Example of unramified abelian extension - MathOverflow

Example of unramified abelian extension

2012-01-23T04:50:25Z

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.

Answer by David Loeffler for Example of unramified abelian extension

2012-01-23T08:19:55Z

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.

Answer by Franz Lemmermeyer for Example of unramified abelian extension

2012-01-23T11:59:35Z

A quick way of seeing what's going on is using the fact that $L = K(i) = K(\sqrt{5})$; the fact that the different above (say of $L/{\mathbb Q}(i)$ divides the different below (e.g. of ${\mathbb Q}(\sqrt{5})/{\mathbb Q}$) shows (take norms) that the discriminant of $L/K$ divides both $-4$ and $5$, hence is trivial. This works for a lot of other examples, too.