### Euclidean rings and generators of ideals

This is to address a point that has just been glazed over in “Topics in Algebra” by Herstein.

In a Euclidean ring, for any two elements $a,b\in R, \exists q,r\in R$ such that $a=bq+r$. Also, there exists a function $d:R\to\Bbb{R}$ such that $d(r).

We also know that the element with the lowest d-value generates the whole ring $R$. The proof of this is elementary.

But what if there are more than one element with the same lowest d-value? Do both these elements generate $R$?

Yes. Proof: Let $d(x)=d(y)$ such that $x$ and $y$ are the elements of $R$ with the lowest d-value. Then for a third element $c\in R$, $c=u_1 x=u_2 y$. Hence, both $x$ and $y$ divide $c$. They also divide each other. Hence, $x$ and $y$ have to be associates. In other words, $x=uy$, where $u$ is a unit in $R$.

Let us approach this from the opposite direction now. If $x=uy$, where $u$ is a unit. Axiomatically, $d(y)\leq d(uy)$. Hence, $d(y)\leq d(x)$. Similarly, $d(x)\leq d(y)$. This shows that $d(x)=d(y)$. Therefore, whenever two elements $a$ and $b$ are associates, their d-values are the same.

Note that if we did not have the axiom that $d(a)\leq d(ab)$, then there would be no reason to believe that if $a$ and $b$ are associates, then $d(a)=d(b)$. Hence, ideals could then potentially be generated  by elements whose d-values would not be the lowest in the ideal, with the restriction that all those elements would be associates of the lowest d-value element.

A summary of the important points is:

1. Associates have the same d-value.

2. An element $a$ generates an ideal iff it has the lowest d-value in the ideal.

3. All associates of the lowest d-value element in an ideal generate the same ideal.

4. If we did not have the axiom $d(a)\leq d(ab)$, then point 1 would not be true, point 2 would not be true (a generator of an ideal wouldn’t have to have the lowest d-value), but point 3 would still be true.

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There’s a couple of things I’d like to add here.

Why is it that a prime element should be such that its factorization does not contain a unit element? Generally, when we think about prime numbers in positive integers, we imagine a number which is absolutely not factorizable except in the form $1.p$ ($p$ being the prime number). A sense of unbreakability is felt. Here, the same prime element $p$ can be broken in at least $n$ ways, where $n$ is the number of unit elements in the Euclidean ring $R$. The sense of absolute unbreakability is lost. I suppose the reason for this is that the concept of ‘unit’ is just an extension of $1$ in natural numbers. As factorization of primes of the form $1.p$ are not counted when dealing with natural numbers, factorizations of the form $u.p_1$ shouldn’t count in $R$, where $p_1$ is an associate of $p$.

Also note that the addition of deletion of axioms would have greatly changed the structure of Euclidean rings. For example, deleting the axiom $d(ab)\geq d(a),d(b)$ would allow infinite prime factorizations of elements, and adding the axiom $d(a+b) would further alter the structure of $R$. One should not forget that the properties of the elements of $R$ are a result of these defining axioms, and the addition and deletion of such would cause substantial alterations. It is just the fact that Euclidean rings mimic many properties of natural numbers that we find them important to study.