Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 (Local Fields)

Let $I(L/K)$ be the inertia group and let $\phi$ denote Frobenius on the residue-field extension. The Relative Weil group isIts Weil-group topology makes $I(L/K)$ an open profinite group with its usual topology and gives $W(L/K)/I(L/K)$ the discrete topology. Thus every inertia coset is an open copy of $I(L/K)$.

Take $L=K^{\mathrm{nr}}$, the maximal unramified extension of $K$. ThenThe subgroup $\{0\}\subset \mathbb{Z}$ is open in the discrete Weil-group topology, but it is not open in the profinite topology inherited from $\hat{\mathbb{Z}}$.

The main theorem of local class field theory gives a continuous Local Artin mapwith dense image, normalized by sending a uniformizer to a chosen Frobenius. For every finite abelian extension $L/K$, it induces the Local Artin reciprocity isomorphismThe existence theorem of local class field theory says that the finite-index open subgroups of $K^{\times }$ are exactly the norm subgroups $N_{L/K}(L^{\times })$ for finite abelian extensions $L/K$, and that the extension is uniquely determined inside $K^{\mathrm{ab}}$.

Write $e=e(L/K)$ and $f=f(L/K)$. The valuation of a field norm satisfiesso the valuation image of the norm subgroup is $f\mathbb{Z}$. The exact sequence obtained from $v_K:K^{\times }\to \mathbb{Z}$ therefore givesThe left side is $[L:K]=ef$ by Local Artin reciprocity. Cancelling $f$ proves

The cyclotomic extension of a p-adic fieldhas degree $p-1$, is Galois with group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, and is totally ramified. Indeed, ${\zeta }_p-1$ is a root of the Eisenstein polynomial

Put $\alpha =\sqrt[p-1]{-p}$. Its polynomial $X^{p-1}+p$ is Eisenstein, so $L_2={\mathbb{Q}}_p(\alpha )$ is totally ramified of degree $p-1$. Every $(p-1)$st root of unity lies in ${\mathbb{Q}}_p$ by the Teichmuller lifts, so every root $\omega \alpha$ of this polynomial lies in $L_2$. Hence $L_2/{\mathbb{Q}}_p$ is also Galois.

For either $L_i/{\mathbb{Q}}_p$, the norm has every possible valuation because the residue-field degree is one. The norm units in a tamely totally ramified extension lie in the principal units $1+p{\mathbb{Z}}_p$: reduction of a unit norm is the $(p-1)$st power of its residue, hence is $1$. Part (b) says that this unit norm subgroup has index $p-1$, exactly the index of $1+p{\mathbb{Z}}_p$ in ${\mathbb{Z}}_p^{\times }$. ConsequentlyThe uniqueness clause in the existence theorem of local class field theory now gives

One strong form of Hensel lemma is this: if $K$ is complete for a discrete valuation $v$, $f\in {\mathcal{O}}_K[X]$, and $a_0\in {\mathcal{O}}_K$ satisfiesthen there is a unique root $\alpha$ satisfying

Define the Newton iteration over a valued fieldTaylor expansion and the ultrametric inequality show inductively that $v(f^{\prime }(a_n))=v(f^{\prime }(a_0))$ andThus the valuations of the corrections $a_{n+1}-a_n$ tend to infinity, so $(a_n)$ is a Cauchy sequence. Completeness gives a limit $\alpha$, and continuity gives $f(\alpha )=0$. If $\beta$ is another root in the stated ball, Taylor expansion of $f(\beta )-f(\alpha )$ shows that the linear term has strictly smaller valuation than all higher terms unless $\beta =\alpha$, proving uniqueness.

Every $x\in {\mathbb{Q}}_2^{\times }$ has a unique form $2^{n}u$ with $n\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_2^{\times }$. Its square class first records $n\mathrm{mod}2$. An odd 2-adic unit is a square precisely when it is congruent to $1$ modulo $8$: necessity follows by squaring an odd integer, and sufficiency follows from Hensel lemma applied in its standard $2$-adic square-root form.

The odd residues $1,3,5,7$ modulo $8$ therefore give four unit square classes. Together with valuation parity this yieldsFor example, the classes of $-1$, $2$, and $5$ form a basis of the square-class group of the 2-adic numbers.

The smallest integer isIndeed, reduction modulo $p$ cannot work: the map $x\mapsto x^p$ is the identity on ${\mathbb{F}}_p^{\times }$, although not every element of ${\mathbb{Z}}_p^{\times }$ is a $p$th power.

For odd $p$, the p-adic unit group decomposes asRaising to the $p$th power is an automorphism on ${\mu }_{p-1}$. On the principal units, the p-adic logarithm identifies it with multiplication by $p$ on $p{\mathbb{Z}}_p$, soHence whether a unit is a $p$th power is determined exactly by its residue modulo $p^2$. Equivalently,This is the pth-power criterion for p-adic units.

Let $G=\mathrm{Gal}(L/K)$. The lower ramification groups are $G_{-1}=G$ and, for $i\ge 0$,

If $L/K$ is totally ramified and ${\pi }_L$ is a uniformizer, then ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. Factoring $F(\sigma {\pi }_L)-F({\pi }_L)$ by $\sigma {\pi }_L-{\pi }_L$ for $F\in {\mathcal{O}}_K[X]$ proves the uniformizer criterion for lower ramification groups

For $\sigma \in G_0$, defineThe inertia group $G_0$ acts trivially on $k_L$, so $\theta (\sigma \tau )=\theta (\sigma )\theta (\tau )$. Its kernel consists exactly of those $\sigma$ for which $\sigma ({\pi }_L)/{\pi }_L\equiv 1$ modulo the maximal ideal, namely $G_1$. The first isomorphism theorem therefore gives an injection

Let $\alpha$ be a root ofThe polynomial is Eisenstein at $3$, so $E={\mathbb{Q}}_3(\alpha )$ is totally ramified of degree three and ${\mathcal{O}}_E={\mathbb{Z}}_3[\alpha ]$. Its discriminant isIts odd valuation makes $\Delta$ nonsquare in ${\mathbb{Q}}_3$, so the Galois group of an irreducible cubic shows that the splitting field $L$ has Galois group $S_3$. The quadratic extension obtained by adjoining $\sqrt{\Delta }$ is ramified, so $L/{\mathbb{Q}}_3$ is totally ramified. Thereforebecause the wild inertia group is the unique Sylow $3$-subgroup of $S_3$.

It remains to find the wild break. Sinceand ${\alpha }^2+1$ is a unit, the different ideal of $E/{\mathbb{Q}}_3$ has exponent $v_E(f^{\prime }(\alpha ))=3$. The extension $L/E$ is a tamely ramified quadratic extension and has different exponent one. The different in a tower therefore gives different exponentOn the other hand, the different exponent from ramification groups iswhere $b$ is the last index for which $G_b=A_3$. Thus $b=1$, and

Krasner's lemma says that if $K$ is complete, $\alpha$ is separable over $K$, andthen $K(\alpha )\subseteq K(\beta )$.

To prove it, let $\tau$ be an automorphism of $\overline{K}$ fixing $K(\beta )$. The absolute value has a unique extension to every finite extension of the complete field $K$, so it is invariant under $\tau$. If $\tau (\alpha )\ne \alpha$, the ultrametric inequality and the displayed strict inequality giveBut $\tau (\beta )=\beta$ and invariance gives $|\tau (\alpha )-\beta |=|\alpha -\beta |$, a contradiction. Every automorphism fixing $K(\beta )$ therefore fixes $\alpha$, which proves the field inclusion by Galois correspondence.

Now let $L/{\mathbb{Q}}_p$ be finite. By the primitive element theorem, write $L={\mathbb{Q}}_p(\alpha )$ with separable minimal polynomial $f\in {\mathbb{Q}}_p[X]$. Approximate the coefficients of $f$ closely by those of a polynomial $g\in \mathbb{Q}[X]$ of the same degree. Continuity of roots over a non-Archimedean field gives a root $\beta$ of $g$ arbitrarily close to $\alpha$. Choose it close enough for Krasner's lemma; thenThe degree bound from $\mathrm{deg}g=\mathrm{deg}f$ forces equality. If $F=\mathbb{Q}(\beta )$ and $\mathfrak{p}$ is the prime selected by the embedding $F\hookrightarrow {\overline{\mathbb{Q}}}_p$, its completion isThus every finite extension of ${\mathbb{Q}}_p$ is the completion of a number field at a prime ideal.

Let $G=\mathrm{Gal}(L/K)$ and let $\mathcal{P},\mathcal{Q}$ lie over $\mathfrak{p}$. These primes define two extensions to $L$ of the $\mathfrak{p}$-adic absolute value on $K$. The conjugacy of extensions of a valuation to a normal extension says that some $\sigma \in G$ carries the first extension to the second. Equivalently, $\sigma \mathcal{P}=\mathcal{Q}$. This proves the transitivity of the Galois action on primes.

The decomposition group is the stabilizerEach such automorphism extends continuously to the completions, and restriction gives the decomposition group of a prime and local Galois group isomorphism

For the splitting field $L=\mathbb{Q}(\sqrt[3]{7},{\zeta }_3)$ of $X^3-7$, the global Galois group is $S_3$. Since $7\equiv 1(\mathrm{mod}3)$, the field ${\mathbb{Q}}_7$ contains ${\zeta }_3$. The local splitting field is therefore ${\mathbb{Q}}_7(\sqrt[3]{7})$, an Eisenstein, totally ramified cyclic extension of degree three. Hence there areprimes of $L$ above $7$, and the decomposition group of each is the normal subgroup

For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

If $K$ is complete and $L/K$ is finite, the unique extension of an absolute value to a finite extension isIt restricts to the given absolute value because $N_{L/K}(a)=a^{[L:K]}$ for $a\in K$. The usual construction through multiplication by $x$ on the finite-dimensional $K$-vector space $L$, together with completeness, proves that this is the only extending absolute value.

For $\sigma \in \mathrm{Gal}(L/K)$, the map $x\mapsto |\sigma (x){|}_L$ is another absolute value extending $|\cdot |$ on $K$. Uniqueness therefore gives

Assume first that $K$ is complete. If $x$ is integral over ${\mathcal{O}}_K$, its monic equation and the ultrametric inequality imply $|x{|}_L\le 1$, so $x\in {\mathcal{O}}_L$. Conversely, if $|x{|}_L\le 1$, part (i) gives $|\sigma (x){|}_L\le 1$ for every $K$-embedding $\sigma$. The coefficients of the minimal polynomial of $x$ are elementary symmetric polynomials in its conjugates, so they all lie in ${\mathcal{O}}_K$. Thus $x$ is integral over ${\mathcal{O}}_K$, proving the integral closure in a finite extension of a complete discretely valued field identity ${\mathcal{O}}_L={\overline{{\mathcal{O}}_K}}^L$.

Completeness is necessary. Give $\mathbb{Q}$ its $5$-adic absolute value, take $L=\mathbb{Q}(i)$, and choose the extension corresponding to the prime $(2+i)$ above $5$. Thenhas nonnegative valuation at $(2+i)$, so it belongs to the chosen valuation ring ${\mathcal{O}}_L$. At the conjugate prime $(2-i)$ it has negative valuation, so it does not lie in the integral closure of ${\mathbb{Z}}_{(5)}$ in $\mathbb{Q}(i)$. Hence the chosen valuation ring can be strictly larger than the integral closure when the base field is not complete.