# Prove that if $R$ is an Integral Domain then $S$ is an Integral Domain

Let $R$ be a ring and $S$ be a subring of $R$. Prove that if $R$ is an Integral Domain then $S$ is an Integral Domain.

I know that an ID is a commutative ring when for $a,b \in R$ if $ab=0$ then $a=0$ $b=0$ so I know I've got to show that if this property is true for a ring that it is also true for a subring but im not sure how.

• @Lepidopterist If I require a proof then I have to write prove that, I am currently trying to do this myself but I thought I could get help so I am putting effort in. Therefore it's not 'polite' to accuse someone of something without knowing all of the facts
Mar 16, 2013 at 20:29
• @Adam: hint: elements of a subring are elements of the ring. If a,b are in S, ab=0, you need to show a =0 or b =0, and you know a,b are also in R.From here, how do you do next to show a =0 or b =0?
– Long
Mar 16, 2013 at 20:31
• @Lepidopterist It wasn't a command I was stating what I needed to do. Go troll somewhere else.
Suppose, for the sake of a contradiction, that $S$ is not an ID. By definition this means that $\exists x,y\in S$ s.t. $xy=0$ but $x,y\not= 0$. But since $S\subset R \implies x,y\in R$ , which means $xy=0 \implies x=0$ or $y=0$. (This follows along the lines of Long Mai...)
• I have one doubt.$2Z$ is subbing of $Z$ but $2Z$ is not integral domain though $Z$ is integral domain.This is a contradiction right? Jun 4, 2019 at 5:02