# Thread: Show that Z[(3)^(1/2)] has infinitely many units

1. ## Show that Z[(3)^(1/2)] has infinitely many units

Hi,

We have started with group theory this week and we have got a set of exercises... I have done the first 10 exercises, but after the 10th one they get so much harder and I have no clue how to solve this one. The next 10 exercises are based on this one...

I'd really appreciate if anyone could give me a hand. Thanks!

"Show that Z[(3)^(1/2)] has infintely many units. (Z is the integer set...)

Call a non-unit x in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)] respectively) irreducible if
0 /= (not equal to) x /= y*z

for any two non-uunit y and z in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)]

Note: I think Z[(-1)^(1/2)] would be the set of all complex numbers of the form a+b(-1)^(1/2). The elements of Z[(-1)^(1/2)] are called Gaussian integers... but I don't know if this has much to do with the actual question being asked :-/

2. Originally Posted by jmgilbert
Hi,

We have started with group theory this week and we have got a set of exercises... I have done the first 10 exercises, but after the 10th one they get so much harder and I have no clue how to solve this one. The next 10 exercises are based on this one...

I'd really appreciate if anyone could give me a hand. Thanks!

"Show that Z[(3)^(1/2)] has infintely many units. (Z is the integer set...)

Call a non-unit x in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)] respectively) irreducible if
0 /= (not equal to) x /= y*z

for any two non-uunit y and z in Z[(-1)^(1/2)] (Z[(-2)^(1/2)], Z[(-3)^(1/2)], Z[(3)^(1/2)]

Note: I think Z[(-1)^(1/2)] would be the set of all complex numbers of the form a+b(-1)^(1/2). The elements of Z[(-1)^(1/2)] are called Gaussian integers... but I don't know if this has much to do with the actual question being asked :-/

Note that if $a+b\sqrt{3}\in\mathbb{Z}\left[\sqrt{3}\right]$ a quick calculation shows that $\frac{1}{a+\sqrt{3}b}=\frac{a-\sqrt{3}{b}}{a^2-3b^2}$. Thus, for $\left(a+b\sqrt{3}\right)\in\mathbb{Z}[\sqrt{3}]$ one would need that $a^2-3b^2\mid a,b$. How often does that hapepn?

3. For questions involving these types of rings, it is often helpful to define a norm map. Define $N:\mathbb{Z}[\sqrt{3}]\rightarrow \mathbb{Z}$ by $N(a+b\sqrt{3})=(a+b\sqrt{3})(a-b\sqrt{3})=a^2-3b^2$. You can check (and this is important) that defined this way, $N$ is a multiplicative map: that is, for any $\alpha ,\beta \in \mathbb{Z}[\sqrt{3}]$, $N(\alpha \beta)=N(\alpha )N(\beta )$. (Check this by straightforward computation.)

Then one can show that $\alpha \in \mathbb{Z}[\sqrt{3}]$ is a unit if and only if $N(\alpha )=\pm 1$.

Once we have that, what we need is a unit whose absolute value is larger than 1. Take the element $\alpha =2+\sqrt{3}$. We see that $N(\alpha )=1$, so by the above statement, $\alpha$ is a unit (indeed, $(2+\sqrt{3})(2-\sqrt{3})=1$). Since $|\alpha |>1$ (the standard absolute value function on $\mathbb{R}$), we know from basic properties of real numbers that $|\alpha^2|=|\alpha|^2>|\alpha|$ so, in particular, $\alpha^2\neq \alpha$.

But since $N$ is a multiplicative map, $N(\alpha^2)=N(\alpha \alpha)=N(\alpha)N(\alpha)=1\cdot 1=1$. Again by the above statement, we conclude that $\alpha^2$ must also be a unit in $\mathbb{Z}[\sqrt{3}]$ (you can actually compute the inverse, if you really want).

Iterating this process gives you an infinite list of units: $\{\alpha^n|n\in \mathbb{N}\}$.