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Let $A$ be a Dedekind domain. Let $P$ be a non-zero prime ideal of $A$. Let $\alpha \in A$. Let $k$ be a non-negative integer. If $\alpha \in P^k$ and $\alpha\notin P^{k+1}$, we write $v_P(\alpha) = k$.

How can we prove the following

Proposition. Let $A$ be a Dedekind domain. Let $P_1,\dots, P_n$ be distinct non-zero prime ideals of $A$. Let $e_1, \dots, e_n$ be non-negative integers. Then there exists $\alpha \in A$ such that $v_{P_i}(\alpha) = e_i$, $i = 1,\dots, n$.

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Are you familiar with the Approximation Theorem? – Bill Dubuque Jun 11 '12 at 1:07
I know one form of approximation theorem(as stated in Serre's Local fields), but I didn't know that in the link. Thanks. – Makoto Kato Jun 11 '12 at 1:22
up vote 3 down vote accepted

For each $i$ choose $\alpha_i\in\ P_i^{e_i}\setminus P_i^{e_i+1}$. Then $v_{P_i}(\alpha_i)=e_i$ for $i=1,\ldots,n$. The ideals $P_1^{e_1+1},\ldots,P_n^{e_n+1}$ are pairwise relatively prime, so the Chinese remainder theorem implies there is $\alpha\in A$ with $\alpha\equiv\alpha_i\pmod{P_i^{e_i+1}}$ for all $i$. Then $\alpha$ has the desired property.

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