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

Statement
Suppose $$R$$ is a commutative unital ring and $$N$$ is a fact about::Euclidean norm on $$R$$ -- in particular, $$R$$ is a fact about::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 fact about::multiplicatively monotone norm and further, $$\tilde{N}(a) \le N(a)$$ for any $$a \ne 0$$. Thus, $$\tilde{N}$$ is a smaller fact about::multiplicatively monotone Euclidean norm on $$R$$.

Related facts

 * Every Euclidean ring has a unique smallest Euclidean norm
 * Euclideanness is localization-closed
 * Euclideanness is quotient-closed

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 $$\tilde{N}(r) < \tilde{N}(b)$$.