# Quotient Rings and Ideals--integral domain question

Could you please verify my proof? I am trying to see if my biconditional logical statements are correct.

the problem is: let $I \neq R$ be an ideal in a commutative ring $R$ with id. show $R/I$ is an integral domain $iff$ whenever $ab \ \in \ I$ then either $a \in \ I$ or $b \in \ I$

the proof:

$R/I$ is an integral domain

$\iff \forall a, b \in R/I$ such that $ab = 0$, either $a=0$ or $b = 0$

$\iff ab \in I$

$\iff a \in I$ or $b \in I$

The other question I have is: the fact that in an ideal, $ab + I =(a+I)(b+I)$ is not contingent on $R$ being a commutative ring, is it? in that case the question would be true in the case of ideals of non commutative rings?

Thank you

• Your second equivalence in the proof should be $(ab \in I) \Rightarrow (a \in I \quad \mathrm{or} \quad b \in I)$. Does this statement relate to $I$ being a prime ideal? Commented Jan 22, 2014 at 0:06
• The equivalence is not valid in noncommutative rings. Your reasoning is incorrect: if $a\in R/I$ then you can't say $a\in R$. An element of $R/I$ is a subset of $R$. Commented Jan 22, 2014 at 0:13
• Thank you Chris. Your remark helped me understand the condition of prime ideals. Commented Jan 22, 2014 at 2:20
• You're welcome, Quester. Commented Jan 22, 2014 at 3:50
• The elements of $R/I$ look like $a+I$. So the product of two elements of $R/I$ being zero would be written as $(a+I)(b+I)=0+I$. Of course, $(a+I)(b+I)$ is $ab+I$ by definition, and $ab+I=0+I$ iff $ab\in I$ (easy to show). Also note you seemed to have dropped the $\forall a,b$.
– anon
Commented Feb 27, 2014 at 1:23