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 December 8th, 2018, 11:38 AM #1 Newbie   Joined: Jul 2018 From: morocco Posts: 23 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 $2 = i(1-i)^2$.

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