Every Euclidean ring has a unique smallest Euclidean norm

From Commalg
Jump to: navigation, search

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:

Related facts

Examples

Facts used

  1. Element of minimum norm in Euclidean ring is a unit
  2. 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, Failed to parse (syntax error): 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.