Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 / 4

The valuation ring is $\mathcal{O}=\{x\in K:|x|\le 1\}$, and $\mathfrak{m}=\{x\in K:|x|<1\}$ is an ideal by the ultrametric inequality. Its complement inside $\mathcal{O}$ consists exactly of elements of field absolute value one, whose inverses also lie in $\mathcal{O}$. Every proper ideal must avoid units and hence lie in $\mathfrak{m}$. Therefore $\mathfrak{m}$ is the unique maximal ideal: $\mathcal{O}$ is a local ring. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.

If $z\in K$ is integral over $\mathcal{O}$, it satisfies a monic equation $z^n+a_{n-1}{z}^{n-1}+\cdots +a_0=0$ with $|a_i|\le 1$. Were $|z|>1$, the leading term would have strictly larger field absolute value than every other term, so the ultrametric inequality would prevent cancellation to zero. Thus $z\in \mathcal{O}$, proving the valuation ring is integrally closed.

For the ideal criterion, assume the valuation is nontrivial and write $v=-\log |\cdot |$. If its value group is discrete, normalize it to $\mathbb{Z}$ and choose a uniformizer $\pi$ of valuation one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge 0$, so $b\in a\mathcal{O}$. Hence $I=(a)=({\pi }^m)$ and $\mathcal{O}$ is a principal ideal domain.

Conversely, if $\mathcal{O}$ is a principal ideal domain, its nonzero maximal ideal is $(\pi )$. For every element $x$ of positive valuation, $x\in \mathfrak{m}$ implies $x=\pi y$ with $y\in \mathcal{O}$, so $v(x)\ge v(\pi )>0$. Thus $\gamma =v(\pi )$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma \rfloor \gamma$. The remainder belongs to the value group and lies in $[0,\gamma )$, so is zero. The value group is therefore $\gamma \mathbb{Z}$, and the valuation is discrete. We have proved the principal ideal criterion for a nontrivial rank-one valuation ring.

The nontriviality qualification matters under the standard definition of discrete valuation, which requires value group isomorphic to $\mathbb{Z}$. With the trivial field absolute value, $\mathcal{O}=K$ is still a principal ideal domain, but the value group is zero and there is no uniformizer. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial value group as a discrete subgroup of $\mathbb{R}$, that exceptional case satisfies the equivalence too.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$, with ${\zeta }_{p^{n+2}}^p={\zeta }_{p^{n+1}}$. Use the slightly faster-converging series permitted by “or otherwise”:The terms have field absolute value $p^{-2n}$, so the partial sums are a Cauchy sequence in ${\overline{\mathbb{Q}}}_p$. We show that an algebraic limit would contradict the growth of cyclotomic extension degrees. The faster weights make the strict conjugate-distance estimate work uniformly also at $p=2$.

Put $L_N={\mathbb{Q}}_p({\zeta }_{p^{N+1}})$. Its degree is $p^N(p-1)$: the translated cyclotomic polynomial ${\Phi }_{p^{N+1}}(1+T)$ has constant term $p$ and reduces modulo $p$ to $T^{p^N(p-1)}$, so is Eisenstein. To verify the reduction, write it as ${\sum }_{j=0}^{p-1}(1+T{)}^{j{p}^N}$ and use characteristic $p$ to reduce this to ${\sum }_{j=0}^{p-1}(1+T^{p^N}{)}^j=T^{p^N(p-1)}$. All roots lie in $L_N$, and its Galois automorphisms are ${\sigma }_a(\zeta )={\zeta }^a$ for units $a$ modulo $p^{N+1}$.

For a nonidentity automorphism, put $r=v_p(a-1)$, choosing a representative with $0\le r\le N$. It fixes the terms with $n<r$. For $n\ge r$, the root ${\zeta }_{p^{n+1}}^{a-1}$ has order $p^{n+1-r}$, so the root-of-unity valuation formula, with $v_p(p)=1$, givesThis is strictly increasing as $n$ increases: successive differences are $2-p^{-(n-r+1)}>0$. Thus the first changed term is the unique term of smallest valuation and cannot cancel. ConsequentlyIn particular no nonidentity automorphism fixes $A_N$, so ${\mathbb{Q}}_p(A_N)=L_N$. Also the distance from $A_N$ to every distinct conjugate is at least $p^{-2N-1/(p-1)}$.

Suppose the Cauchy sequence converged to $A\in {\overline{\mathbb{Q}}}_p$. The tail estimate givesApply Krasner's lemma with $\alpha =A_N$, $\beta =A$. It forces $L_N={\mathbb{Q}}_p(A_N)\subseteq {\mathbb{Q}}_p(A)$ for every $N$. The right-hand side has finite degree, while $[L_N:{\mathbb{Q}}_p]=p^N(p-1)$ is unbounded, a contradiction. ThereforeThis proves algebraic closure of the p-adic numbers is not complete by an explicit algebraic Cauchy sequence and a fully quantified conjugate-separation argument.