Finding Units in $\mathbb{Z}[i]$: Is There More Than $1, -1, i$, and $-i$?

[evinda]

Hello! :)
Let $d \in \mathbb{Z}$ and the set $\mathbb{Z}[\sqrt{d}]=\{ a+b \sqrt{d},a,b \in \mathbb{Z}\}$.
I have to find all the units of the ring $\mathbb{Z}=\{ a+bi , a,b \in \mathbb{Z}\}$
I thought that these: $1,-1,i,-i$ are units of this ring..But are there also others?I think not..Am I right? (Thinking)

 

[I like Serena]

Hello! :)
Let $d \in \mathbb{Z}$ and the set $\mathbb{Z}[\sqrt{d}]=\{ a+b \sqrt{d},a,b \in \mathbb{Z}\}$.
I have to find all the units of the ring $\mathbb{Z}=\{ a+bi , a,b \in \mathbb{Z}\}$
I thought that these: $1,-1,i,-i$ are units of this ring..But are there also others?I think not..Am I right? (Thinking)

Hi!

Which one is it? Is it $\mathbb{Z}[\sqrt{d}]$? Or is it $\mathbb{Z}$??

 

[evinda]

Hi!

Which one is it? Is it $\mathbb{Z}[\sqrt{d}]$? Or is it $\mathbb{Z}$??

It is $\mathbb{Z}[\sqrt{d}]$ with $d=i^2$..

 

[I like Serena]

It is $\mathbb{Z}[\sqrt{d}]$ with $d=i^2$..

Heh.

The only units in $\mathbb{Z}$ are indeed $\pm 1, \pm i$.

To verify, consider the Gaussian integer norm given by $N(a+bi)=a^2+b^2$.
When two numbers are multiplied, their norms are multiplied as well.
Since no non-zero number has a norm of less than 1, all units have to have norm 1.

 

[Deveno]

The key to understanding WHY this works is:

Theorem: the norm $N$ on $\Bbb Z$ is multiplicative:

$N(uv) = N(u)N(v)$, for $u,v \in \Bbb Z$.

Proof: Let $u = a+bi,v = c+di,\ a,b,c,d \in \Bbb Z$.

Then $N(uv) = N((a+bi)(c+di)) = N(ac - bd + (ad + bc)i)$

$= (ac - bd)^2 + (ad + bc)^2$

$= a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2$

$= a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2$

$= (a^2 + b^2)c^2 + (a^2 + b^2)d^2$

$= (a^2 + b^2)(c^2 + d^2) = N(a+bi)N(c+di) = N(u)N(v)$. QED.

Therefore, if $u$ is a unit in $\Bbb Z$, we have some $v \in \Bbb Z$ with $uv = 1$.

So $N(u)N(v) = N(uv) = N(1) = 1^2 + 0^2 = 1$.

Now, $N(u),N(v)$ are INTEGERS, so $N(u)$ is a unit in $\Bbb Z$. Moreover, since $N(u)$ is always non-negative, it follows that:

$N(u) = 1$

From this, we obtain, if $u = a+bi$, that:

$a^2 + b^2 = 1$

This means that $|a|,|b| \leq 1$. Possible solutions are then...?

 

[evinda]

Heh.

The only units in $\mathbb{Z}$ are indeed $\pm 1, \pm i$.

To verify, consider the Gaussian integer norm given by $N(a+bi)=a^2+b^2$.
When two numbers are multiplied, their norms are multiplied as well.
Since no non-zero number has a norm of less than 1, all units have to have norm 1.

The key to understanding WHY this works is:

Theorem: the norm $N$ on $\Bbb Z$ is multiplicative:

$N(uv) = N(u)N(v)$, for $u,v \in \Bbb Z$.

Proof: Let $u = a+bi,v = c+di,\ a,b,c,d \in \Bbb Z$.

Then $N(uv) = N((a+bi)(c+di)) = N(ac - bd + (ad + bc)i)$

$= (ac - bd)^2 + (ad + bc)^2$

$= a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2$

$= a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2$

$= (a^2 + b^2)c^2 + (a^2 + b^2)d^2$

$= (a^2 + b^2)(c^2 + d^2) = N(a+bi)N(c+di) = N(u)N(v)$. QED.

Therefore, if $u$ is a unit in $\Bbb Z$, we have some $v \in \Bbb Z$ with $uv = 1$.

So $N(u)N(v) = N(uv) = N(1) = 1^2 + 0^2 = 1$.

Now, $N(u),N(v)$ are INTEGERS, so $N(u)$ is a unit in $\Bbb Z$. Moreover, since $N(u)$ is always non-negative, it follows that:

$N(u) = 1$

From this, we obtain, if $u = a+bi$, that:

$a^2 + b^2 = 1$

This means that $|a|,|b| \leq 1$. Possible solutions are then...?

I understand..Thank you both very much! (Nod)