In his notes in Algebraic Number theory, J S Milne gives the following as an example of an unramified Abelian extension :


$ K = \mathbb Q (\sqrt{-5})$ having a quadratic extension $L = \mathbb Q (\sqrt{-1}, \sqrt{-5})$. 
Then, $L/K$ has discriminant a unit, so it ramifies. 


My question is, considering the simple extension $L = K(i)$ gives the discriminant to be $-4$, which clearly isn't a unit in $\mathcal O\_K$. Am I committing any mistake? 


Can you suggest other examples of unramified extensions?


Since I am a beginner in Class Field theory, related examples (of Abelian / non-Abelian extensions), counter-examples and other insights are more than welcome. 



Example of unramified abelian extension
You are slipping up because $i$ does not generate the ring of integers of $L$ as an $\mathcal{O}\_K$-algebra: we have $\mathcal{O}\_K = \mathbb{Z}[\sqrt{-5}]$, but $\mathcal{O}\_L = \mathbb{Z}\left[i, \frac{1 + \sqrt{5}}{2}\right] \ne \mathbb{Z}[i, \sqrt{-5}]$. Hence the discriminant of $L/K$ is not the same as the discriminant of the order $\mathbb{Z}[i, \sqrt{-5}]$, which is what you've calculated.


