2

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:

  1. If $R \subset \mathbb{Q}(\sqrt{-1})$, then $R^{\times} = [\pm1, \pm \sqrt{-1}] \cap R.$
  2. If $R \subset \mathbb{Q}(\sqrt{-3})$, then $R^{\times} = [\pm1, \pm \alpha, \pm \alpha^2] \cap R,$ where $\alpha = \frac{-1 + \sqrt{-3}}{2}$.
  3. 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?

Riley
  • 1,113
  • 8
  • 23
  • 2
    Hint: In $R$ the algebraic conjugate is the usual complex conjugate. So units have unit length. – Jyrki Lahtonen Aug 25 '15 at 17:52
  • Assuming $R$ consists of algebraic integers, if $u\in R^\times$, then $N(u)=\pm1$, because $N(u)=\pm 1$ for units of all rings of algebraic integers. This time we have the further piece of information that $N(u)=u\overline{u}$. – Jyrki Lahtonen Aug 25 '15 at 18:55
  • @JyrkiLahtonen I've made an attempt for the first case. Would you be kind enough to check if it's that what you mean? – Riley Aug 25 '15 at 19:12
  • @JyrkiLahtonen That logic does give some problems for the second case I'm afraid: I don't see how $a^2 + 3b^2 = \pm 1$ and $a^2 - 9 b^2 = \pm 1$ results in the corresponding solution.. – Riley Aug 25 '15 at 19:16
  • In the second case the ring $R$ consists of numbers of the form $a+b(\dfrac{1+i\sqrt{3}}2).$ Note that the complex norm $||z||^2=z\overline{z}$ is always non-negative. The algebraic norm of a unit could also be $-1$, but not in these cases. – Jyrki Lahtonen Aug 25 '15 at 19:21
  • 1
    And in your attempt at the first case you only need to consider $a^2+b^2=1$ because $$N(a+bi)=(a+bi)(a-bi)=a^2+b^2.$$ – Jyrki Lahtonen Aug 25 '15 at 19:22
  • @JyrkiLahtonen, how do you know that it are those numbers? I just took $\sqrt{-3}$ as generator.. – Riley Aug 25 '15 at 19:22
  • Those are the algebraic integers of $\Bbb{Q}(\sqrt{-3})$. I assumed that your R is the maximal order, i.e. consists of exactly the algebraic integers. This begs the question: What is your definition of a quadratic imaginary ring? – Jyrki Lahtonen Aug 25 '15 at 19:24
  • @JyrkiLahtonen, the same definition as you I believe. Indeed, $\alpha ^3 = 1$, so $\alpha$ is an algebraic integer. However, $x^2 + 3 = 0 $ is also a monic polynomial in $\mathbb{Z}[x]$ and moreover, it's quadratic. – Riley Aug 25 '15 at 19:28
  • That polynomial shows that $\sqrt{-3}$ is an algebraic integer. $x^2+x+1=0$ shows that $(-1\pm i\sqrt{3})/$ are also algebraic integers. So they are all in $R$. – Jyrki Lahtonen Aug 25 '15 at 19:31
  • 1
    Have you seen the general result? If $R$ is the ring of algebraic integers of $\Bbb{Q}(\sqrt d)$, $d$ squarefree, then $$R={a+b\sqrt{d}\mid a,b\in\Bbb{Z}},$$ if $d\equiv 2\pmod 4$ or $d\equiv 3\pmod 4$. And $$R={a+b\frac{1+\sqrt{d}}2\mid a,b\in\Bbb{Z}},$$ if $d\equiv1\pmod4$. In this question we always have $d<0$, but that does not change the above description of $R$. – Jyrki Lahtonen Aug 25 '15 at 19:33
  • Actually, no. That makes a lot more sense now, thanks! This leads me however to a final issue: why is it that its $\pm 1$ in all the other cases? – Riley Aug 25 '15 at 19:35
  • 3
    In the last case even the half integer coefficients lead to too high complex absolute values. It may be a good idea to draw a picture, and plot some of the small elements of $R$ there (in all 3 cases). This is a unique thing about imaginary quadratic integers - they form a discrete set on the complex plane. So staring at a picture makes many a thing clear. The integers of a real quadratic integers form an unorderly dense mess on the number line (which leads to other fun phenomena, but let's not worry about that now). – Jyrki Lahtonen Aug 25 '15 at 19:39
  • Ok, got it (I think). Thanks for all the help, I really appreciate it. If you post those remarks as an answer, I'll be happy to accept it! – Riley Aug 25 '15 at 19:41
  • The usual way to prove this is with norms, which in the imaginary cases are never negative. Are you trying to forge a different path here? –  Aug 27 '15 at 13:30
  • @AlonsodelArte, yes I know the methods; I was only confused because I didn't had the general result which Jyrki stated, once I proved that, everything fell into place :) – Riley Aug 27 '15 at 19:47

1 Answers1

1

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

qwr
  • 11,470
  • I gave hints like those with a view if inducing the asker to work it out, and post an answer. Thanks (+1) for answering. This question will never appear in the unanswered queue :-) – Jyrki Lahtonen Jan 19 '22 at 12:12