# Euclideanness is quotient-closed

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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## Statement

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.

## 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 $R/I$.