let $m$ be an integer with $m\equiv 1 \pmod4$ and $m<-3$.
$U\left(\mathbb{Z}+\mathbb{Z}(\frac{1+\sqrt{m}}{2})\right)=\{\pm1\}$ How can I prove that?
let $m$ be an integer with $m\equiv 1 \pmod4$ and $m<-3$.
$U\left(\mathbb{Z}+\mathbb{Z}(\frac{1+\sqrt{m}}{2})\right)=\{\pm1\}$ How can I prove that?
If you've been introduced to enough machinery of Algebraic Number Theory, you can recognize:
So it boils down to looking for roots of unity. However:
Since $[ \mathbb{Q}(\sqrt{m}) : \mathbb{Q} ] = 2$, you just have to check that neither of $\sqrt{-1}$ and $\sqrt{-3}$ are in $\mathbb{Q}(\sqrt{m})$.
In particular, you probably meant to assume $m$ is squarefree, since $\sqrt{-3} \in \mathbb{Q}(\sqrt{-27})$ (and thus so is $\zeta_3$).