Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 / 5

For the discrete valuation $v$ corresponding to $|\cdot |$, the valuation ring and its maximal ideal areThe first is a subring of the field $K$, hence an integral domain. An element of ${\mathcal{O}}_K$ is a unit exactly when its valuation is zero, so every nonunit lies in $\mathfrak{m}$ and $\mathfrak{m}$ is the unique maximal ideal.

Choose a uniformizer $\pi$ with $v(\pi )=1$. If $I\ne 0$ is an ideal, the set of valuations of its nonzero elements has a least member $n$. Choose $x\in I$ with $v(x)=n$. Then $x=u{\pi }^n$ for a unit $u$, so $(x)=({\pi }^n)$. Every $y\in I$ has $v(y)\ge n$, hence $y/x\in {\mathcal{O}}_K$, and therefore $I=(x)$. Thus ${\mathcal{O}}_K$ is a discrete valuation ring, in particular a principal ideal domain.

If $K$ is complete and $L/K$ is finite, the unique extension of an absolute value to a finite extension isIt restricts to the given absolute value because $N_{L/K}(a)=a^{[L:K]}$ for $a\in K$. The usual construction through multiplication by $x$ on the finite-dimensional $K$-vector space $L$, together with completeness, proves that this is the only extending absolute value.

For $\sigma \in \mathrm{Gal}(L/K)$, the map $x\mapsto |\sigma (x){|}_L$ is another absolute value extending $|\cdot |$ on $K$. Uniqueness therefore gives

Assume first that $K$ is complete. If $x$ is integral over ${\mathcal{O}}_K$, its monic equation and the ultrametric inequality imply $|x{|}_L\le 1$, so $x\in {\mathcal{O}}_L$. Conversely, if $|x{|}_L\le 1$, part (i) gives $|\sigma (x){|}_L\le 1$ for every $K$-embedding $\sigma$. The coefficients of the minimal polynomial of $x$ are elementary symmetric polynomials in its conjugates, so they all lie in ${\mathcal{O}}_K$. Thus $x$ is integral over ${\mathcal{O}}_K$, proving the integral closure in a finite extension of a complete discretely valued field identity ${\mathcal{O}}_L={\overline{{\mathcal{O}}_K}}^L$.

Completeness is necessary. Give $\mathbb{Q}$ its $5$-adic absolute value, take $L=\mathbb{Q}(i)$, and choose the extension corresponding to the prime $(2+i)$ above $5$. Thenhas nonnegative valuation at $(2+i)$, so it belongs to the chosen valuation ring ${\mathcal{O}}_L$. At the conjugate prime $(2-i)$ it has negative valuation, so it does not lie in the integral closure of ${\mathbb{Z}}_{(5)}$ in $\mathbb{Q}(i)$. Hence the chosen valuation ring can be strictly larger than the integral closure when the base field is not complete.