My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum

LinkBack Thread Tools Display Modes
December 8th, 2018, 11:38 AM   #1
Joined: Jul 2018
From: morocco

Posts: 26
Thanks: 0

Math Focus: algebraic number theory
unramification in a quadratic extension

Let $K$ be an extension, $\varepsilon$ be unit of $K$ and $\mathfrak p$ a prime ideal of $K$ above a rational prime $p$ such that $p$ is unramified in $K$. is $\mathfrak p$ unramified in $K(\sqrt \varepsilon)$?
Chems is offline  
December 29th, 2018, 10:01 PM   #2
Joined: Dec 2018
From: Euclidean Plane

Posts: 7
Thanks: 3

Depends on which unit we choose. For instance, let $K = \mathbb{Q}$ and let $\mathfrak{p}$ be the prime ideal $(2)$ above the rational prime $p=2$.

If we choose the unit $\varepsilon = 1$, then $K(\sqrt{\varepsilon}) = \mathbb{Q}$ and $\mathfrak{p}$ remains unramified.

If we choose the unit $\varepsilon = -1$, then $K(\sqrt{\varepsilon}) = \mathbb{Q}(i)$ and $\mathfrak{p}$ is the square of the ideal $(1-i)$, since $2 = i(1-i)^2$.
Eucler is offline  

  My Math Forum > College Math Forum > Number Theory

extension, extenson, quadratic, qudratic, unramification

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
An extension to FLT magicterry Number Theory 1 March 31st, 2017 09:00 PM
extension fields over Q Tom19 Abstract Algebra 2 November 13th, 2013 05:20 AM
About Field Extension yuesining Abstract Algebra 7 April 20th, 2011 08:59 AM
Extension Fields Scooter Number Theory 1 October 19th, 2010 08:08 PM
Degree of an extension Francisco Abstract Algebra 1 June 24th, 2008 09:58 PM

Copyright © 2019 My Math Forum. All rights reserved.