Welcome to MathOverflow

Question

2

1

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.

flag	asked Jan 23 2012 at 4:50 Abhishek Parab 701●5●15

2

The discriminant of $\mathbb{Q}(i)/\mathbb{Q}$ is (4) -- the discriminant of $K(i)/K$ is (1). – Cam McLeman Jan 23 2012 at 4:59

2

For other examples, I'd probably start by looking up the Hilbert class field. – Cam McLeman Jan 23 2012 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 2012 at 5:41

1

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 2012 at 8:14

Thanks, professor Dalawat – Abhishek Parab Jan 23 2012 at 16:59

David Loeffler · Accepted Answer · 2012-01-23 08:19:55Z UTC

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.

Franz Lemmermeyer · Answer 2 · 2012-01-23 11:59:35Z UTC

7

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.

answered Jan 23 2012 at 11:59

Franz Lemmermeyer
17.2k●1●38●101

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 2012 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 2012 at 17:06

1

and be sure to check out math.stackexchange.com/questions/33573/… – Franz Lemmermeyer Jan 23 2012 at 17:14

+1 for the useful stackexchange link! – Abhishek Parab Jan 23 2012 at 22:25

Example of unramified abelian extension

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.