# 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 ${\mathrm{p}}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{\mathrm{p}}_{k}$ of primes of $A$ and integers ${a}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{a}_{k}$, there is $x\mathrm{\in}{K}^{\mathrm{\star}}$ such that ${\nu}_{{\mathrm{p}}_{i}}\mathit{}\mathrm{(}\mathrm{(}x\mathrm{)}\mathrm{)}\mathrm{=}{a}_{i}$ and for all other prime ideals^{} $\mathrm{p}$, ${\nu}_{\mathrm{p}}\mathit{}\mathrm{(}\mathrm{(}x\mathrm{)}\mathrm{)}\mathrm{\ge}\mathrm{0}$. Here ${\nu}_{\mathrm{p}}$ is the $\mathrm{p}$-adic valuation associated with a prime ideal $\mathrm{p}$.

###### Proof.

Assume first that all ${a}_{i}\ge 0$. By the Chinese Remainder Theorem,

$$A/{\U0001d52d}_{1}^{{a}_{1}+1}\times \mathrm{\cdots}A/{\U0001d52d}_{k}^{{a}_{k}+1}\cong A/{\U0001d52d}_{1}^{{a}_{1}+1}\mathrm{\cdots}{\U0001d52d}_{k}^{{a}_{k}+1}$$ |

Thus the map

$$A\to A/{\U0001d52d}_{1}^{{a}_{1}+1}\times \mathrm{\cdots}A/{\U0001d52d}_{k}^{{a}_{k}+1}$$ |

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 $({x}_{1},\mathrm{\dots},{x}_{k})$. Clearly ${\nu}_{{\U0001d52d}_{i}}((x))={a}_{i}$. But $x\in A$, so all other valuations are nonnegative.

In the general case, assume wlog that we are given a set ${\U0001d52d}_{1},\mathrm{\dots},{\U0001d52d}_{r}$ of primes of $A$ and integers ${a}_{1},\mathrm{\dots},{a}_{r}\ge 0$, and a set ${\U0001d52e}_{1},\mathrm{\dots},{\U0001d52e}_{t}$ of primes with integers $$. First choose $y\in {K}^{\star}$ (using the case already proved above) so that

$$\{\begin{array}{cc}{\nu}_{\U0001d52d}((y))=0\hfill & \U0001d52d={\U0001d52d}_{i}\hfill \\ {\nu}_{\U0001d52d}((y))=-{b}_{i}\hfill & \U0001d52d={\U0001d52e}_{j}\hfill \\ {\nu}_{\U0001d52d}((y))\ge 0\hfill & \text{otherwise}\hfill \end{array}$$ |

Now, there are only a finite number of primes ${\U0001d52d}_{k}^{\prime}$ such that ${\U0001d52d}_{k}^{\prime}$ is not the same as any of the ${\U0001d52e}_{j}$ and ${\nu}_{{\U0001d52d}_{k}^{\prime}}((y))>0$. Let ${\nu}_{{\U0001d52d}_{k}^{\prime}}((y))={c}_{k}>0$. Again using the case proved above, choose $x\in {K}^{\star}$ such that

$$\{\begin{array}{cc}{\nu}_{\U0001d52d}((x))={a}_{i}\hfill & \U0001d52d={\U0001d52d}_{i}\hfill \\ {\nu}_{\U0001d52d}((x))=0\hfill & \U0001d52d={\U0001d52e}_{j}\hfill \\ {\nu}_{\U0001d52d}((x))={c}_{k}\hfill & \U0001d52d={\U0001d52d}_{k}^{\prime}\hfill \\ {\nu}_{\U0001d52d}((x))\ge 0\hfill & \text{otherwise}\hfill \end{array}$$ |

Then $x/y$ is the required element. ∎

Title | weak approximation theorem |
---|---|

Canonical name | WeakApproximationTheorem |

Date of creation | 2013-03-22 18:35:21 |

Last modified on | 2013-03-22 18:35:21 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 13F05 |

Classification | msc 11R04 |

Related topic | IndependenceOfTheValuations |

Related topic | ChineseRemainderTheoremInTermsOfDivisorTheory |