Units of Gaussian Integers
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:
\(\ds \cmod \alpha^2 \cdot \cmod \beta^2\) | \(=\) | \(\ds \cmod {\alpha \beta}^2\) | Modulus of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod 1^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
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
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $3$