Euclidean rings and prime factorization

Now we will talk about the factorization of elements in Euclidean rings. On pg.146 of “Topics in Algebra” by Herstein, it says:

“Let $R$ be a Euclidean ring. Then every element in $R$ is either a unit in $R$ or can be written as the product of a finite number of prime elements in $R$ .”

This seems elementary. Take any element $a\in R$ . If it is prime, then we’re done. If it is not, then keep on splitting it into factors. For example, let $a=bc$ . If $b$ is prime, then we leave it as it is. If it is not (if $b=fg$ ), we split it as $fgc$ . The same with $c$ , and so on.

The theorem says $a$ can be represented as the product of a finite number of primes. But what if this splitting is a never-ending process? We don’t face this problem with $\Bbb{N}$ , as splitting causes positive integers to decrease in magnitude and there’s a lower limit to how much a positive integer can decrease. But we might face this problem with other Euclidean rings.

Circumventing this problem throws light on a fact that is often forgotten. $d(a)\leq d(ab)$ . When we take $a$ and start splitting it, we’re decreasing the d-values of the individual factors as we continue to split them. Note that we will not split an associate into a unit and its associate. For example, if $f$ and $g$ are associates, and $a=ft$ , then we will not split $f=u_1 g$ where $u_1$ is a unit, as we’re only looking to split elements that are not prime. Hence, if only splittings involving units are possible for $f$ , then we know that $f$ is prime, and leave it as it is. Let us suppose $f=pq$ where neither $p$ nor $q$ s a unit. Then $d(p),d(q)<d(f)$ , as they’re not associates. This shows that as we keep splitting non-prime elements into factors that are not units, then the d-value of each individual factor keeps strictly decreasing. This has a lower bound as d-values are positive real numbers, and we’re bound to arrive upon a finite factorization in a finite number of steps.

What’s the deal with units then? We’ll return to this after a short discussion on $d(0)$ and $d(1)$ .

If $a\neq 0$ , then $d(1)\leq d(1.a)=d(a)$ for every $a\in R$ . Hence, $d(1)=d(u)$ for all non-zero units $u\in R$ (proof: let $1=u_1 b$ , where $u_1$ is a unit. Then $b$ has to be $u_{1}^{-1}$ , which is also a unit. The same can be said of all units. Hence, $1$ is associate with all units only), and $d(1)<d(a)$ for all non-zero non-units in $R$ .

Now what about $d(0)$ ? If the axiom of a Euclidean ring was $a=qb+r\implies d(r)<d(b)$ rather than $d(r)<d(b)$ provided $r\neq 0$ , then we could conduct some investigation into this. Let us imagine $d(0)$ exists. Then $a=1.a+0$ . Hence, $d(0)<d(a)$ for all $a\in R$ . But $0=0.0+0$ . Hence, $d(0)<d(0)$ , which is impossible as $d:R\to \Bbb{R}^+$ is a well-defined mapping. Hence, in order to facilitate the well-defined existence of $d$ and also keep $a=qb+r\implies d(r)<d(b)$ as an axiom used to define a Euclidean ring, we forego defining $f(0)$ .

Now returning to our discussion, we’ve already stated that as $d(0)$ is not defined, and $1$ has the lowest d-value in $R$ . Moreover, as all associates of $1$ are units, $d(u)=d(1)$ , where $u$ is a unit. If it were possible to split $u$ into factors $ab$ such that both $a$ and $b$ are not units, then $d(a),d(b)<d(u)$ , which is not possible. Hence, every unit $u$ is prime in a Euclidean ring. Note that this is not a natural property of such structures, but a result of the arbitrary axiom that $d(ab)\geq d(a),d(b)$ .