Field of Fractions of Number Ring is Corresponding Number Field

In the case of monogenic number fields we have simply that \(K(\mathbb{Z}[\alpha]) = \mathbb{Q}(\alpha)\).

Proof

Let \(\mathbb{L} = K(\mathcal{O}_\mathbb{K})\). Then, since \(\mathbb{L}\) is the smallest field containing \(\mathcal{O}_\mathbb{K}\), and \(\mathbb{K}\) is a field containing \(\mathcal{O}_\mathbb{K}\), we have that

Now let \(\alpha \in \mathbb{K}\) be arbitrary. Since \(\mathbb{K}\) is a number field, there exists an integer \(k\) such that \(k\alpha \in \mathcal{O}_\mathbb{K}\) by this result.

So by the above inclusion, we have that \(k\alpha \in \mathbb{L}\), and since \(k \in \mathbb{Z} \subseteq \mathcal{O}_\mathbb{K} \subseteq \mathbb{L}\), \(k^{-1} \in \mathbb{L}\) from the properties of a field. Hence \(k^{-1}k\alpha = \alpha \in \mathbb{L}\). Since we started by assuming \(\alpha \in \mathbb{K}\) arbitrarily, we have that \(\mathbb{K} \subseteq \mathbb{L}\) and therefore