Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 1

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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