unramification in a quadratic extension

Jul 2018
26
0
morocco
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 $K(\sqrt \varepsilon)$?
 
Dec 2018
7
3
Euclidean Plane
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 $2 = i(1-i)^2$.