Gaussian integers

Overview

Variants include any square-free integers and extension fields of rationals

Gaussian integers

The Gaussian integers $\mathbb{Z}[i]\le \mathbb{C}$ is the commutative ring with unity with elements

$$\mathbb{Z}[i]=\{a+bi:a,b\in \mathbb{Z}\};$$

that is; $\mathbb{Z}[i]$ is the set of complex numbers where both real and imaginary parts are integers. Addition and multiplication on the ring $\mathbb{Z}[i]$ are induced by the usual operations on complex numbers.

Group of units for the Gaussian integers

The Gaussian integers $\mathbb{Z}[i]$ are not a field, and the set of elements with inverses are

$$(\mathbb{Z}[i]{)}^{\ast }=\{\pm 1,\pm i\}=⟨i⟩,$$

a cyclic group of order 4.

Relevant theorems:

• (Theorem) Irreducible elements in the Gaussian integers

Related notes:

• Finite extensions of the rationals

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Properties

• Algebraic rings: The Gaussian integers are not a field, and the group of units (i.e., set of element with inverses) is

$$(\mathbb{Z}[i]{)}^{\ast }=\{\pm 1,\pm i\}=⟨i⟩,$$

a cyclic group of order 4.

• Euclidean domains: The Gaussian integers are a Euclidean domain with norm $N:\mathbb{Z}[i]\backslash \{0\}\to \mathbb{Z}$ defined by $N(\alpha )=\alpha \overline{\alpha }$, where $\overline{\alpha }$ denotes complex conjugation, i.e., $N(a+bi)=a^2+b^2.$ This implies that $\mathbb{Z}[i]$ is also a principal ideal domain and a unique factorization domain.
• Euclidean domains: There is natural extension of the norm $N$ defined above to a norm $N:\mathbb{Q}(i)\to \mathbb{Q}$, the Finite extensions of the rationals obtained by adjoining $i$.

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Variants of the ring $\mathbb{Z}[i]$

Complex numbers with rational coefficients

$$\mathbb{Q}(i)=\{a+bi:a,b\in \mathbb{Q}\}$$

• We have $\mathbb{Z}\le \mathbb{Z}[i]\le \mathbb{Q}(i)\le \mathbb{C}$.
• Multiplicative inverses: If at least one of a or b is not $0$, we can find a multiplicative inverse for $a+bi$ by rationalizing the denominator:

$$\frac{1}{a+bi}=\frac{1}{a+bi}\cdot \frac{a-bi}{a-bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i.$$

$\sqrt{2}$ as the imaginary part

Consider the set with integer coefficients:

$$\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in \mathbb{Z}\}.$$

• Closure under multiplication: $(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}.$
• Group of units: We have wip

We can also define the set using rational coefficients:

$\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}.$

$\sqrt[3]{2}$ as the imaginary part

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Code snippets

Gaussian integer ring $\mathbb{Z}[i]$:

\mathbb Z [i]

Multiplicative inverses $\frac{1}{a+bi}$:

\frac{1}{a + bi}