What are the reduced elements of $\mathbb Q(\sqrt{30})$?

Question

What are the reduced elements of $\mathbb Q(\sqrt{30})$ ?

From the definition here(on page $32$);

An element $\beta\in\mathbb Q(\sqrt{d})$ is said to be reduced, if $\beta>1$ and $-1<\beta'<0$ and $\beta'$ is the image of $\beta$ under the nontrivial automorphism which sends $\sqrt{d}$ to $-\sqrt{d}$

If $\beta$ is of the form $a+b\sqrt{30}$, then $0>\beta'=a-b\sqrt{30}>-1$

this looks too complicated to determine all reduced elements

So if $\beta$ is reduced, then clearly it is irrational, and we've a theorem which states: For $\beta\in\mathbb Q(\sqrt{d})\setminus\mathbb Q$, there are only finitely many reduced elements for any $D\in\mathbb N_{>0}$, where $D$ is the discriminant $(B^2-4AC)$ of the minimal polynomial $(AT^2+BT+C=0$ and $ \gcd(A,B,C)=1)$ of $\beta$. Is this useful ?

For example if we define $\beta_0=\sqrt{30}+\lfloor\sqrt{30}\rfloor,\quad\beta_{n+1}=\frac{1}{\beta_n-\lfloor\beta_n\rfloor}$ than $(\beta_n)_n$ is a periodic sequence of reduced elements of $\mathbb Q(\sqrt{30})$, but these are not all.

Do you have an idea ?

Answer 1

If I understand the definition correctly, all you need to do is solve the system of inequalities $$a+b\sqrt{30}>1\text{ and } 0>a-b\sqrt{30}>-1$$ for rational numbers. You can't of course expect to list the solutions -there are infinitely many. It is not difficult to describe the set though:

The first inequality is a half plane; to see that notice that you can rewrite is as $(a,b)\cdot (1,\sqrt{30})>1$ which means the projection of $(a,b)$ on $(1,\sqrt{30})$ is more than $\frac{1}{\sqrt{31}}$. This is a half-plane, with boundary orthogonal to the vector $(1,\sqrt{30})$ cutting it at $(\frac{1}{\sqrt{31}},\sqrt{\frac{30}{31}})$.

The second inequality is similarly an infinite strip vertical to the vector $(1,-\sqrt{30})$. The picture would look like this:

Now, to give a maybe more precise answer you can definitely parametrize this domain. Notice that the lines $a-b\sqrt{30}=-1$ and $a+b\sqrt{30}=1$ intersect at point $(0,\frac{1}{\sqrt{30}})$ and the other two intersect at $(\frac{1}{2},\frac{1}{2\sqrt{30}}$).

The domain is then parametrized as the union of sets $A$ and $B$ where $$A=\big\{(a,b):a\in[0,1/2]\text{ and }b\in[\frac{1-a}{\sqrt{30}},\frac{1+a}{\sqrt{30}}]\big\}\text{ and }$$ $$B=\big\{(a,b):a\in[1/2,\infty)\text{ and }b\in[\frac{a}{\sqrt{30}},\frac{a+1}{\sqrt{30}}]\big\}$$

Of course, you want the points with rational coordinates there...

Answer 2

For any $D>0$ there are only finitely many reduced elements $\beta$ whose minimal polynomial has discriminant $D$

The proof relies on the fact that there are only finitely many possible values for the coefficients $A,B$ in the minimal polynomial $A x^2+B x+C$ of $\beta$

Apply this result to $\mathbb{Q}\left(\sqrt{30}\right)$

Now $D=30m^2$ for some $m\in \mathbb{Z}$.

Choose $A,B$ and calculate $\beta =\frac{-B+\sqrt{D}}{2A}$