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

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

Question:

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?

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  • $\begingroup$ 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
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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.

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