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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.

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The discriminant of $\mathbb{Q}(i)/\mathbb{Q}$ is (4) -- the discriminant of $K(i)/K$ is (1). – Cam McLeman Jan 23 '12 at 4:59
For other examples, I'd probably start by looking up the Hilbert class field. – Cam McLeman Jan 23 '12 at 5:00
I don't follow how the discriminant of $K(i)/K$ is 1. Since $i$ is a primitive element satisfying $f(x) = x^2 + 1$, the discriminant is $(-1)N_{L/K}(f'(i))$ so by my calculations, is still -4. – Abhishek Parab Jan 23 '12 at 5:41
Think of it locally. $\mathbf{Q}_2(\sqrt 5)$ is the unramified quadratic extension of $\mathbf{Q}_2$, and $\mathbf{Q}_2(\sqrt{-1})$ and $\mathbf{Q}_2(\sqrt{-5})$ are ramified quadratic extensions, so the compositum $\mathbf{Q}_2(\sqrt 5, \sqrt{-1})=\mathbf{Q}_2(\sqrt{-5}, \sqrt{-1})$ is an unramified extension of $\mathbf{Q}_2(\sqrt{-1})$ and of $\mathbf{Q}_2(\sqrt{-5})$. – Chandan Singh Dalawat Jan 23 '12 at 8:14
Thanks, professor Dalawat – Abhishek Parab Jan 23 '12 at 16:59
up vote 15 down vote accepted

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.

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Yeah, thanks! I now get it! – Abhishek Parab Jan 23 '12 at 16:54

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.

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Thanks for your explanation but I will take time to understand what different is, how to do computations and how it is used. Hopefully before my CFT course ends! – Abhishek Parab Jan 23 '12 at 16:54
You can replace them safely by discriminants in this example: if L = F_1F_2$ is a compositum of independent number fields, then disc(L/F_1) divides disc(F_2/K). – Franz Lemmermeyer Jan 23 '12 at 17:06
and be sure to check out… – Franz Lemmermeyer Jan 23 '12 at 17:14
+1 for the useful stackexchange link! – Abhishek Parab Jan 23 '12 at 22:25

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