Here is another example of a quadratic extension:
Let $K=\mathbb{Q}(\sqrt{-5})$ and let $L=K(i)$, where $i^2=-1$. As it turns out, $L$ is the Hilbert Class Field of $K$, and therefore $L/K$ is everywhere unramified.
Exercise 2.42 (pages 51 and 52), in five parts, determines the ring of integers of $\mathbb{Q}[\sqrt{m},\sqrt{n}]$ for $m$ and $n$ distinct squarefree integers different from $1$. (See this question and response.) In particular, $L=\mathbb{Q}(\sqrt{5},\sqrt{-5})$ and therefore $\mathbb{O}_L$ has an integral basis $B=\left\{1,\frac{1+\sqrt{5}}{2},\sqrt{-5},\frac{\sqrt{-5}+i}{2}\right\}.$ Let $\alpha = \frac{\sqrt{-5}+i}{2}$. Then I claim that $\mathcal{O}_K[\alpha] = \mathcal{O}_L$. Clearly, $\mathcal{O}_K[\alpha]\subseteq \mathcal{O}_L$, so it suffices to show that $B\subseteq \mathcal{O}_K[\alpha]$.
We know that $\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]$. Thus, $1,\sqrt{-5}$, and $\alpha$ belong to $\mathcal{O}_K[\alpha]$. Also, $i=2\alpha-\sqrt{-5}\in \mathcal{O}_K[\alpha]$. It remains to show that $\frac{1+\sqrt{5}}{2}\in \mathcal{O}_K[\alpha]$, but $\frac{1+\sqrt{5}}{2} = 1+i\cdot\frac{i+\sqrt{-5}}{2}=1+i\cdot \alpha\in \mathcal{O}_K[\alpha].$ Hence, $\mathcal{O}_K[\alpha]=\mathcal{O}_L$.