Let $R$ be an imaginary quadratic ring. Then, the unit group $R^{\times}$ is finite. To prove this, I worked with normal forms, algebraic integers and the fact that $R \not \subset \mathbb{R}$. But I also want to prove the following cases:
- If $R \subset \mathbb{Q}(\sqrt{-1})$, then $R^{\times} = [\pm1, \pm \sqrt{-1}] \cap R.$
- If $R \subset \mathbb{Q}(\sqrt{-3})$, then $R^{\times} = [\pm1, \pm \alpha, \pm \alpha^2] \cap R,$ where $\alpha = \frac{-1 + \sqrt{-3}}{2}$.
- In all other cases: $R^{\times} = [\pm 1]$.
I'm lost here. Anyone who knows what to do? Thanks!
Thanks to the hint of Jyrki, I make…
AN ATTEMPT FOR CASE $1$: Let $x = a + bi \in R$. Then we have two conditions: $a^2+b^2 = \pm 1 $ and $a^2 – b^2 = \pm 1$. So $\pm 1$ is a solution because of the first equation and is repeated in the second equation, $\pm \sqrt{-1}$ is a consequence of the second equation. Correct?
Best Answer
As pointed out in the comments by Jyrki Lahtonen, the main idea is that integers of a quadratic field has norm the usual complex conjugate $N(\alpha) = \alpha \overline \alpha$, and an element is a unit iff it has norm $\pm 1$.
We can represent all quadratic integers $x \in \mathbb Q(\sqrt d)$ as $x = a + b \sqrt d$, where $a,b$ are integers for $d \equiv 2,3 \pmod 4$ and half-integers for $d \equiv 1 \pmod 4$.
Thus for an imaginary quadratic field, $d < 0$, so $N(x) = a^2 - db^2$ quickly becomes too big ($> 1$) for any $d < -3$, and the small cases $d = -1, -2, -3$ can be verified by hand. More info: https://en.wikipedia.org/wiki/Quadratic_integer#Norm_and_conjugation