Let $I$ be an ideal of $R$, it is known that

$I$ is prime $\iff$ $R/I$ is an integral domain.


If we assume that $R$ is a UFD with $1$, is it true that

$I$ is prime $\iff$ $R/I$ is a UFD?

what if UFD is replaced by PID?

Update For the last part, I mean:

If $R$ is PID, then

$I$ is prime $\iff$ $R/I$ is PID?

  • $\begingroup$ In the last question, do you mean "when R is a PID"? $\endgroup$ – lhf Mar 25 '19 at 1:01
  • $\begingroup$ Maybe take a look at math.stackexchange.com/questions/403565/… , you'll see that a quotient of a UFD is not always a IFD even when it is integral. $\endgroup$ – Captain Lama Mar 25 '19 at 1:08
  • $\begingroup$ And, no, in general if $R$ is a UFD and $I$ is prime, it is not necessarily the case that $R/I$ is a PID. For example, $\mathbb{Z}[x,y]$ is a UFD, $(y)$ is a prime ideal, but $\mathbb{Z}[x,y]/(y) \cong \mathbb{Z}[x]$ is not a PID. $\endgroup$ – Arturo Magidin Mar 25 '19 at 1:15
  • $\begingroup$ @lhf Yes. That is what I mean $\endgroup$ – qinr Mar 25 '19 at 1:15
  • $\begingroup$ (Ehr... where are you “replacing” UFD by PID? In the premise, or in the “if and only if” statement?) $\endgroup$ – Arturo Magidin Mar 25 '19 at 1:16

First question is hard; see for example this prior question. In general, it is difficult to know if a quotient of a UFD is a factorial ring (every element can be written as a product of a unit and irreducible elements, though perhaps not uniquely), let alone a unique factorization ring.

Second question is, alas, trivial. If $R$ is a principal ideal ring, then any quotient of $R$ is a principal ideal ring: if $I\triangleleft R$, and $K$ is an ideal of $R/I$, then $K=J/I$ for some (unique) ideal $J$ of $R$ that contains $I$. Since $R$ is a PIR, $J=(a)$ for some $a\in R$. Every element of $K$ is of the form $j+I$ for some $j\in J$, and hence $j$ is of the form $na + ra + as + \sum_{i=1}^m(r_ias_i) + I$ for some $n\in\mathbb{Z}$, $m\geq 0$, $r,s,r_i,s_i\in R$ (not assuming $R$ is commutative or has a $1$). But $$na+ra+as+\sum_{i=1}^m(r_ias_i)+I$$ is the same as $$n(a+I) + (r+I)(a+I)+(a+I)(s+I) + \sum_{i=1}^m((r_i+I)(a+I)(s_i+I)),$$ which is clearly in the principal ideal $(a+I)$ of $R/I$; hence $K\subseteq (a+I)$. And since $a\in J$, trivially $(a+I)\subseteq K$, giving equality. Thus, $K=(a+I)$ is principal.

In particular, for a principal ideal commutative ring with unity $R$, $I$ is prime if and only if $R/I$ is a PID; since “principal ideal” holds just because $R$ is a principal ideal ring, and then this equivalent is all about whether it is a domain or not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.