Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 (Algebraic Number Theory)

The prime ideal factorization of $pB$ iswhich defines the ramification index of a prime ideal $e_{q/p}=v_q(pB)$. The residue-field degree isLocalizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the residue field $R/p$. The Chinese remainder theorem and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the fundamental identity for prime decomposition gives

The completion of a number field at a prime ideal does not change either invariant. More precisely, ${\hat{B}}_q$ is the valuation ring of $L_q$, its maximal ideal is $q{\hat{B}}_q$, andThus the completed extension has the same ramification index and the same residue-field extension, hence the same residue-field degree $f_{q/p}$. This is completion preserves ramification and residue data.

Write $G=\mathrm{Gal}(L/K)$. The decomposition group and inertia group at $q$ areThe transitivity of the Galois action on primes gives $G/D_{q/p}$ as the set of primes above $p$. Completion induces the canonical isomorphismand reduction gives the exact sequenceConsequently $|I_{q/p}|=e_{q/p}$, $|D_{q/p}/I_{q/p}|=f_{q/p}$ and, if $g$ is the number of primes above $p$,

LetOver ${\mathbb{Q}}_2$, the Newton polygon has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from Hensel lemma. There is one unit root because $F(1)$ is even and $F^{\prime }(1)$ is odd. For the two remaining roots, putwhose parenthesized polynomial has a simple root $Y\equiv 1(\mathrm{mod}2)$, andwhose parenthesized polynomial has a simple root $Z\equiv 1(\mathrm{mod}2)$. Thus $F$ splits completely over ${\mathbb{Q}}_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one

Over ${\mathbb{Q}}_5$, $F$ is an Eisenstein polynomial. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root thenwith $e=3$ and $f=1$.

The polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $5$. Its polynomial discriminant iswhich is not a square number, so the Galois group of an irreducible cubic isAt $2$, all three roots already lie in ${\mathbb{Q}}_2$, so the local splitting field is trivial andAt $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv 3(\mathrm{mod}5)$, a nonsquare unit, so the quadratic resolvent field of a cubic is the unramified quadratic extension of ${\mathbb{Q}}_5$. The local splitting field therefore has degree six, withThese calculations are summarized by local factorization of X3 plus 25X2 minus 50X plus 40.

Krasner's lemma states that if $K$ is complete, $\alpha$ is separable over $K$, and an algebraic element $\beta$ satisfiesfor every other $K$-conjugate ${\alpha }^{\prime }$ of $\alpha$, then $K(\alpha )\subseteq K(\beta )$.

Let $f\in {\mathbb{C}}_p[X]$ be nonconstant and let $\alpha$ be a root in an algebraic closure of ${\mathbb{C}}_p$. Because the characteristic is zero, replace $f$ by the separable minimal polynomial of $\alpha$. Approximate its coefficients arbitrarily closely by elements of ${\overline{\mathbb{Q}}}_p$. By continuity of roots over a non-Archimedean field, the approximating polynomial has a root $\beta$ arbitrarily close to $\alpha$. Since ${\overline{\mathbb{Q}}}_p$ is algebraically closed, $\beta \in {\overline{\mathbb{Q}}}_p\subset {\mathbb{C}}_p$.

Choose the approximation so that $\beta$ is closer to $\alpha$ than every other ${\mathbb{C}}_p$-conjugate of $\alpha$. Krasner's lemma givesso $\alpha \in {\mathbb{C}}_p$. Hence completion of an algebraic closure of a p-adic field is algebraically closed proves that ${\mathbb{C}}_p$ is algebraically closed.

For the coefficient $a_n=1/n!$, the p-adic absolute value gives $|a_n{|}_p=p^{v_p(n!)}$. By Legendre formula,where $s_p(n)$ is the sum of the base-$p$ digits of $n$. Thus $v_p(n!)/n\to 1/(p-1)$. The Cauchy-Hadamard theorem now yieldsand therefore the radius of convergence of the p-adic exponential is

A finite extension $L/K$ of non-Archimedean local fields is an unramified extension when its ramification index is one and its residue-field degree equals $[L:K]$. Equivalently, its maximal ideal is generated by a uniformizer of $K$ and its residue-field extension is separable.

A finite extension of number fields is an everywhere unramified extension of number fields when every nonzero prime ideal of $K$ is unramified in $L$. Under the convention that includes infinite places, one also requires every real embedding of $K$ to remain real.

The Hilbert class field $\tilde{K}$ is the maximal everywhere unramified abelian extension of the number field $K$, with complete splitting at real places if infinite places are included. Global class field theory gives the canonical Artin reciprocity mapso $[\tilde{K}:K]$ is the class number of $K$.

Put $K=\mathbb{Q}(\sqrt{-6})$. Its fundamental discriminant is $-24$. The Minkowski bound for ideal classes isso every ideal class contains an integral ideal of norm $1$, $2$ or $3$. The primes above $2$ and $3$ are ramified, their classes have order at most two, andshows that they represent the same class. This class is nontrivial because the norm form $a^2+6b^2$ does not represent $2$. Hence

Now let $H=K(\sqrt{2})=\mathbb{Q}(\sqrt{2},\sqrt{-3})$. The discriminant of a biquadratic field is the product of the discriminants of its three quadratic subfields, soThe relative discriminant ${\mathfrak{D}}_{H/K}$ therefore has norm one, proving that $H/K$ is unramified at every finite prime. Since $[H:K]=2=|\mathrm{Cl}(K)|$, the Hilbert class field of Q of square root minus six is

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 ${\mathbb{Q}}_2({\xi }_2)={\mathbb{Q}}_2$.