Every Euclidean ring has a unique smallest Euclidean norm

Statement

Every Euclidean ring (and in particular, every Euclidean domain) has a unique smallest Euclidean norm $N$. More specifically, given a commutative unital ring $R$ that possesses a 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 universal side divisors.
• $N$ is a 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 automorphism-invariant 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, $0 \} \cup \bigcup_{m=0}^n-1$ . 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.