Euclideanness is quotient-closed

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This article gives the statement, and possibly proof, of a commutative unital ring property (i.e., Euclidean ring) satisfying a commutative unital ring metaproperty (i.e., quotient-closed property of commutative unital rings)
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Suppose R is a commutative unital ring that possesses a 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.

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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.