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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?

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The quotient of a Euclidean domain by a non-zero prime ideal is a field, isn't it? – Hurkyl Jan 26 '12 at 4:10
@user15464: That a quotient of a PID by a prime ideal is PID follows for the same reason as it does for Euclidean; on the other hand, with a PID you have that a quotient by any ideal is still a principal ideal ring (though it won't be a domain if the ideal is not prime), and I suspect that's really what inspired the question. – Arturo Magidin Jan 26 '12 at 15:31
up vote 14 down vote accepted

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 $1$.

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Congratulations on hitting 100k! – J. M. Jan 26 '12 at 5:59
@J.M. Thanks! But according to the audit I just ran, I'm actually a bit short. – Arturo Magidin Jan 26 '12 at 6:52
How about now? 2200 answers is also very impressing. Congrats :-) – Asaf Karagila Jan 26 '12 at 16:51

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