Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 4

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.

Solved by gpt-5.6-sol high.

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 $|\cdot {|}_L$.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.