Gaussian integers

You have already seen a few domain extensions of integers \mathbb{Z} of the form \mathbb{Z}[\sqrt{D}]=\{a+b\cdot\sqrt{D}|a,b\in\mathbb{Z}\}, where D\neq 0,1 is a square-free integer (i.e., n^2\nmid D\ \forall\ n\geq 2). The case of our interest in this post are Gaussian integers \mathbb{Z}[i], that are obtained by setting D=-1.

Sources:

Theory:

We will now study prime elements of \mathbb{Z}[i] and we will see whether it has some interesting consequences. The following statement is very useful in this situation:

\textbf{Lemma:} Let p\in\mathbb{Z} be a prime number. If p is NOT of the form p=a^2+b^2 for some a,b\in\mathbb{Z}, then p is a prime element in \mathbb{Z}[i]. On the other hand, if p=a^2+b^2, then a+b\cdot i and a-b\cdot i are prime elements. Moreover, up to multiplication by units, all prime elements of \mathbb{Z}[i] are of one of these types.

A very important consequence of this lemma is the following theorem:

\textbf{Theorem:} A prime number p\in\mathbb{N} can be written as p=a^2+b^2\iff p=2 or p\equiv 1 mod 4. All prime elements in \mathbb{Z}[i] (up to multiplication by units) are

  • 1+i;
  • p\in\mathbb{N} prime numbers, such that p\equiv 3 mod 4;
  • a\pm b\cdot i, where p=a^2+b^2\equiv 1 mod 4; a,b\in\mathbb{N}.

Properties:

  • \mathbb{Z}[\sqrt{D}] is a Euclidian domain for D=-2,-1,2,3,\ldots, but it is not Euclidian for D=-3,5.
  • There is an operation of conjugation in \mathbb{Z}[i] that takes an element \alpha=a+b\cdot i and sends it to the element \overline{\alpha}=a-b\cdot i.
  • We have a Euclidian norm on \mathbb{Z}[i]: if \alpha=a+b\cdot i, then N(\alpha)=a^2+b^2=\alpha\cdot\overline{\alpha}.
  • It is a general fact that N(\alpha)=1\iff\alpha is invertible. Therefore, units (invertible elements) of Gaussian integers are 1,-1,i, and -i

Examples:

Example 1:

Decide whether 2,3 or 5 are prime elements in \mathbb{Z}[i].

Solution:

We will use Lemma from above:

  • 2=1^2+1^2, hence, 2 is not a prime element \big(2=-i\cdot(1+i)^2\big), but 1+i and 1-i are prime. 
  • 3\neq a^2+b^2, since x^2\not\equiv 0,1 mod 4. Therefore, 3 is a prime element.
  • 5=2^2+1^2, so 5 is not a prime element, but 2+i and 2-i are.

Example 2:

Show that 1,-1,i and -i are the only invertible elements in \mathbb{Z}[i].

Solution:

We use the general algebraic fact mentioned above: \alpha=a+b\cdot i is invertible \iff N(\alpha)=1. So, we are trying to solve an equation a^2+b^2=1, where a,b\in\mathbb{Z}. Since the square of any integer is non-negative, the only possible solutions are a=\pm 1 & b=0, or a=0 & b=\pm 1, and we are done.

Problems:

Problem 1:

Prove that 1+i and 1-i are associated in \mathbb{Z}[i].

Problem 2:

Compute the norm of the following elements: 3, 3\cdot i, 3-i, and 3+3\cdot i.

Leave a Reply

Your email address will not be published. Required fields are marked *