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.

Related facts

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 $b+I\neq I$ . 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 ${\overline {N}}(r+I)\leq N(r)$ 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 $R/I$ .