PlanetMath: 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.

Proof. Assume first that all $a_i\geq 0$ . By the Chinese Remainder Theorem, $$ A/\smp_1^{a_1+1}\times\cdots A/\smp_k^{a_k+1}\cong A/\smp_1^{a_1+1}\cdots\smp_k^{a_k+1} $$ Thus the map $$ A\to A/\smp_1^{a_1+1}\times\cdots A/\smp_k^{a_k+1} $$ 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 $(x_1,\dotsc,x_k)$ . Clearly $\nu_{\smp_i}((x))=a_i$ . But $x\in A$ , so all other valuations are nonnegative.

In the general case, assume wlog that we are given a set $\mathfrak{p}_1,\dotsc,\mathfrak{p}_r$ of primes of $A$ and integers $a_1,\dotsc,a_r\geq 0$ , and a set $\mathfrak{q}_1,\dotsc,\mathfrak{q}_t$ of primes with integers $b_1,\dotsc,b_t<0$ . First choose $y\in K^{\star}$ (using the case already proved above) so that

$\begin{displaymath} \begin{cases} \nu_{\mathfrak{p}}((y)) = 0 & \mathfrak{p}= \m... ... \nu_{\mathfrak{p}}((y)) \geq 0 & \text{otherwise} \end{cases} \end{displaymath}$

Now, there are only a finite number of primes $\smp'_k$ such that $\smp'_k$ is not the same as any of the $\smq_j$ and $\nu_{\smp'_k}((y))>0$ . Let $\nu_{\smp'_k}((y)) = c_k>0$ . Again using the case proved above, choose $x\in K^{\star}$ such that

$\begin{displaymath} \begin{cases} \nu_{\mathfrak{p}}((x)) = a_i & \mathfrak{p}= ... ... \nu_{\mathfrak{p}}((x)) \geq 0 & \text{otherwise} \end{cases} \end{displaymath}$

Then $x/y$ is the required element. $\qedsymbol$