2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ The discriminant of $\mathbb{Q}(i)/\mathbb{Q}$ is (4) -- the discriminant of $K(i)/K$ is (1). $\endgroup$ – Cam McLeman Jan 23 '12 at 4:59
  • 2
    $\begingroup$ For other examples, I'd probably start by looking up the Hilbert class field. $\endgroup$ – Cam McLeman Jan 23 '12 at 5:00
  • $\begingroup$ 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. $\endgroup$ – Abhishek Parab Jan 23 '12 at 5:41
  • 1
    $\begingroup$ 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 $\mathbf{Q}_2(\sqrt{-5})$. $\endgroup$ – Chandan Singh Dalawat Jan 23 '12 at 8:14
15
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Abhishek Parab Jan 23 '12 at 16:54
  • $\begingroup$ 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). $\endgroup$ – Franz Lemmermeyer Jan 23 '12 at 17:06
  • 1
    $\begingroup$ and be sure to check out math.stackexchange.com/questions/33573/… $\endgroup$ – Franz Lemmermeyer Jan 23 '12 at 17:14
  • $\begingroup$ +1 for the useful stackexchange link! $\endgroup$ – Abhishek Parab Jan 23 '12 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.