Strong approximation theorem for Dedekind Domains

Question

This is a theorem in "Maximal Orders" by Reiner. Page 48 stated without proof. And is said to be an easy consequence of The Chinese remainder Theorem. I am attempting to prove the theorem and need a little help.

R is a Dedekind domain with fraction field $K$.

Let $P_{1},\cdots ,P_{n}$ be distinct non zero prime ideals of the Dedekind Domain $R$, and let $a_{1},\cdots ,a_{n}\in K$ be given as well as positive integers $r_{1},\cdots,r_{n}$. Then there exists $b\in K$ such that $$\begin{cases}b-a_{i}\in(P_{i})^{r_{i}}\cdot R_{P_{i}}& 1\leq i\leq n,\\ b\in R_{P}&\text{for all } P\neq P_{1},\cdots,P_{n}, \end{cases}$$ Where $P$ denotes a variable nonzero prime ideal of $R$.

My attempt is as follows.

First assume that each $a_{i}\in R$. Then by chinese remainder theorem there exists $b\in R$ such that $b\equiv a_{i}\mod P_{i}^{r_{i}}$ for all $i$. Therefore $b-a_{i}\in P_{i}^{r_{i}}\subset P_{i}^{r_{i}}\cdot R_{P_{i}}$. Also since $b\in R$, $b\in R_{P}$ for all $P$ prime. Now let $a_{i}\in K$. Then for each $i$, $a_{i}=x_{i}/s_{i}$ for $x_{i},s_{i}\in R$ and $s_{i}\neq 0$. Define $a'_{i}=a_{i}\prod_{i=1}^{n}s_{i}$. Then each $a'_{i}\in R$. There exists $b\in R$ such that $b\equiv a'_{i}\mod P_{i}^{r_{i}}$ for each $i$. I claim that $b'=\tfrac{b}{\prod s_{i}}$ is the element that does the above.

The issue that I am running into is I have no controll over the $s_{i}$. Therefore I have no way to ensure that $b'-a_{i}=(b-a'_{i})/\prod s_{i}$ is in $P_{i}^{r_{i}}\cdot R_{P_{i}}$. Any ideas or suggestions? Thanks in advance.

Answer 1

Rewrite to reflect the context of the text the op is using:

Let's prove we can do it for one, then using the CRT we will bootstrap to the others.

Then if we want to match some $b_i$ with the right property relative to $P_i$, and then it's easy to make sure $b_i\in R_P$ for $P\ne P_i$ because it won't change the $P_i$ valuation of $b_i$.

But then this is just a matter of choosing a $b_i\in K$ as any element from $a_i+P_i^{r_i}$, then by definition $b_i$ has the requisite property, and you can reconstruct $b$ from the $b_i$ as you do when you piece together CRT things. The existence of the appropriate multipliers follows from weak-approximation (not a small part of the proof, despite brevity).

Answer 2

You're almost there. Let $S$ be the finite set of prime ideals that divide $s = \prod s_i$. For a prime ideal $P$, denote $e(P)$ the exponent of $P$ in the prime factorization of $s$. Then use the CRT to find $b \in R$ with $b \equiv a_i' \pmod {P_i^{r_i + e(P_i)}}$ for all $i$, and $b \equiv 0 \pmod {P^{e(P)}}$ for all $P \in S$. Then $b' = b/s$ does the job.