weak approximation theorem

The weak approximation theorem allows selection, in a Dedekind ring, of an element having specific valuations at a specific finite set of primes, and nonnegative valuations at all other primes. It is essentially a generalization of the Chinese Remainder theorem, as is evident from its proof.

Theorem 1 (Weak ).

Let $A$ be a Dedekind domain with fraction field $K$ . Then for any finite set $\\mathfrak{p}_{1},\\ldots,\\mathfrak{p}_{k}$ of primes of $A$ and integers $a_{1},\\ldots,a_{k}$ , there is $x\\in K^{\\star}$ such that $\u_{\\mathfrak{p}_{i}}((x))=a_{i}$ and for all other prime ideals $\\mathfrak{p}$ , $\u_{\\mathfrak{p}}((x))\\geq 0$ . Here $\u_{\\mathfrak{p}}$ is the $\\mathfrak{p}$ -adic valuation associated with a prime ideal $\\mathfrak{p}$ .

Proof.

Assume first that all $a_{i}\\geq 0$ . By the Chinese Remainder Theorem,

A/\\mathfrak{p}_{1}^{a_{1}+1}\\times\\cdots A/\\mathfrak{p}_{k}^{a_{k}+1}\\cong A/%\\mathfrak{p}_{1}^{a_{1}+1}\\cdots\\mathfrak{p}_{k}^{a_{k}+1}

Thus the map

A\\to A/\\mathfrak{p}_{1}^{a_{1}+1}\\times\\cdots A/\\mathfrak{p}_{k}^{a_{k}+1}

is surjective. Now choose $x_{i}\\in p_{i}^{a_{i}},x_{i}\otin p_{i}^{a_{i}+1}$ ; this is possible since these two ideals are unequal by unique factorization. Choose $x\\in A$ with image $(x_{1},\\ldots,x_{k})$ . Clearly $\u_{\\mathfrak{p}_{i}}((x))=a_{i}$ . But $x\\in A$ , so all other valuations are nonnegative.

In the general case, assume wlog that we are given a set $\\mathfrak{p}_{1},\\ldots,\\mathfrak{p}_{r}$ of primes of $A$ and integers $a_{1},\\ldots,a_{r}\\geq 0$ , and a set $\\mathfrak{q}_{1},\\ldots,\\mathfrak{q}_{t}$ of primes with integers $b_{1},\\ldots,b_{t}<0$ . First choose $y\\in K^{\\star}$ (using the case already proved above) so that

\\begin{cases}\u_{\\mathfrak{p}}((y))=0&\\mathfrak{p}=\\mathfrak{p}_{i}\\\\\u_{\\mathfrak{p}}((y))=-b_{i}&\\mathfrak{p}=\\mathfrak{q}_{j}\\\\\u_{\\mathfrak{p}}((y))\\geq 0&\\text{otherwise}\\end{cases}

Now, there are only a finite number of primes $\\mathfrak{p}^{\\prime}_{k}$ such that $\\mathfrak{p}^{\\prime}_{k}$ is not the same as any of the $\\mathfrak{q}_{j}$ and $\u_{\\mathfrak{p}^{\\prime}_{k}}((y))>0$ . Let $\u_{\\mathfrak{p}^{\\prime}_{k}}((y))=c_{k}>0$ . Again using the case proved above, choose $x\\in K^{\\star}$ such that

\\begin{cases}\u_{\\mathfrak{p}}((x))=a_{i}&\\mathfrak{p}=\\mathfrak{p}_{i}\\\\\u_{\\mathfrak{p}}((x))=0&\\mathfrak{p}=\\mathfrak{q}_{j}\\\\\u_{\\mathfrak{p}}((x))=c_{k}&\\mathfrak{p}=\\mathfrak{p}^{\\prime}_{k}\\\\\u_{\\mathfrak{p}}((x))\\geq 0&\\text{otherwise}\\end{cases}

Then $x/y$ is the required element.∎