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

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.

Solved by gpt-5.6-sol high.

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 $|x-y|=|x|$.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.