Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm

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Statement

Suppose R is a commutative unital ring and N is a Euclidean norm on R -- in particular, R is a Eulidean ring. We can define a new Euclidean norm \tilde{N} on R as follows (for a \ne 0):

\tilde{N}(a) = \min \{ N(ax) \mid x \in R, ax \ne 0 \}.

In other words, it is the minimum of the norms of nonzero elements in the principal ideal generated by a.

\tilde{N} is a multiplicatively monotone norm and further, \tilde{N}(a) \le N(a) for any a \ne 0. Thus, \tilde{N} is a smaller multiplicatively monotone Euclidean norm on R.

Related facts

Proof

Given: A ring R with a Euclidean norm N. Define, for a \ne 0, \tilde{N}(a) = \min \{ N(ax) \mid x \in R, ax \ne 0 \}.

To prove: \tilde{N} is Euclidean, \tilde{N}(ab) \ge \max \{ \tilde{N}(a), \tilde{N}(b) \} for ab \ne 0, and \tilde{N}(a) \le N(a) for a \ne 0.

Proof:

  1. Proof that \tilde{N}(a) \le N(a) for all a \ne 0: This is direct since \tilde{N}(a) is the minimum over a collection of numbers that includes N(a).
  2. Proof that \tilde{N}(ab) \ge \max \{ \tilde{N}(a), \tilde{N}(b)\} for ab \ne 0: This follows from the fact that the set of nonzero multiples of ab is contained in the set of nonzero multiples of a, as well as in the set of nonzero multiples of b.
  3. Proof that \tilde{N} is Euclidean: Suppose a,b \in R with b \ne 0. Then, \tilde{N}(b) = N(bx) for some x \in R. Euclidean division of a by bx with respect to the original norm N yields a = (bx)q + r where r = 0 or N(r) < N(bx). In particular, r = 0 or \tilde{N}(r) \le N(r) \le N(bx) = \tilde{N}(b). Thus, we have a = b(xq) + r where r = 0 or \tilde{N}(r) < \tilde{N}(b).