Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R$. Prove that if $I$ is prime then $R/I$ is an integral domain.

Asked Nov 03 '19 at 02:16

Active Nov 03 '19 at 02:25

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Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R$. Prove that if $I$ is prime then $R/I$ is an integral domain.

This makes sense, but I'm not sure how to prove it using only the axioms. I know an ideal of $R$ is prime iff $\forall a,b\in R,$ if $ab\in I,$ then $a\in I$ or $b\in I.$ Also, integral domains are commutative, unital, and have no zero divisors. As well, $R/ I$ is the set $\{[a] : a\in R\},$ where $\forall a,b, a\sim b$ iff $a-b \in I.$

How do I show that commutativity preserved by the quotient ring? Also, how do I show that $R/ I$ is still unital?

Is commutativity preserved because $\forall a, b\in R/I, [a][b]=[ab]=[ba]=[b][a]$?

How do I show that if $a,b\in R$ such that $ab=0,$ then $a=0$ or $b=0$?

edited Nov 03 '19 at 02:25

asked Nov 03 '19 at 02:16

Do you know how multiplication in the quotient is defined? This makes all this a trivial check. – Randall Nov 03 '19 at 02:18
$R\backslash I$ means the set of elements that are in $R$ but not in $I$. What you mean is $R/I$, the quotient of $R$ by $I$. – Arturo Magidin Nov 03 '19 at 02:20
Right. So can't you check commutativity? – Randall Nov 03 '19 at 02:20
Also, in your last edit, you want to show that either $a=0$ or $b=0$, not $\neq$. And you need to show this in the quotient, not in $R$. – Randall Nov 03 '19 at 02:22
You are correct. – Randall Nov 03 '19 at 02:24

Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R$. Prove that if $I$ is prime then $R/I$ is an integral domain.

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