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.