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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.

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    $\begingroup$ @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 $\endgroup$
    – Adam
    Mar 16, 2013 at 20:29
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    $\begingroup$ @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? $\endgroup$
    – Long
    Mar 16, 2013 at 20:31
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    $\begingroup$ @Lepidopterist It wasn't a command I was stating what I needed to do. Go troll somewhere else. $\endgroup$
    – Adam
    Mar 16, 2013 at 20:36
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    $\begingroup$ @Adam: what Lepidopterist was trying to say is that it is more polite to ask "how would I do this" rather than saying "Do this". To some, this may seem a small thing, but it really grates on others. $\endgroup$
    – robjohn ♦
    Mar 16, 2013 at 21:58
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    $\begingroup$ @robjohn Fair enough but to be honest I don't see the difference and I have no idea why he felt the need to comment on the question. If he didn't like it he could have just not answered it. There was no need to start having a go at me $\endgroup$
    – Adam
    Mar 17, 2013 at 14:20

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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...)

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  • $\begingroup$ 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? $\endgroup$
    – ASHWINI SANKHE
    Jun 4, 2019 at 5:02

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