Units of Gaussian Integers/Proof 2

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

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


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 $\alpha$.


\(\displaystyle \cmod \alpha^2 \cdot \cmod \beta^2\) \(=\) \(\displaystyle \cmod {\alpha \beta}^2\) 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 $\cmod b = 0$


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

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