\mathbb{Q}}_2({\xi }_{2^n})$. The extension is a cyclotomic extension of a p-adic field and is totally ramified. Under Local Artin reciprocity, the action on ${\xi }_{2^n}$ is the reduction of a 2-adic unit modulo $2^n$, up to the harmless inverse determined by the Frobenius convention. A uniformizer acts trivially because $L_n/{\mathbb{Q}}_2$ has no unramified part. The kernel is thereforeBy the norm subgroup of a local field extension, this kernel is exactly the required norm group. Hence, for every $n>0$,For $n=1$, the factor $1+2{\mathbb{Z}}_2$ is all of ${\mathbb{Z}}_2^{\times }$, consistently with ${\mathbb{Q}}_2({\xi }_2)={\mathbb{Q}}_2$.