Exploration 5.1: Norm

December 6, 2023.

Questions:

What is a norm?

Recall that Euclidean domains have a division algorithm. This division algorithm required a notion of degree, which is defined as the degree of the polynomial in the case of polynomial rings. In non-polynomial rings, this notion of degree generalizes to a function $R\setminus\{0\}\to\NN$ mapping each nonzero to a natural number $\ge 0$ called the Euclidean norm.

In this exploration, we’ll be looking at the Euclidean domain $\ZZ[i]$ . This is a fine example of an Euclidean domain that is not a field – it contains $2$ and not $1/2$ so it lacks multiplicative inverses, thus not a field, but you can define a division algorithm on it.

Theorem: The Gaussian integers

\ZZ[i]

are a Euclidean domain.

We can define an explicit division for Gaussian integers, thus proving the condition for being a Euclidean domain.
We can just use division in the complex numbers $\CC$ except with remainder. Given two Gaussian integers $a+bi$ and $c+di$ , transform $\frac{a+bi}{c+di}$ as we do in $\CC$ : $\begin{aligned} \frac{a+bi}{c+di} &=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\ &=\frac{ac-adi+bci+bd}{c^2+d^2}\\ &=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i \end{aligned}$
The remaining divisions by $c^2+d^2$ are in the integers $\ZZ$ , a Euclidean domain. Therefore we can use the division algorithm in the integers to obtain $q_1,r_1,q_2,r_2$ , and rewrite the above to $q_1+q_2i$ with remainder $r_1+r_2i$ . These satisfy the property that $q_1=\text{round}\left(\frac{ac+bd}{c^2+d^2}\right)\quad\text{and}\quad q_2=\text{round}\left(\frac{bc-ad}{c^2+d^2}\right)$ and due to the rounding, $|r_1|,|r_2|<\frac{\sqrt{c^2+d^2}}{2}$ Thus $a+bi=(q_1+q_2i)(c+di)+(r_1+r_2i)$ This completes the division algorithm.

This division algorithm differs from the one for polynomial rings — instead of enforcing that the remainder has degree less than that of the divisor, we enforce that the magnitude of the remainder is less than $\frac{\sqrt{c^2+d^2}}{2}$ , which is the magnitude of $c+di$ divided by $2$ . If we define the Euclidean norm $N(\cdot)$ in the Gaussian integers to be $N(a+bi)=a^2+b^2$ , then we can make a statement about the norm of the remainder $r_1+r_2i$ :

$N(r_1+r_2i)=r_1^2+r_2^2\le\left(\frac{\sqrt{c^2+d^2}}{2}\right)^2+\left(\frac{\sqrt{c^2+d^2}}{2}\right)^2=\frac{c^2+d^2}{2}\le N(c+di)$

This shows $N(r_1+r_2i)\le N(c+di)$ , which is the generalized form of $\deg r<\deg g$ in the polynomial version. Indeed, $N(\cdot)=\deg\cdot$ is the Euclidean norm in polynomial rings.

So in general, in particular for non-polynomial rings, a Euclidean domain defines a Euclidean norm $N:R\setminus\{0\}\to\NN$ , and admits a division algorithm that lets you divide $a$ by nonzero $b$ to obtain a quotient $q$ and remainder $r$ such that $N(r)<N(b)$ .

TODO proof that norm is all that is required to define a division algorithm

In this section, we explore how to define PIDs via norms.

We just saw that an integral domain $R$ is a Euclidean domain if and only if it admits a Euclidean norm. Can we extend this kind of definition to other types of domains?

The answer is yes: for PIDs, we can similarly define such a norm.

We will use this to prove that the Gaussian integers are a PID, without using the fact that it is a Euclidean domain.

Let $N(a+bi)=a^2+b^2$

$R\setminus\{0\}\to\NN$

Previously we only discussed irreducibility in the context of polynomial rings. However, elements in non-polynomial rings can also be irreducible – we only require that the element be not factorable into two nonunits.

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