Units in $\mathbb Z[\sqrt{2}]$

Question

I am trying to find all units in $\mathbb Z[\sqrt{2}]$. Suppose $x=a+b\sqrt{2}$ is a unit. Then there is $y=c+d\sqrt{2}$ such that $$xy=(a+b\sqrt{2})(c+d\sqrt{2})=1$$ So $$ac+2bd+(ad+bc)\sqrt{2}=1$$

From here one deduces $$ad+bc=0,$$$$ac+2bd=1$$

I've tried to find $c$ and $d$ from here or to get restrictions for $x=a+b\sqrt{2}$ but I couldn't. I would appreciate some help.

By (1), take $c=at$ and $d=-at.$ Substitute them in to (2) and find $t.$ Then you will get the multiplicative inverse $c+d \sqrt{2}$ of $a+b \sqrt{2}$ in $\mathbb Z[\sqrt{2}].$ — Bumblebee, Sep 12 '14 at 08:13

Dietrich Burde · Accepted Answer · 2014-09-12T08:24:40.847

3

The unit group of $\mathbb{Z}[\sqrt{2}]$ has rank $1$ according to Dirchlet's theorem, so we need to find a fundamental unit. Let $a=x+\sqrt{2}y$. Then the norm equation $N(a)=\pm 1$ says that $x^2-2y^2=\pm 1$. Here $x=y=1$ is a solution. We see that $\epsilon=1+\sqrt{2}$ is a fundamental unit, and all units are given by $\pm (1+\sqrt{2})^n$ for $n\in \mathbb{Z}$. Because of $-(1+\sqrt{2})^{-1}=(1-\sqrt{2})$ we can rewrite this as the set of $\pm(1\pm \sqrt{2})^n$ for all postive integers $n$.

edited Sep 12 '14 at 08:24

answered Sep 12 '14 at 08:15

Dietrich Burde

130,978

How can we tell this unit is fundamental? – Amitai Yuval Sep 12 '14 at 08:17
2

Because every unit is of the form $\pm \epsilon^n$, see here: http://math.stackexchange.com/questions/280878/the-units-of-mathbb-z-sqrt2. – Dietrich Burde Sep 12 '14 at 08:20
OK! The link is quite useful, indeed. Actually, I would just give it in the first place, and skip all other explanations. – Amitai Yuval Sep 12 '14 at 08:26
Yes, exactly. I just wrote the result in a concise way for the OP. You can skip all other explanations of course. – Dietrich Burde Sep 12 '14 at 08:29

score 1 · Answer 2 · answered Sep 12 '14 at 08:11

1

You can show that $|a^2-2b^2|$ is an Euclidean norm, and then you can imply that $x$ is a unit iff it's norm is 1

answered Sep 12 '14 at 08:11

Snufsan

2,125

Units in $\mathbb Z[\sqrt{2}]$

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