The Ring of Gaussian Integers Z(i)

Recall from the Rings page that if $+$ and $*$ are binary operations on the set $R$, then $R$ is called a ring under $+$ and $*$ denoted $(R, +, *)$ when the following are satisfied:

• 1. For all $a, b \in R$ we have that $(a + b \in R)$ (Closure under $+$).

• 2. For all $a, b, c \in R$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $R$ under $+$).

• 3. There exists an $0 \in R$ such that for all $a \in R$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $R$ under $+$).

• 4. For all $a \in R$ there exists a $-a \in R$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $R$ under $+$).

• 5. For all $a, b \in R$ we have that $a + b = b + a$ (Commutativity of elements in $R$ under $+$).

• 6. For all $a, b \in R$ we have that $a * b = b * a$ (Closure under $*$).

• 7. For all $a, b, c \in R$, $a * (b * c) = (a * b) * c$ (Associativity of elements in $R$ under $*$).

• 8. There exists a $1 \in R$ such that for all $a \in R$ we have that $a * 1 = a$ and $1 * a = a$ (The existence of an identity element $1$ of $R$ under $*$).

• 9. For all $a, b, c \in R$ we have that $a * (b + c) = (a * b) + (b * c)$ and $(a + b) * c = (a * c) + (b * c)$ (Distributivity of $*$ over $+$).

We will now look at the ring of Gaussian integers, $\mathbb{Z} (i)$. We first define a Gaussian integer.

Definition: A Gaussian integer is a number of the form $a + bi$ where $a, b \in \mathbb{Z}$ and $i^2 = -1$.

For example, the numbers $2 + 3i$ is a Gaussian integer, however, $\frac{1}{2} + 2i$ is not a Gaussian integer.

We will now prove that $(\mathbb{Z}(i), +, *)$ forms a ring where $+$ denotes standard addition and $*$ denotes standard multiplication.

Let $x, y \in \mathbb{Z}(i)$ where $x = a + bi$ and $y = c + di$ where of course, $a, b, c, d \in \mathbb{Z}$.

We first show that $\mathbb{Z} (i)$ is closed under $+$. Noting that the sum of two integers is an integer, we have that:

(1)

\begin{align} \quad x + y = (a + bi) + (c + di) = (a+c) + (b + d)i \in \mathbb{Z}(i) \end{align}

The associativity of $\mathbb{Z}(i)$ is inherited from the associativity of the complex numbers $\mathbb{C}$ under the same operation, so we will skip showing this outright.

The additive identity for $+$ is $0 = 0 + 0i$ since:

(2)

\begin{align} \quad x + 0 = (a + bi) + (0 + 0i) = (a + 0) + (b + 0)i = a + bi = x \end{align}

And similarly:

(3)

\begin{align} \quad 0 + x = (0 + 0i) + (a + bi) = (0 + a) + (0 + b)i = a + bi = x \end{align}

The inverse element under $+$ for $x$ is $-x = -a - bi$ since:

(4)

\begin{align} \quad x + (-x) = (a + bi) + (-a - bi) = (a - a) + (b - b)i = 0 + 0i = 0 \end{align}

And similarly:

(5)

\begin{align} \quad (-x) + x = (-a - bi) + (a + bi) = (-a + a) + (-b + b)i = 0 + 0i = 0 \end{align}

The commutativity of $+$ is also inherited from the commutativity of $+$ on $\mathbb{C}$ so we will not reverify it here.

We now show that $\mathbb{Z}(i)$ is closed under $*$. Noting that the product of two integers is an integer, we have that:

(6)

\begin{align} \quad x * y = (a + bi) * (c + di) = (ac - bd) + (ad + bc)i \in \mathbb{Z}(i) \end{align}

The associativity of $*$ is inherited from the associativity of $*$ on $\mathbb{C}$ (once again, verify this if you need to)!

The identity element of $*$ is $1 + 0i \in \mathbb{Z}(i)$.

Lastly, it's not hard to see that the distributivity of $*$ over $+$ is… you guessed it… inherited from the distributivity of $*$ over $+$ on $\mathbb{C}$.

Therefore, we have verified all of the ring axioms and so $(\mathbb{Z}(i), +, *)$ is a ring.