Let $R$ be a Euclidean domain, i.e., a ring with a norm $N : R \rightarrow \mathbb N$ such that for any $a,b\in R$ with $b\not=0$, we may write $a = qb + r$ for some $q,r \in R$ with $N(r) < N(b)$. Let $I$ be a prime ideal of $R$, so $R/I$ is a domain. Must $R/I$ in fact be a Euclidean domain?
I know that this is true if we replace "Euclidean domain" with "PID", and false if we replace it with "UFD", so there is no clear intuition to be gained from similar concepts. My first attempts at a proof was to borrow a construction from analysis and define a norm on $R/I$ by
$N(a + I) = \inf\{N(a+i) : i\in I\}$
but this didn't really get me anywhere.
So, this fact true? If so, how can I prove it? If not, whats a counterexample?
Best Answer
It is, but for silly reasons.
In a Euclidean domain, every nonzero prime ideal is maximal: for if $(p)$ is a prime ideal, and $(p)\subseteq (q)$, then $q|p$. Thus, $p=qx$ for some $x$. Since $p$ is prime, either $p|q$ and so $(p)=(q)$ (since $p$ and $q$ are associates), or $p|x$, in which case $q$ is a unit and $(q)=R$.
Since every nonzero prime ideal is maximal, if $I$ is a prime ideal then you have either $R/I\cong R$ and so the quotient is Euclidean; or you have that $R/I$ is a field, in which case it is a Euclidean domain with the trivial norm map that sends every nonzero element to $1$.