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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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    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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I suppose that $A(P')$ = {$x \in K$; $v_P(x) \geq 0$ for all $p \in P'$}, and that $U$ is the group of units of $A(P')$.

Let $\psi$ be the map $U \rightarrow \mathbb{Z}^{s}$ assigning $x$ to $\left(v_{p_{1}}(x), v_{p_{2}}(x),\ldots, v_{p_{|S|}}(x)\right)$.

Suppose that $\psi$ is surjective. Take $(1, 0,\dots, 0) \in \mathbb{Z}^{s}$. Then there exists $x \in U$ such that $\psi(x) = (1, 0,\dots, 0)$. Then $v_{p_1}(x) = 1$, and $v_p(x) = 0$ for all $p \neq p_1$. Hence $p_1 = xA$. Since $p_1$ is not necessarily principal, $\psi$ is not necessarily surjective.

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    @BrandonCarter $A(P')$ = {$x \in K$; $v_P(x) \geq 0$ for all $p \in P' = S - P$}. So if $x \in U$ and $p \in S - P$, $v_p(x) = 0$.2012-11-27