Timeline for Example of unramified abelian extension

8 events

when toggle format	what		by	license	comment
Jan 23, 2012 at 16:55	vote	accept	Abhishek Parab
Jan 23, 2012 at 11:59	answer	added	Franz Lemmermeyer		timeline score: 9
Jan 23, 2012 at 8:19	answer	added	David Loeffler		timeline score: 19
Jan 23, 2012 at 8:14	comment	added	Chandan Singh Dalawat		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 $\mathbf{Q}_2(\sqrt{-5})$.
Jan 23, 2012 at 5:41	comment	added	Abhishek Parab		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.
Jan 23, 2012 at 5:00	comment	added	Cam McLeman		For other examples, I'd probably start by looking up the Hilbert class field.
Jan 23, 2012 at 4:59	comment	added	Cam McLeman		The discriminant of $\mathbb{Q}(i)/\mathbb{Q}$ is (4) -- the discriminant of $K(i)/K$ is (1).
Jan 23, 2012 at 4:50	history	asked	Abhishek Parab	CC BY-SA 3.0