# weak approximation theorem

###### 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 $\nu_{\mathfrak{p}_{i}}((x))=a_{i}$ and for all other prime ideals    $\mathfrak{p}$, $\nu_{\mathfrak{p}}((x))\geq 0$. Here $\nu_{\mathfrak{p}}$ 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 $x_{i}\in p_{i}^{a_{i}},x_{i}\notin p_{i}^{a_{i}+1}$; this is possible since these two ideals are unequal by unique factorization  . Choose $x\in A$ with image $(x_{1},\ldots,x_{k})$. Clearly $\nu_{\mathfrak{p}_{i}}((x))=a_{i}$. 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

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

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 $\nu_{\mathfrak{p}^{\prime}_{k}}((y))>0$. Let $\nu_{\mathfrak{p}^{\prime}_{k}}((y))=c_{k}>0$. Again using the case proved above, choose $x\in K^{\star}$ such that

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

Then $x/y$ is the required element. ∎

Title weak approximation theorem WeakApproximationTheorem 2013-03-22 18:35:21 2013-03-22 18:35:21 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 13F05 msc 11R04 IndependenceOfTheValuations ChineseRemainderTheoremInTermsOfDivisorTheory