# Unique units of Gaussian integers

How can we show that $\pm 1, \pm i$ are the only units in the ring of Gaussian integers, the $\mathbb Z[i]$?

Thank you.

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Do you know about norms, and that the norm of a product is the product of the norms? Don't need it, but nicer. –  André Nicolas Feb 11 '12 at 7:53

## 2 Answers

Units are those elements in a ring that are invertible. Assume $a+bi$ is a unit.

Then. $\exists c+di \in \mathbb Z[i]$ such that $(a+bi) \cdot(c+di)=1$. This implies $$\begin{cases}ac-bd=1\\bc+ad=0 \end{cases}$$

Now solve this system and remember that $a,b,c,d \in \mathbb Z$.

We get from the second equation that, $bc=-ad$, This means that, from the first equation, $$\text{If b \neq0, then, } d(a^2+b^2)=-b$$

This leaves you with two possibilities: $d=-1$ 0r $d=-b$

But, if $d=-b$ then, we have that $c=a$ and $a^2+b^2=1$. Now since, we have been given that, $b \neq 0$. So, the only solution for $a^2+b^2=1$ with $a,b \in \mathbb Z$ will be, $a=0$ and $b=\pm 1$.

Now, if $d=-1$, then, note that if $c=0$, we immediately have that, $b=1$.

More generally, the last equation we derived from the system gives you that, $b(1-b)=a^2$ which has no integral solutions if $b \neq 0$.

For the last case, if $b=0$, then, the Gaussian integer is just $a$ and $a$ has an inverse in $\mathbb Z[i]$ iff $a=\pm 1$. This is because, the multiplicative inverse is $\frac{1}{a}$, which will be in $\mathbb Z$ iff $a=\pm 1$.

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If $z,w\in\mathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or

$$(a^2+b^2)(c^2+d^2)=1, \quad z=a+bi,\; w=c+di.$$

Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2\le1$. Check by hand the only solutions here correspond to $(a,b)=(\pm1,0)$ or $(0,\pm1)$.

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Equivalently, by taking conjugates we conclude that $\bar{z}\bar{w}=1$ and therefore $(z\bar z)(w\bar{w})=1$ and therefore $z\bar{z}=1$. –  André Nicolas Feb 11 '12 at 8:14