$R/I$ is commutative iff $rs-sr \in I$

Question

How to go with proving this : Let $R$ be a ring and let $I$ be an Ideal of $R$.

Prove that the factor ring $R/I$ is commutative iff $rs-sr \in I$ $\forall r$ and $s$ in $R$

any hint how to go with this...

Answer 1

Let $R/I$ is commutative, then $(r+I)(s+I)=(s+I)(r+I) \implies (rs+I)=(sr+I) \implies rs-sr \in I$.

Conversely, let for any $r,s \in R$, $rs-sr \in I$ then reversing the above implications, you will get it.

Answer 2

You should really look at the definition of a quotient again! (you will see the result immediately)

Basically you have equivalence classes $[x] \in R/I$ which are zero precisely when one (or every) representative $x \in [x]$ lies in $I$. Hence your statement follows then easy by:

$$ [rs]=[r][s]=[s][r]=[sr] $$