Units of Gaussian Integers

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Theorem

Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.

The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$.


Proof 1

Let $a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.

Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the zero of $\struct {\Z \sqbrk i, +, \times}$.

Then:

$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$

This leads (after algebra) to:

$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b} {a^2 + b^2}$

Let $a^2 + b^2 = n$.

We have that $n \in \Z, n > 0$.

If $c$ and $d$ are integers, then $a$ and $b$ must both be divisible by $n$.

Let $a = n x, b = n y$.

Then:

$n^2 x^2 + n^2 y^2 = n$

and so:

$n \paren {x^2 + y^2} = 1$

Thus $n = a^2 + b^2 = 1$ and so as $a, b \in \Z$ we have:

$a^2 = 1, b^2 = 0$

or:

$a^2 = 0, b^2 = 1$

from which the result follows.

$\blacksquare$


Proof 2

Let $\alpha = a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.

Then by definition of unit:

$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$

Let $\cmod \alpha$ denote the modulus of $\alpha$.


Then:

\(\displaystyle \cmod \alpha^2 \cdot \cmod \beta^2\) \(=\) \(\displaystyle \cmod {\alpha \beta}^2\) Modulus of Product
\(\displaystyle \) \(=\) \(\displaystyle \cmod 1^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

By Divisors of One:

$\cmod a^2 = 1$ or $-1$

Since $\cmod \alpha$ and $\cmod \beta$ are positive integers:

$\cmod \alpha^2 = a^2 + b^2 = 1$

and so either:

$\cmod a = 1$ and $\cmod b = 0$

or:

$\cmod b = 1$ and $\cmod a = 0$.

Hence the set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {\pm 1, \pm i}$.

$\blacksquare$


Sources