Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 2 / b

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.

Solved by gpt-5.6-sol high.