Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 28 / 3

Choose $a\in K^{\times }$ with $0<|a{|}_1<1$. For any $x\ne 0$, the powers $x^n$ tend to zero precisely when $|x{|}_1<1$. Equality of the two topologies therefore impliesIn particular $0<|a{|}_2<1$. For positive integers $n$ and arbitrary integers $m$, apply this equivalence to $x^n{a}^{-m}$. It says thatThe inequality reverses on division by either negative logarithm. The rational cuts of the two real numbers are equal, so density of $\mathbb{Q}$ makes the numbers equal. Hence the equivalent absolute values satisfyThe formula also holds at zero. Nontriviality is essential to choosing $a$; no Archimedean or non-Archimedean restriction was used in the proof.

The idele group is the restricted productwith componentwise multiplication and the restricted product topology on the idele group relative to the unit groups at finite places. We need weak approximation for number fields only at the finitely many specified coordinates, not density in the full restricted-product topology.

Here is a proof including the Archimedean places. At each finite place in $S$, choose $a_v\in K$ sufficiently close to $x_v$, using the definition of the completion. Choose a nonzero $d\in {\mathcal{O}}_K$ clearing their denominators. Pick powers ${\mathfrak{p}}_v^{m_v}$ so that $|z/d{|}_v<\epsilon$ whenever $z\in {\mathfrak{p}}_v^{m_v}$. Put $I={\prod }_{v\in S,v\text{ finite}}{\mathfrak{p}}_v^{m_v}$. For each positive rational integer $M$ coprime to the rational primes below these places, the Chinese remainder theorem gives $b_M\in {\mathcal{O}}_K$ with $b_M\equiv Md{a}_v(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$. All of $b_M+I$ satisfy the same congruences.

Under the Minkowski embedding of a number field, $I$ is a full lattice in ${\mathbb{R}}^{r_1}\times {\mathbb{C}}^{r_2}$. A fundamental parallelepiped is bounded, so every real vector is within a fixed distance of this lattice, uniformly over its translates. Choose a target vector whose coordinates at the specified Archimedean places are those of $Md{x}_v$, assigning arbitrary values at the other Archimedean places. There is $b\in b_M+I$ within that fixed distance of the target. Dividing by $Md$ makes all specified Archimedean errors tend to zero as $M\to \infty$ through the allowed integers. At finite places $M$ is a unit, so the earlier congruence bounds persist. By taking the initial approximations sufficiently close, the ultrametric inequality gives the requested errors there too. Thus some $\boxed{y=b/(Md)\in K}$ satisfies every required inequality. If there are no finite places, set $I={\mathcal{O}}_K$ and omit the congruences.

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.

The square classes are open in $K_v^{\times }$. Indeed the strong form of Hensel lemma applied to $X^2-u$ at $1$ shows that $u$ is a square whenever $v(u-1)>2v(2)$. Thus any sufficiently close approximation $a$ to $d_v$ has $a/d_v$ square. By weak approximation for number fields, choose $a\in K$ satisfying this at every finite $v\in S$. Then the local extensions $K_v(\sqrt{a})$ are all unramified of degree two. At least one such condition ensures $a$ is not a square in $K$. If $S$ contains no finite place, impose the same condition at one auxiliary finite place.

Set $L=K(\sqrt{a})$. At every specified finite prime there is one prime of $L$ above it, with ramification index one and residue degree two. Hence $\boxed{\mathfrak{p}{\mathcal{O}}_L\text{ is prime for every finite }\mathfrak{p}\in S}$. The dyadic case is covered by the unramified construction, rather than by a nonexistent nonsquare residue unit in characteristic two.