The ring of integers of a number field is finitely generated. 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.
 A: 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. 
