Every Euclidean ring has a unique smallest Euclidean norm

Statement
Every fact about::Euclidean ring (and in particular, every fact about::Euclidean domain) has a unique smallest Euclidean norm $$N$$. More specifically, given a commutative unital ring $$R$$ that possesses a fact about::Euclidean norm, there is a Euclidean norm $$N$$ with the following properties:


 * For any Euclidean norm $$N_1$$ on $$R$$, and any $$a \in R \setminus \{ 0 \}$$, $$N(a) \le N_1(a)$$.
 * The elements of norm zero are precisely the units and the elements of norm one are precisely the fact about::universal side divisors.
 * $$N$$ is a fact about::multiplicatively monotone Euclidean norm, i.e., it is multiplicatively monotone: if $$ab \ne 0$$, $$N(ab) \ge \max \{ N(a), N(b) \}$$.
 * $$N$$ is an fact about::automorphism-invariant Euclidean norm.

Related facts

 * Element of minimum norm in Euclidean ring is a unit
 * Element of minimum norm among non-units in Euclidean ring is a universal side divisor
 * Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm

Examples

 * Over the particular example::ring of rational integers, the smallest Euclidean norm is the greatest integer of the logarithm to base two of the absolute value..
 * Over the particular example::polynomial ring over a field, the smallest Euclidean norm is the usual degree function.

Facts used

 * 1) uses::Element of minimum norm in Euclidean ring is a unit
 * 2) uses::Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm

Proof
Given: A commutative unital ring $$R$$ with a Euclidean norm $$N_1$$.

Construction of the sloewst growing Euclidean norm
We define the norm inductively as follows.

For $$n \ge 0$$, define $$R_n$$ as follows:


 * $$R_0$$ is the set of units.
 * For $$n \ge 1$$, $$R_n$$ is defined as the set of those $$x$$ such that $$x \ne 0$$, $$x \notin \bigcup_{m=0}^{n-1} R_m$$, and for every $$y \in R$$, there exists $$r \in \{ 0 \} \cup \bigcup_{m=0}^{n-1} R_m$$ such that $$x | y - r$$.

For any $$x \in R \setminus \{ 0 \}$$, $$N(x)$$ is defined as the unique $$n$$ such that $$x \in R_n$$.

Note that $$N(x)$$ is well-defined for all $$x$$, $$N$$ is Euclidean by definition: for every $$x,y \in R$$ with $$x \ne 0$$, there exists $$r$$ such that $$x | y -r$$ with $$r = 0$$ or $$r \in \bigcup_{m=0}^{N(x)-1} R_m$$, which is precisely saying that $$r = 0$$ or $$N(r) < N(x)$$. Thus, we can write:

$$y = xq + r$$

with $$r = 0$$ or $$N(r) < N(x)$$.

Proof that this Euclidean norm exists for a Euclidean ring and is smaller than any other Euclidean norm
We prove that if $$N_1(a) = n$$ for the Euclidean norm $$N_1$$, then $$a \in \bigcup_{m=0}^n R_m$$. This will prove that $$N$$ is well-defined and $$N(x) \le N_1(x)$$ for all nonzero $$x$$.

Given: $$a \in R \setminus \{ 0 \}$$ with $$N_1(x) = n$$.

To prove: $$a \in \bigcup_{m=0}^n R_m$$.

Proof: We prove this claim by induction on $$n$$.

Base case: When $$n = 0$$, we have $$N_1(x) = 0$$. By fact (1), $$a$$ is a unit, so $$a \in R_0$$ by definition.

Induction step: Suppose $$n > 1$$, and the statement is true for $$0 \le m \le n - 1$$. We have $$N_1(x) = n$$.


 * If $$x \in \bigcup_{m=0}^{n-1} R_m$$, we are done.
 * Suppose $$x \notin \bigcup_{m=0}^{n-1} R_m$$. For any $$y \in R$$, the division algorithm yields:

$$y = xq + r$$

where $$r = 0$$ or $$N_1(r) < N_1(x) = n$$. By the inductive assumption, $$r \in \{ 0 \} \cup \bigcup_{m=0}^n-1} R_m$$. Thus, for any $$y \in R$$, $$x | y - r$$ for some $$r \in \{ 0 \} \cup \bigcup_{m=0}^{n-1} R_m$$.

Thus, $$x \in R_n$$.

Proof that the norm is multiplicatively monotone
Given: $$a,b \in R$$ such that $$ab \ne 0$$.

To prove: $$N(ab) \ge \max \{ N(a), N(b) \}$$.

Proof: One way of proving this is to invoke fact (2), which states that given any Euclidean norm, we can obtain a smaller multiplicatively monotone Euclidean norm. Since the norm $$N$$ is the smallest Euclidean norm, it must be equal to the multiplicatively monotone Euclidean norm obtained from it using fact (2).

We can also check the condition directly. If $$ab \in R_n$$, then either $$a \in R_m$$ for some $$m < n$$, or $$a$$ satisfies all the conditions for being in $$R_n$$. In either case, $$N(ab) \ge N(a)$$. Similarly, $$N(ab) \ge N(b)$$.

Proof that the norm is automorphism-invariant
The way the norm is defined is clearly automorphism-invariant, since it depends on no choices.