📂Abstract Algebra

Integral Domain Norm

Definition ¹

An multiplicative norm $N : D \to \mathbb{Z}$ on an integral domain $D$ with respect to all $\alpha , \beta \in D$ is defined by the following conditions:

• (i): $N (\alpha) = 0 \iff \alpha = 0$
• (ii): $N ( \alpha \beta ) = N ( \alpha ) N ( \beta )$

Theorem

Let $p \in \mathbb{Z}$ be a prime.

• [1]: If a multiplicative norm $N$ is defined on $D$, then $N(1) = 1$ and for all units $u \in D$, $| N ( u ) | = 1$
• [2]: If all $\alpha \in D$ that satisfy $| N ( \alpha )| =1$ are units in $D$, then any $\pi \in D$ that satisfies $| N ( \pi ) | = p$ is an irreducible element in $D$.

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• A unit is an element that has a multiplicative inverse.

Explanation

While the term norm typically presupposes $N (\alpha) \ge 0$, the conditions such as $\nu ( \alpha) = N ( \alpha)$ added for $\alpha \ne 0$ often make it a multiplicative as well as a Euclidean norm. General guarantees cannot be made; however, it’s rare to find motivation to study algebraic structures that do not even conform to this level of common sense. If a norm is defined, it is almost certainly safe to assume $N : D \to \mathbb{N}_{0}$.

The definition of a norm provides significant aid in understanding the arithmetic structure of an integral domain $D$. In algebraic number theory, various norms suited to the domain are defined, allowing algebraic structures that don’t seem to fall within the realm of number theory to be ‘pulled in’ for study. Needless to say, they are directly applicable to number theory. Interesting examples include Gaussian integers $\mathbb{Z} [i]$$i$, $\omega$ and Eisenstein integers $\mathbb{Z} [\omega]$(../1291).

According to theorem [2], even if we know little about an element $\pi$ of that $D$, merely knowing that $N ( \pi )$ is prime ensures that $\pi$ is an irreducible element in $D$. As is known, a prime $p$ is an irreducible element in $\mathbb{Z}$, and $N$ through condition (ii) can be viewed as preserving the properties of an irreducible element from $D$ to $\mathbb{Z}$.

Proof

[1]

Strategy: Ripping apart an element of $D$ via condition (ii) naturally leads to deduction.

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For the identity element $1 \in D$, computing $N(1)$ yields $$N(1) = N \left( 1 \cdot 1 \right) = N (1) N (1)$$ Therefore, $N(1)$. Furthermore, if $u \in D$ is a unit, its inverse $u^{-1} \in D$ exists by definition, hence $$1 = N ( 1) = N ( u u^{-1} ) = N (u ) N (u^{-1})$$ Of course, since $N (u)$ is an integer, it must be $| N ( u) | =1$.

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[2]

Given that all $u \in D$ satisfying $| N(u) | = 1$ are units in $D$. If $\pi \in D$ is such that $| N ( \pi ) | = 1$ and $\pi = \alpha \beta$, then $$p = | N ( \pi ) | = | N ( \alpha ) N ( \beta ) |$$ Since $p$ is prime, it must be that $| N ( \alpha ) | = 1$ or $| N ( \beta ) | = 1$. Given the assumption, one of $\alpha$ or $\beta$ must be a unit in $D$, thus $\pi$ becomes an irreducible element in $D$.

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