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For a number field $K$, we define the ring of integers of $K$ to be $$\mathcal{O}_K:=\{x\in K\big|\ (\exists f\in\mathbb{Z}[X])(f\ \text{ is monic and } f(x)=0)\}.$$ Is there any easy way to see from this definition that $\mathcal{O}_K$ is finitely generated? I'm able to show that this is true, but I'm looking for something like a one-line proof or at least a clever remark clarifying this fact.

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I don't think a one-line proof exists. But the classical proof, albeit not trivial, is not that hard. Cf. for example Milne's lecture notes at page 36. – PseudoNeo Aug 20 '12 at 23:02
up vote 15 down vote accepted

The shortest proof I know is the following, and it doesn't generalize to a more abstract setting: let $\sigma_1, ... \sigma_n$ denote the $n$ embeddings $K \to \mathbb{C}$. Consider the map $(\sigma_1, ..., \sigma_n) : K \to \mathbb{C}^n$. Then this map embeds $\mathcal{O}_K$ as a discrete subgroup of $\mathbb{C}^n$, which is necessarily a finitely-generated free abelian group.*

The important step here is to check that the image of $\mathcal{O}_K$ is actually discrete, but this follows e.g. from the fact that if the $\sigma_i$ are too small then the elementary symmetric polynomials in the $\sigma_i$ (the coefficients of the characteristic polynomial) cannot take integer values (other than $0$).

*Sketch: given a discrete subgroup $G$ of $\mathbb{R}^d$, let $g_1, ..., g_r$ be a subset of $G$ which is a basis of $\text{span}(G)$ as a vector space. Then $\mathbb{Z} g_1 \oplus ... \oplus \mathbb{Z} g_r$ has finite index in $G$ because there are finitely many elements of $G$ in the parallelogram determined by the $g_i$ by discreteness.

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Nice. BTW, whatever proof you give has to use something special about number rings. See e.g. the wiki – Stephen Aug 21 '12 at 1:32
@Steve: thanks for the info. I had in mind the Dedekind case but wasn't aware of the restriction on the field of fractions. – Qiaochu Yuan Aug 21 '12 at 1:58

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