Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 (Local Fields)

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfying $|x|=0\iff x=0$, $|xy|=|x||y|$, and $|x+y|\le |x|+|y|$. It is non-Archimedean when the stronger inequality $|x+y|\le \max (|x|,|y|)$ holds.

If $\mathrm{char}K=p>0$, thenThe ordinary triangle inequality givesLetting $r\to \infty$ proves the strong triangle inequality.

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This can be false: take $y=-x\ne 0$. Then $|x+y|=0<|x|=\min (|x|,|y|)$. The upper inequality is always the ultrametric inequality.

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This is always true. If, say, $|x|>|y|$, then the ultrametric inequality applied to $x=(x+y)-y$ forces $|x+y|=|x|$; applying it to $x=(x-y)+y$ similarly gives $|x-y|=|x|$.

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This can be false. In a valued field with value group $\mathbb{Q}$, such as the Hahn series field $k((t^{\mathbb{Q}}))$, the maximal idealof the valuation ring is not principal: if $v(a)=q>0$, an element of valuation $q/2$ belongs to $\mathfrak{m}$ but not to $(a)$. Thus the valuation ring need not be a principal ideal domain.

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This is always true. Every open ball in an ultrametric space is also closed: a point outside a ball has a disjoint ball of the same radius around it. Distinct points can therefore be separated by clopen sets, so every connected subset is a singleton and $K$ is totally disconnected.

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Two absolute values are equivalent when $|\cdot {|}_2=|\cdot {|}_1^c$ for some $c>0$; equivalently, they induce the same topology. The nontrivial non-Archimedean absolute values on $\mathbb{Q}$ are, up to equivalence, exactly the p-adic absolute value $|\cdot {|}_p$.

Indeed $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$. If $(m,p)=1$, choose $a_r,b_r\in \mathbb{Z}$ with $a_{r}m+b_r{p}^r=1$. For large $r$, $|b_r{p}^r|<1$, so the ultrametric inequality forces $|m|=1$. HenceIf no prime has absolute value below one, the absolute value is trivial. This proves the non-Archimedean part of Ostrowski theorem.

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The restriction to $\mathbb{Q}$ cannot be trivial: otherwise the infinitely many integers would be pairwise distance one inside a bounded ball, contradicting local compactness. By part (c), it induces the $p$-adic topology for one prime $p$. Since a locally compact valued field is complete, the embedding $\mathbb{Q}\hookrightarrow K$ extends to a closed embedding ${\mathbb{Q}}_p\hookrightarrow K$.

Now $K$ is a locally compact topological vector space over the nondiscrete local field ${\mathbb{Q}}_p$. Such a vector space is finite-dimensional: a compact neighbourhood, together with a maximal linearly independent subset chosen at a fixed separation scale, is totally bounded only if that subset is finite, and its span is then open and closed; maximality makes it all of $K$. Thus $[K:{\mathbb{Q}}_p]<\infty$.

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One form of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[X]$, and $f(a_1)\equiv 0(\mathrm{mod}\pi )$ while $f^{\prime }(a_1)\not\equiv 0(\mathrm{mod}\pi )$, then there is a unique $\alpha \in R$ with $f(\alpha )=0$ and $\alpha \equiv a_1(\mathrm{mod}\pi )$.

Inductively, if $f(a_n)\equiv 0(\mathrm{mod}{\pi }^n)$, choose $t$ modulo $\pi$ so thatand put $a_{n+1}=a_n+{\pi }^{n}t$. The unit $f^{\prime }(a_n)$ makes $t$ unique. The resulting sequence is Cauchy, so completeness gives a root $\alpha$. Applying the same first-order congruence to two roots proves uniqueness.

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For $f(X)=X^2-X-2m$, $f(0)\equiv 0(\mathrm{mod}2)$ and $f^{\prime }(0)\equiv 1(\mathrm{mod}2)$. Hensel's lemma gives $x\in {\mathbb{Z}}_2$ with $x^2-x=2m$, hence

For $v=\infty$, signs give two square classes and $4/|2{|}_{\infty }=2$. For odd $p$, parity of $v_p$ gives two classes and ${\mathbb{F}}_p^{\ast }/({\mathbb{F}}_p^{\ast }{)}^2$ gives two more, while Hensel makes every unit congruent to $1$ modulo $p$ a square; hence there are four. For $p=2$, valuation parity gives two classes and odd units modulo squares are represented by $1,3,5,7\mathrm{mod}8$, giving eight. Thus in every case

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The Laurent series field $k((t))$ consists of series ${\sum }_{n\ge N}{a}_n{t}^n$. The mapis a discrete valuation. A Cauchy sequence has each coefficient eventually constant, and these stabilized coefficients define its limit, proving completeness.

For $K={\mathbb{F}}_p((t))$, write $K^{\ast }=t^{\mathbb{Z}}\times {\mathbb{F}}_p^{\ast }\times (1+t{\mathbb{F}}_p[[t]])$. If $p$ is odd, Hensel's lemma makes squaring an automorphism of the last factor, so the square-class group has order four. If $p=2$, Frobenius sends $\sum a_n{t}^n$ to $\sum a_n^2{t}^{2n}$, and the classes of units with arbitrarily placed odd-degree terms give infinitely many square classes. Hence the group is finite exactly for odd $p$.

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For sufficiently large $r$, the convergent p-adic logarithm and p-adic exponential series are inverse homomorphismsTheir identities $\log (xy)=\log x+\log y$ and $\exp (x+y)=\exp x\exp y$ follow first formally and then by convergence. Multiplication by ${\pi }^r$ identifies $({\mathcal{O}}_K,+)$ with $({\pi }^r{\mathcal{O}}_K,+)$. Since $(1+{\pi }^r{\mathcal{O}}_K)$ has finite index in ${\mathcal{O}}_K^{\ast }$, the conclusion follows.

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Since ${\bar{a}}^{q-1}=1$ in the residue field, $u=a^{q-1}\in 1+\pi {\mathcal{O}}_K$. Powers $u^{q^n}$ tend to one, by the binomial theorem initially and the $p$-adic logarithm once they enter its convergence domain. Therefore $a^{q^n}$ is Cauchy; let its limit be $\omega$. Reduction modulo $\pi$ gives $\bar{\omega }=\bar{a}$, whileThis is the Teichmuller representative of $\bar{a}$.

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For odd $p$, ${\mathbb{Q}}_p$ contains exactly the $p-1$ Teichmuller roots of unity; for $p=2$ it contains $\{\pm 1\}$, so there are two. For odd $p$, adjoining ${\zeta }_p$ adds the $p$ roots of unity of $p$-power order and no primitive $p^2$th root, because the latter would enlarge the degree by $p$. Combining the coprime-order groups givesFor $p=2$, ${\zeta }_2=-1$ already lies in ${\mathbb{Q}}_2$, so the answer remains two.

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The residue field $k_K={\mathbb{F}}_q$ has a unique degree-$n$ extension ${\mathbb{F}}_{q^n}$. Choose a monic irreducible polynomial $\bar{f}$ defining it and lift $\bar{f}$ to a monic $f\in {\mathcal{O}}_K[X]$. Hensel's lemma shows that a root generates an unramified extension $L/K$ of degree $n$ with that residue field. Any two such extensions embed into a common algebraic closure and have the same Teichmuller lifts of ${\mathbb{F}}_{q^n}$, which generate them; hence they coincide. This proves existence and uniqueness of the unramified extension.

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For $G=\mathrm{Gal}(L/K)$ defineThese are the lower ramification groups. If $\sigma$ lies in every $G_i$, then $\sigma ({\pi }_L)={\pi }_L$; the equivalent definition using all $a\in {\mathcal{O}}_L$ then gives $\sigma (a)=a$, so $\sigma =1$.

For $\sigma \in G_0$, setBecause inertia acts trivially on $k_L$, $\theta$ is a homomorphism. Its kernel is exactly $G_1$, so it induces an injection $G_0/G_1\hookrightarrow k_L^{\ast }$.

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The polynomial $X^3-2$ is Eisenstein over ${\mathbb{Q}}_2$, hence irreducible. Its discriminant is $-108=36(-3)$, and $-3\equiv 5(\mathrm{mod}8)$ is not a square in ${\mathbb{Q}}_2$. Therefore its splitting field has Galois group $S_3\cong D_6$.

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Suppose $G\cong D_{10}$. For residue characteristic two, $G_1$ is a normal $2$-group, $G_0/G_1$ embeds in $k_L^{\ast }$, and $G/G_0\cong \mathrm{Gal}(k_L/{\mathbb{F}}_2)$ is cyclic. The only proper nontrivial normal subgroup of $D_{10}$ is its rotation subgroup $C_5$, and it has no nontrivial normal $2$-subgroup.

Thus either $G_0=G$ or $G_0=C_5$. In the first case $G_1=1$, so part (b) would embed the noncyclic group $D_{10}$ in the cyclic group $k_L^{\ast }$, impossible. In the second case the residue degree is $|G/G_0|=2$, so $k_L={\mathbb{F}}_4$ and $k_L^{\ast }$ has order three; it cannot contain the injected group $G_0/G_1=C_5$. Both cases contradict the ramification constraints, so no such extension exists.

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