Let $k = \mathbb{Q}(\sqrt{m})$, where $m \equiv 1 \pmod{8}$. Let $\epsilon$ be the fundamental unit of $k$ satisfying $\epsilon > 1$. 


A paper I'm reading involves studying the 2-torsion fields of elliptic curves $E/k$ which have global minimal models and everywhere good reduction. 


The fields $k(\sqrt{\pm \epsilon})$ come up a lot, and the paper constantly considers the cases $N\_{k/\mathbb{Q}}(\epsilon) = 1$ and $N\_{k/\mathbb{Q}}(\epsilon) = -1$ separately.


I don't really get how $N(\epsilon)$ controls the structure of $k(\sqrt{\pm \epsilon})/\mathbb{Q}$. For example, the paper states that if $N(\epsilon) = 1$, then the field extensions are Galois. Why is this true, and in general what does the sign of the norm of an element tell you? Since the sign goes away when you consider norms of ideals, I've never really thought about this.


Sorry if there's just a basic Galois theory / algebraic number theory fact I'm missing! 



Significance of the sign of the field norm for units in real quadratic fields
See whether this argument holds water:


Let $\epsilon=a+b\sqrt m$, let $\epsilon'=a-b\sqrt m$, so the norm of $\epsilon$ is $\epsilon\epsilon'$. If the norm is minus one, then $\sqrt\epsilon$ is real, but $\sqrt{\epsilon'}$ is not, so it's not in $k(\sqrt\epsilon)$, so that extension is not Galois. If the norm is plus one, then $\sqrt{\epsilon'}$ is the inverse of $\sqrt\epsilon$, so the extension is Galois. 


