Fields from Euclidean Domains: When do operations modulo an irreducible in an Euclidean Domain $R$ generate a field?

Question

Somehow searching for this question I only find "all fields are euclidean domains" and "euclidean domains with unique quotient-remainder are either $F$ or the polynomial ring $F[X]$ for some field $F$".

Things I think I know:

• irreducible are primes in GCD domains and thus in Euclidean Domains, so all irreducibles generate prime ideals
• Euclidean domains are commutative rings so the "modulo irreducible" creates a quotient ring by a prime ideal which is always at least an integral domain
• the quotient ring is a field iff the ideal is maximal
• therefore the question translates to asking either "when does the irreducible generates a maximal ideal?" or "in which euclidean domains do all irreducibles generate maximal ideals (like for the integers and polynomials over finite fields)?"
• This seems to be slightly different than asking when are all the prime ideals maximal (which is answered in If every prime ideal is maximal, what can we say about the ring?) because in this case the zero ideal is not required to be maximal.
• For the polynomial ring $F[x]$ of a finite field $F$ it always generates a field.

Examples

• For the polynomials over complex numbers only $x-c$ generate maximal ideals and the other irreducible polynomials do not, so being an euclidean domain and even being $F[X]$ is clearly not enough unless $F$ is finite.
• In polynomial rings over fields the quotient-remainder are unique so uniqueness is not sufficient.
• The euclidean domain of integers generates a field, so the uniqueness of quotient-remainder is not necessary.

Things that I am confused about

• I've seen multiple times than in principal ideal domains, and thus euclidean domains, all non-zero prime ideals are maximal, but this seems to contradict the above
• If $R$ is a field then it has no irreducibles because all elements all elements are divisible and either way the remainder is always zero, so in this case the answer is either "yes, but trivial" or "the question needs state first that an irreducible exists, thus excluding fields"
• Euclidean domains are unique-factorization domains which themselves are atomic domains which are defined as having factorization into irreducibles for any non-zero non-unit element, therefore an irreducible always exists in Euclidean domains, but if fields do not have irreducibles then they shouldn't be euclidean domains ???

Related questions:

• Is restricting to finite euclidean domains sufficient? Are there even finite Euclidean domains that are not finite fields? For example, can the rings of matrices over finite fields be euclidean domains?
• Does adding a discreteness structure like the integers help?

I am particularly interested in separate answers for finite and non-finite euclidean domains.

Answer 1

I think you've gotten yourself a bit confused; not to worry, that's part of doing mathematics. Here's what you need to know, hopefully laid out in a way that's clarifying.

1. In a commutative ring $R$, an element $a$ is irreducible if it is a non-zero non-unit such that whenever $a = cd$, either $c$ or $d$ is a unit. A non-zero non-unit element $a$ is prime if whenever $a \mid cd$, either $a \mid c$ or $a \mid d$.

2. A PID is a UFD, hence an element $a$ is irreducible if and only if it is prime.

3. In a PID, all non-zero prime ideals are maximal.

4. For $R$ a commutative ring and $I$ an ideal of $R$, $R/I$ is a field if and only if $I$ is maximal.

5. A Euclidean domain is a PID.

With these facts in hand, let $R$ be a Euclidean domain and $a$ an irreducible element of $R$. Then $a$ is prime, so the ideal $(a)$ a maximal ideal. Thus, $R/(a)$ is a field. In other words, if you have a Euclidean domain, taking the quotient by the ideal generated by any irreducible element of $R$ gives you a field.