2
$\begingroup$

I looked in other questions but I didn't find any answers regarding quadratic integer rings. Apologies if I missed it.

Given the ring $\mathbb{Z}[\sqrt{n}]$ where $n$ is a square-free positive integer, I would like to find the fundamental unit (i.e. some $a + b\sqrt{n}$ such that $\langle a + b\sqrt{n}\rangle = \mathbb{Z}[\sqrt{n}]^\times$, the units of $\mathbb{Z}[\sqrt{n}]$). I do know if I take the smallest $y$ such that $ny^2$ is of the form $x^2 \pm 1$, I get a unit, $x + y\sqrt{n}$. After all, we have $ny^2 = x^2 \pm 1$, so $ny^2 - x^2 = \pm 1$, meaning that $N(x + y\sqrt{n}) = x^2 - ny^2 = \mp 1$ ($N$ is just a standard Euclidean function). Since this particular Euclidean function is also multiplicative we know that any power of $x + y\sqrt{n}$ is also a unit. However, this is where I get stuck. Is this $x + y\sqrt{n}$ indeed a fundamental unit? If so, how do I show it can generate all units? If not, how do I find the actual fundamental unit?

$\endgroup$
  • $\begingroup$ See here on MSE, for $n=2$. For the algorithm in general, see here. $\endgroup$ – Dietrich Burde Sep 21 '16 at 12:57
  • $\begingroup$ By the way, try $n=9199$ in the algorithm. This is from Milne's lecture notes in algebraic number theory. $\endgroup$ – Dietrich Burde Sep 21 '16 at 13:18
  • $\begingroup$ Thanks very much for those links. My lack of knowledge shows, as I assumed that the method and/or result to find it for $\mathbb{Z}[\sqrt{n}]$ would differ (however slightly) from finding the solution for $\mathbb{Q}[\sqrt{n}]$. Both those links are very illuminating. $\endgroup$ – rwmak Sep 21 '16 at 17:33
  • $\begingroup$ Note that $\mathbb{Z}[\sqrt n]$ is only the ring of integers of a real quadratic field when $n\equiv_41$. $\endgroup$ – j0equ1nn Aug 25 '17 at 8:19
2
$\begingroup$

It certainly has been discussed on MSE how to find a fundamental unit for the unit group of the ring of integers of a real quadratic number field $\mathbb{Q}(\sqrt{n})$. A nice survey, how to use continued fractions, and a table with examples for squarefree $n\le 21$ is given here.

Remark: The ring of integers in $\mathbb{Q}(\sqrt{n})$ is not always $\mathbb{Z}(\sqrt{n})$. This is only true for $n\equiv 2,3\bmod 4$. For the remaining case $n\equiv 1 \bmod 4$ (note that $n$ is squarefree), it is slightly different. An example here, with $n=141$ is discussed here on MSE. The result is as follows: since $\sqrt{141}=[11,\overline{1,6,1,22}]$, we have that $95+8\sqrt{141}$ is a fundamental unit!

$\endgroup$
  • $\begingroup$ I had assumed that the ring of integers would always be trivial. Thanks for pointing out that it isn't! $\endgroup$ – rwmak Sep 21 '16 at 17:36
  • $\begingroup$ Sorry, that's an absolutely terrible way of wording it. I was referring to the fact that the ring is only $\mathbb{Z}[\sqrt{n}]$ for $n \equiv 2, 3 \mod 4$, not trivial in the normal sense. In my head I just called that trivial. My bad! $\endgroup$ – rwmak Sep 21 '16 at 20:06
  • $\begingroup$ No problem, I understood it now. $\endgroup$ – Dietrich Burde Sep 21 '16 at 20:09
3
$\begingroup$

A very practical way to obtain the fundamental unit of a real quadratic field is as follows:

for $\Bbb Q(\sqrt d)$ we have $$a_n+b_n\sqrt d=(a_1+b_1\sqrt d)^n\Rightarrow b_{n+1}=a_1b_n+b_1a_n$$ This implies that the sequence $\{b_n\}$ is strictely increasing because $a_1,b_1,a_n,b_n$ are positive. Hence we can see at the sequence $d,2^2d,3^2d,4^2d,....$ and stop at the first term for which $db^2$ is such that $a^2-db^2=\pm1$ for some integer $a$. In this case the $a$ and $b$ are the searched $a_1$ and $b_1$ of the fundamental unit. This method as far as I know is due to Pierre Samuel.

Examples.- (1) For $\Bbb Q(\sqrt6)$ we have $6\cdot1=6;\space6\cdot2^2=24=5^2-1$ then the fundamental unit of $\Bbb Q(\sqrt6)$ is $5+2\sqrt6$.

(2) For $\Bbb Q(\sqrt7)$ we have $7\cdot3^2=63=8^2-1$ so the f. u. is $8+3\sqrt7$

$\endgroup$
2
$\begingroup$

In the case of a real quadratic field, the fundamental unit is the smallest unit of the form $ x + y \sqrt{d} $ such that $ x \geq 0 $ and $ y \geq 1 $. To see this, note that if $ x + y \sqrt{d} > 1 $ is a unit, we have that $ x^2 - dy^2 = \pm 1 $. Assume that $ x $ and $ y $ had different signs, then we would have

$$ x + y \sqrt{d} = \frac{\pm 1}{x - y \sqrt{d}} $$

and $ |x - y \sqrt{d}| \geq 1 $ since $ x $ and $ -y $ have the same sign. This is a contradiction, therefore $ x $ and $ y $ are both nonnegative in $ x + y \sqrt{d} $. Since the fundamental unit is the smallest unit greater than $ 1 $, it follows that we may simply look at units of the form $ x + y \sqrt{d} $ where $ x, y $ are nonnegative, which reduces the problem to a necessarily finite brute force search.

While Dirichlet's unit theorem necessarily implies that the above found unit must be the fundamental unit, there is a more elementary proof of this fact. Assume that $ \varepsilon $ is the smallest unit greater than $ 1 $, and $ x $ is any unit greater than $ 1 $. We want to show that $ x $ is a power of $ \varepsilon $. Since the sequence $ a_n = \varepsilon^n $ diverges, there is a greatest integer $ n $ such that $ a_n = \varepsilon^n \leq x $. Then, $ x/\varepsilon^n $ is a unit $ \geq 1 $, but it is less than $ \varepsilon $ since $ x < \varepsilon^{n+1} $. By the definition of $ \varepsilon $, the only unit in the interval $ [1, \varepsilon) $ is $ 1 $, so it follows that $ x = \varepsilon^n $.

A more sophisticated algorithm to find the fundamental unit involves continued fractions, see these lecture notes for further information.

$\endgroup$
  • 1
    $\begingroup$ I just want to say that as somebody who's still working on fully grasping the Dirichlet unit theorem I really appreciate your simple proof! $\endgroup$ – rwmak Sep 21 '16 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.