Euclideanness is quotient-closed

Statement
Suppose $$R$$ is a commutative unital ring that possesses a fact about::Euclidean norm $$N$$. Suppose $$I$$ is an ideal inside $$R$$. Then, $$R/I$$ is a Euclidean ring, with norm given by:

$$\overline{N}(x + I) = \min \{ N(x + r) \mid r \in I \}$$.

In other words, the norm of a coset is defined as the minimum of normso f all elements in the coset. (Note that this minimum is well-defined since it is the minimum over a nonempty subset of a well-ordered set.

Related facts

 * Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm
 * Euclideanness is localization-closed

Proof
Given: A commutative unital ring $$R$$ with Euclidean norm $$N$$. An ideal $$I$$ of $$R$$. $$\overline{N}$$ is defined on $$R/I$$ by:

$$\overline{N}(x + I) = \min \{ N(x + r) \mid r \in I \}$$.

To prove: $$\overline{N}$$ is a Euclidean norm on $$R/I$$.

Proof: Suppose $$a + I, b + I$$ are two elements of $$R/I$$ with $$b + I \ne I$$. Now, there exists $$c \in a + I, d \in b + I$$ with $$N(c) = \overline{N}(a)$$ and $$N(d) = \overline{N}(b)$$. By the Euclidean division in $$R$$ we have:

$$c = dq + r$$

where $$r = 0$$ or $$N(r) < N(d)$$. Going modulo $$I$$, we get:

$$c + I = (d + I)(q + I) + (r + I)$$

which can be rewritten as:

$$a + I = (b + I)(q + I) + (r + I)$$

where $$r + I = I$$ or $$N(r) < N(d)$$. Note that $$\overline{N}(r + I) \le N(r)$$ by definition, and $$N(d) = \overline{N}(b + I)$$ by the choice of $$d$$. Thus, we have $$r + I = I$$ or $$\overline{N}(r + I) < \overline{N}(b + I)$$, which is precisely the condition for Euclidean division in $$R/I$$.