### Euclidean rings and prime factorization

#### by ayushkhaitan3437

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

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

This seems elementary. Take any element . If it is prime, then we’re done. If it is not, then keep on splitting it into factors. For example, let . If is prime, then we leave it as it is. If it is not (if ), we split it as . The same with , and so on.

The theorem says 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 , 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. . When we take 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 and are associates, and , then we will not split where is a unit, as we’re only looking to split elements that are not prime. Hence, if only splittings involving units are possible for , then we know that is prime, and leave it as it is. Let us suppose where neither nor s a unit. Then , 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 and .

If , then for every . Hence, for all non-zero units (proof: let , where is a unit. Then has to be , which is also a unit. The same can be said of all units. Hence, is associate with all units only), and for all non-zero non-units in .

Now what about ? If the axiom of a Euclidean ring was rather than provided , then we could conduct some investigation into this. Let us imagine exists. Then . Hence, for all . But . Hence, , which is impossible as is a well-defined mapping. Hence, in order to facilitate the well-defined existence of and also keep as an axiom used to define a Euclidean ring, we forego defining .

Now returning to our discussion, we’ve already stated that as is not defined, and has the lowest d-value in . Moreover, as all associates of are units, , where is a unit. If it were possible to split into factors such that both and are not units, then , which is not possible. Hence, every unit 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 .

Summarising the above arguments:

1. A unit is prime in a Euclidean ring .

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

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

4. Removing the axiom would nullify a lot of these properties.