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).

Summarising the above arguments:

1. A unit is prime in a Euclidean ring R.

2. Every element in R can be split into a finite number of prime factors.

3. In order to avoid contradictions, d(0) is not defined. Also, 1 has the lowest d-value.

4. Removing the axiom d(ab)\geq d(a),d(b) would nullify a lot of these properties.

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Graduate student

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