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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 $U$? – Ted Oct 23 '11 at 2:40
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. – Arturo Magidin Oct 23 '11 at 4:15
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 details – Anna Oct 23 '11 at 5:41
@Anna: And presumably, $S=\{p_1,\ldots,p_{|S|}\}$? – Arturo Magidin Oct 23 '11 at 5:56
To determine the image, use the Chinese remainder theorem (sometimes also called an "approximation theorem" in this context). – Ted Oct 23 '11 at 6:10

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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This isn't right. You only have $v_p(x) = 0$ for all $p \in S \setminus \{p_1\}$, and so you can't conclude that $p_1 = xA$. – Brandon Carter Nov 27 '12 at 0:09
@BrandonCarter Since $x \in U, v_p(x) = 0$ for all $p \in P - S$. – Makoto Kato Nov 27 '12 at 0:37
No. For instance, 3 is a unit in $\mathbb{Z}[\sqrt {-1}]_{(i+1)}$ (here $S = \{(i+1)\}$), but $v_3(3) \neq 0$. – Brandon Carter Nov 27 '12 at 0:42
@BrandonCarter Let $x \in U$. Let $p \in P'$. Since $1/x \in A(P')$, $v_p(1/x) \ge 0$, i.e. $v_p(x) \le 0$. On the other hand, since $x \in A(P')$, $v_p(x) \ge 0$. Hence $v_p(x) = 0$. – Makoto Kato Nov 27 '12 at 0:58
That only applies for primes in $(P')^{-1}A$, which is to say exactly the primes contained in $S$. You can't say anything about the others. – Brandon Carter Nov 27 '12 at 1:50

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