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Let $K$ be a number field, let $A$ be the ring of integers of $K$, and let $P$ denote the set of maximal ideals of $A$. For $p \in P$ and $x \in K^{\times}$ write $v_{p}$ for the exponent of $p$ in the factorization of the $Ax$ into a product of prime ideals. Put $v_{p}(0) = + \infty$. Take for $P'$ the complement of a finite set $S \subset P$. Show that the group of units of $A(P')$ is of finite type and that the quotient $U/A^{\times}$ is a free $\mathbb{Z}$-module of rank the cardinality of $S$.

My idea is to work with the map $x \rightarrow \left(v_{p_{1}}(x), v_{p_{2}}(x),\ldots, v_{p_{|S|}}(x)\right)$ of $U$ to $\mathbb{Z}^{s}$ that has kernel $A^{\times}$. I'm having trouble determining its image? Is it something obvious that I am missing?

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    What is $A(P')$? What is $U$?2011-10-23
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    *Please* don't post as if you were assigning homework. You might want to give background (as in, explain your notation), and what your thoughts so far are.2011-10-23
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    You are right. $A'(P)$ denotes the set of elements $x$ of $K$ such that $v_{p}(x) \geq 0$ for all $p \in P'$. I included some ideas; I was in a rush when I wrote the post - that explains the lack of details2011-10-23
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    @Anna: And presumably, $S=\{p_1,\ldots,p_{|S|}\}$?2011-10-23
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    To determine the image, use the Chinese remainder theorem (sometimes also called an "approximation theorem" in this context).2011-10-23
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    @Ted: I was also thinking about that, so you can please be more precise? Thanks!2011-10-23
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    @Arturo: Yes, sorry.2011-10-23
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    @Anna: I think you mean Ted; for any $(a_1,\ldots,a_{|S|})$, find $x_i\in p_i^{a_i}-p_i^{a_i+1}$. Then use the CRT to find $x\in A$ such that $x\equiv x_i$ modulo $p_i^{a_{i+1}}$ for each $i$, so that $v_{p_i}(a) = a_i$ for each $i$.2011-10-23
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    So the map is indeed surjective, thanks! What about $A(P')^{\times}$?2011-10-23
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    This is really late, but this is just a statement about what is often called the "$S$-units" of a number ring. It's actually a fairly common theorem. You can find it on Neukirch on pages 71-722013-10-06

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