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 December 8th, 2018, 11:38 AM #1 Newbie   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 $K(\sqrt \varepsilon)$? December 29th, 2018, 10:01 PM #2 Newbie   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 $2 = i(1-i)^2$. Tags extension, extenson, quadratic, qudratic, unramification Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post magicterry Number Theory 1 March 31st, 2017 09:00 PM Tom19 Abstract Algebra 2 November 13th, 2013 05:20 AM yuesining Abstract Algebra 7 April 20th, 2011 08:59 AM Scooter Number Theory 1 October 19th, 2010 08:08 PM Francisco Abstract Algebra 1 June 24th, 2008 09:58 PM

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