# Quotient of UFD

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?

• In the last question, do you mean "when R is a PID"? – lhf Mar 25 '19 at 1:01
• 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. – Captain Lama Mar 25 '19 at 1:08
• 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. – Arturo Magidin Mar 25 '19 at 1:15
• @lhf Yes. That is what I mean – qinr Mar 25 '19 at 1:15
• (Ehr... where are you “replacing” UFD by PID? In the premise, or in the “if and only if” statement?) – Arturo Magidin Mar 25 '19 at 1:16

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.