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Let $K$ be a number field, $\mathcal{O}_k$ its ring of integers, $\operatorname{Cl}(K)$ its class group and $h_K = \lvert \operatorname{Cl}(K)\rvert$ its class number. Let $X = \operatorname{Spec}(\mathcal{O}_K)$ and let $U\subseteq X$ be an open subset of $X$. Apparently (ex.3.19 Liu), using the fact that $h_K$ is finite it can be shown that $U$ is in fact a principal open subset of $X$ i.e. $U = D(\alpha) = \left\{[P]\in X: \alpha\notin P\right\}$ for some element $\alpha\in\mathcal{O}_K$. I've run into a possibly trivial roadblock in my attempt at a proof and would like it if someone could fix it for me or tell me where I went wrong. I've written out my attempt in full here:

Proof: Since $U$ is open we have $U = X\backslash V(I)$ for some ideal $I\subseteq \mathcal{O}_K$. Since $\mathcal{O}_K$ is a Dedekind domain we have $I = P_1\dots P_n$ for unique prime (hence maximal) ideals $P_i\in\mathcal{O}_K$. If $U=X$ then we can take $U = D(1)$. Therefore suppose $U\neq X$; we have $V(I) = \bigcup_{i=1}^n V(P_i)$. Since each $P_i$ is maximal, the point $x_i = [P_i]\in X$ is closed. Hence $U$ is the complement in $X$ of the finite set $\left\{x_1, \dots, x_n\right\}$.

Now we want to find an element $\alpha\in\mathcal{O}_K$ such $U = D(\alpha)$. This element $\alpha$, if it exists, has to be contained in each $P_i$ since if $\alpha\notin P_i$ for some $i$ then $x_i = [P_i]\in D(\alpha)$, but $U = D(\alpha) = X\backslash\left\{x_1, \dots, x_n\right\}$. Therefore we need an element $\alpha\in J:= \bigcap_{i=1}^n P_i$. Since $I\subseteq J$ it would be nice to pick an element of $I$ for this $\alpha$, but since $I$ may not be principal there is no natural choice.

Instead, note that since the class number $h_K$ is finite there exists an integer $m\leq h_K$ such that $I^m$ is a principal ideal, say $I^m = (\alpha)$. Now $I^m \subseteq I\subseteq J$ and therefore $D(\alpha)\subseteq U$. Now take a point $x = [Q]\in U$ for some prime ideal $Q\subseteq\mathcal{O}_K$; then $x\notin V(I)$ so $Q\nsupseteq I$. Therefore there exists $\beta\in I$ such that $\beta\notin Q$. The aim of the game is to show $\alpha\notin Q$.

Suppose $\alpha\in Q$; then it follows that since $I^m = (\alpha)$ there exists $\gamma\in I^{m-1}, \delta\in\mathcal{O}_K$ such that $\beta \gamma =\alpha\delta$. By the assumption on $\alpha$ we have $\beta\gamma\in Q$, and by assumption on $\beta$ we know $\gamma\in Q$. I want to use this to derive a contradiction because ultimately I wanted to show this would imply $\beta\in Q$, and thus the assumption $\alpha\in Q$ was wrong. But I can't figure out what to do here. Any help would be appreciated.

Alex Saad
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  • You can also just use that $V(IJ) = V(I \cap J) = V(I) \cup V(J)$ (set-theoretically, not scheme-theoretically, since the ideals $IJ$ and $I \cap J$ may differ). This immediately implies that $V(I^m) = V(I)$. – Remy Dec 14 '15 at 03:42
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    Here's a relevant thread: http://math.stackexchange.com/questions/174210/the-prime-spectrum-of-a-dedekind-domain/174244#174244 – Viktor Vaughn Dec 14 '15 at 07:15

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So the last bit with $\beta$ isn't needed. I'd forgotten a key feature from algebraic number theory; an ideal $P$ divides an ideal $Q$ (written $P\vert Q$) iff $Q\subseteq P$. Ideal divisibility works in exactly the same way as for numbers.

Using this with the integer $m$ above we see that $D(\alpha) = X\backslash V(I^m)$, and $V(I^m)= \left\{ [P]\in X: P\vert I^m\right\}$. Now since there is unique factorisation of ideals in $\mathcal{O}_K$ into primes we get $V(I^m) = V(I)$ and therefore $D(\alpha) = X\backslash V(I) = U$.

Alex Saad
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