Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 (Local Fields)

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

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Let $G=\mathrm{Gal}(L/K)$ and normalize $v_L$. The lower ramification groups arewith $G_{-1}=G$. If $K\subseteq M\subseteq L$, the same inequality defines $G_s(L/M)$ inside $\mathrm{Gal}(L/M)$, so directly

Now suppose $M/K$ is Finite Galois extension. The inertia group is the kernel of the action on the residue field. Restriction sends $G_0(L/K)$ into $G_0(M/K)$. Conversely, the maximal unramified subextension of $M/K$ is the intersection of $M$ with the maximal unramified subextension of $L/K$. The Galois correspondence therefore shows that the restriction image is all of $G_0(M/K)$.

For the explicit extension, take ${\alpha }^3=3$ and a primitive cube root of unity $\zeta$. The polynomial $X^3-3$ is Eisenstein over ${\mathbb{Q}}_3$, while ${\mathbb{Q}}_3(\zeta )/{\mathbb{Q}}_3$ is a ramified quadratic extension. Its splitting fieldis therefore a totally ramified extension of degree six with Galois group $S_3$. With $v_L(3)=6$, one has $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$, sois a uniformizer. Let $\tau (\alpha )=\zeta \alpha$, $\tau (\zeta )=\zeta$, and let $\sigma (\alpha )=\alpha$, $\sigma (\zeta )={\zeta }^{-1}$. ThenThe two nonidentity elements of $⟨\tau ⟩=A_3$ have ramification number four, whereas each transposition has ramification number one. HenceIn particular, $A_3$ is the wild inertia group.

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The common topology determines which elements have absolute value below one, becausein that topology. It therefore also determines all comparisons: $|x{|}_i<|y{|}_i$ exactly when $|x/y{|}_i<1$.

Choose $a\in K^{\times }$ with $|a{|}_1>1$; then also $|a{|}_2>1$. For every $x\in K^{\times }$ and positive integers $m,n$, the preceding observation givesThus the two real numbershave the same rational upper cuts and are equal. Settinggives $|x{|}_1=|x{|}_2^c$ for every $x$. Hence the two absolute values are equivalent absolute values.

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Suppose $x\notin {\mathcal{O}}_K$, so $v(x)<0$. The ultrametric inequality gives $v(1+x)=v(x)$. If $1+x=y^n$, thenFor a discrete valuation, only finitely many positive integers $n$ divide the fixed nonzero integer $v(x)$. Therefore $x$ cannot belong to $S$, proving $S\subseteq {\mathcal{O}}_K$.

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Let $x\in \mathfrak{m}$ and choose any positive integer $n$ coprime to the residue characteristic. Forone has $f(1)\equiv 0(\mathrm{mod}\mathfrak{m})$ and $f^{\prime }(1)=n\not\equiv 0(\mathrm{mod}\mathfrak{m})$. The simple-root form of Hensel lemma produces $y\in K$ with $y^n=1+x$. Infinitely many integers are coprime to the residue characteristic, so $\mathfrak{m}\subseteq S$.

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The valuation ring and its maximal ideal areIf ${\mathcal{O}}_K$ is Noetherian, its maximal ideal is finitely generated. Ideals in a valuation ring are totally ordered, so every finitely generated ideal is generated by one of its generators; write $\mathfrak{m}=(\pi )$. Then $v(\pi )$ is the smallest positive element of the value group. Subtracting integral multiples of this value shows that every value is an integral multiple of $v(\pi )$, so $v$ is discrete.

Now assume $K$ is complete and discretely valued. Parts i and ii, applied to $v$ and to the other discrete valuation $v^{\prime }$, giveIf $v(a)=0$ and $\pi$ is a uniformizer for $v$, then $\pi a^r\in {\mathfrak{m}}_v$ for every $r\in \mathbb{Z}$. Hencefor every integer $r$, forcing $v^{\prime }(a)=0$ and in particular $v^{\prime }(a)\ge 0$. Every nonzero $x$ has the form $x={\pi }^{m}u$ with $v(u)=0$, soThe proportionality constant is positive because $v^{\prime }$ is nontrivial. Thus the valuations, and their associated absolute values, are equivalent.

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One useful form of Hensel lemma is this: if $R$ is a complete discrete valuation ring, $f\in R[X]$, andthen there is a unique $a\in R$ such that $f(a)=0$ and $a\equiv a_1(\mathrm{mod}\pi )$. Indeed, after constructing $a_r$ with $f(a_r)\equiv 0(\mathrm{mod}{\pi }^r)$, choose the unique $t$ modulo $\pi$ for whichand put $a_{r+1}=a_r+{\pi }^{r}t$. The resulting Cauchy sequence converges by completeness, and the same first-order congruence proves uniqueness.

Apply this to $X^{p-1}-1$. Every nonzero class in ${\mathbb{F}}_p$ is a simple root, so it has a unique Teichmuller representative in ${\mathbb{Z}}_p$. These give all roots of unity of order prime to $p$. For odd $p$, the group $1+p{\mathbb{Z}}_p$ has no nontrivial torsion: if $u\ne 1$, then the binomial theorem gives $v_p(u^p-1)=v_p(u-1)+1$, which is incompatible with finite $p$-power order. HenceFor $p=2$, the subgroup $1+4{\mathbb{Z}}_2$ is torsion-free by the same argument, while $-1$ supplies the extra torsion element. Thus $\mu ({\mathbb{Q}}_2)=\{\pm 1\}\cong C_2$. This describes the roots of unity in a p-adic field for $K={\mathbb{Q}}_p$.

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Suppose first that $x^p+y^p=z^p$ with $x,y,z\in {\mathbb{Z}}_p^{\times }$. Divide by $y^p$ and put $u=x/y$, $w=z/y$. Reduction modulo $p$ gives $w\equiv u+1(\mathrm{mod}p)$. Choose an integer $a\equiv u(\mathrm{mod}p)$; then $p\nmid a(a+1)$. A unit's $p$th power modulo $p^2$ depends only on its residue modulo $p$, because $(a+pt{)}^p\equiv a^p(\mathrm{mod}{p}^2)$. Therefore

Conversely, suppose such an integer $a$ exists. The congruence saysFor odd $p$, the $p$th-power map sends $1+p{\mathbb{Z}}_p$ onto $1+p^2{\mathbb{Z}}_p$: under the p-adic logarithm it becomes multiplication by $p$. Hence the displayed ratio is $t^p$ for some $t\in 1+p{\mathbb{Z}}_p$. Takingproduces the required solution in ${\mathbb{Z}}_p^{\times }$.

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One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

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Choose a $K$-basis of $L$. Each extended absolute value is a norm on the finite-dimensional $K$-vector space $L$, and equivalence of norms in finite dimensions shows that every such norm induces the same topology when $K$ is complete. Consequently any two extended absolute values induce the same topology on $L$. By the result of Question 2a, one is a positive real power of the other. Their restrictions to the nontrivially valued field $K$ are both $|\cdot |$, so that power is one. In particular every extension is equal, and therefore equivalent, to $|\cdot {|}_L$.

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By local factorization and extended absolute values, extensions of $|\cdot {|}_p$ to the number field $L=\mathbb{Q}(\alpha )$ correspond to the irreducible factors of $X^3-2$ over ${\mathbb{Q}}_p$.

For $p=2$, the polynomial is Eisenstein, hence irreducible, so there is one extension. For $p=3$, a root would be a unit $x$ with $x\equiv -1(\mathrm{mod}3)$, but then $x^3\equiv -1\equiv 8(\mathrm{mod}9)$, not $2(\mathrm{mod}9)$. A reducible cubic has a root, so the polynomial is again irreducible and there is one extension.

For $p=5$, reduction givesThe factors are coprime, and the quadratic has discriminant $3$, a nonsquare modulo $5$. Hensel lemma lifts this as one linear and one irreducible quadratic factor over ${\mathbb{Q}}_5$, giving two extensions. The requested numbers are thereforefor $p=2,3,5$, respectively.

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Restriction to the maximal unramified extension gives a surjectionThe abelian Weil group is the inverse image of the dense subgroup $\mathbb{Z}\subset \hat{\mathbb{Z}}$ generated by Frobenius:It contains the full inertia kernel. Since $\mathbb{Z}$ is dense in $\hat{\mathbb{Z}}$, the inverse image is dense in $\mathrm{Gal}({\mathbb{Q}}_p^{\mathrm{ab}}/{\mathbb{Q}}_p)$ with its profinite topology.

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The translated cyclotomic polynomialis Eisenstein at $p$. Hence ${\zeta }_{p^n}-1$ generates a totally ramified extension of degreeEvery automorphism sends ${\zeta }_{p^n}$ to ${\zeta }_{p^n}^a$ for a unique $a\in (\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$. This gives an injectionand equality of orders makes it an isomorphism.

Normalize the Local Artin map so that $p$ maps to arithmetic Frobenius on the maximal unramified extension. For $u\in {\mathbb{Z}}_p^{\times }$, define its action on $K_n$ bywhere using $u$ rather than $u^{-1}$ gives the opposite Frobenius convention. These compatible maps, together with the image of $p$, define reciprocity on $p^{\mathbb{Z}}\times {\mathbb{Z}}_p^{\times }={\mathbb{Q}}_p^{\times }$ and hence on every finite abelian extension by the Local Kronecker-Weber theorem.

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Local Artin reciprocity states that the continuous maphas dense image and induces, for every finite abelian extension $L/K$, an isomorphismThus the norm subgroup of a local field extension is the kernel of the Artin map restricted to $L$.

For $L=K_n$, the uniformizer $p$ maps to the unramified Frobenius and therefore acts trivially on the totally ramified cyclotomic extension. By part b, a unit $u$ acts trivially on $K_n$ exactly when $u\equiv 1(\mathrm{mod}{p}^n)$. Sincethe kernel, and hence the norm subgroup, is

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