Euclideanness is localization-closed

Statement
Suppose $$R$$ is a Euclidean ring and $$S$$ is a fact about::multiplicatively closed subset of $$R$$ not containing any zero divisors (without loss of generality, we may assume that $$S$$ is a fact about::saturated multiplicatively closed subset. Let $$Q = S^{-1}R$$ be the localization of $$R$$ at $$S$$. Then, $$Q$$ is also a Euclidean ring. Further, if $$N$$ is a Euclidean norm on $$R$$, we can define a new Euclidean norm $$\tilde{N}$$ on $$Q$$ as follows:

$$\tilde{N}(q) = \min \{ N(qx) \mid x \in Q, qx \in R \setminus \{ 0 \} \}$$.

To see that this is well-defined, observe that any $$q \in Q$$ can be expressed as $$s^{-1}r$$ for some $$s \in S$$, $$r \in R$$, and if $$q \ne 0$$, $$r \ne 0$$. Thus, $$qs \in R \setminus \{ 0 \}$$. Hence, the set on the right side is nonempty. A minimum over a nonempty well-ordered set is well-defined, so the expression is well-defined.

Note that doing this operation for $$S = \{ 1 \}$$ does not necessarily give back the same Euclidean norm as we started with. It gives back the same norm only if the original norm was multiplicatively monotone.

Examples

 * Using the fact that the particular example::polynomial ring over a field is Euclidean, we can prove that the particular example::Laurent polynomial ring over a field is also Euclidean. The Euclidean norm of a Laurent polynomial is defined as the difference between the highest and lowest degrees among the constituent monomials with nonzero coefficients.

Related facts

 * Euclideanness is quotient-closed
 * Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm: In fact, this is precisely the new norm we get when we do the process outlined above, setting $$S = \{ 1 \}$$.

Proof
==