# Prove the quotient ring is commutative?

Can someone write a small proof that is being omitted by the book? I'm sure it's easy but for some reason I'm a little confused.

It says:

"Let R be a ring (not necessarily commutative). Let J be an ideal of R containing all the differences xy-yx as x and y range over R. It is quite easy to show that the quotient ring R/J is commutative."

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We can actually prove something a bit stronger. Namely: Let $R$ be a ring and $J$ an ideal of $R$. $R/J$ is commutative if and only if $xy - yx\in J$ for all $x,y\in R$.
Proof: To show that a ring is commutative, we need to show that $xy = yx$ for all $x,y$ in the ring. But this is the same as showing that $xy - yx = 0$. To be $0$ in a quotient ring is the same as being in the ideal that is being modded by. So this shows precisely that $R/J$ is commutative exactly when $xy - yx \in J$ for all $x,y \in R$.
I did not say we needed $ab = ba$ for elements in $J$. I said that we need $ab - ba \in J$ for all $a,b \in R$. – Tobias Kildetoft Dec 18 '12 at 18:11
When does $xy \equiv yx \pmod J$?