Let $R$ be a commutative ring with $1$, and let $[R,R]$ be the ideal generated by the set $$ \{ \ xy-yx \ : \ x,y \in R \ \} $$ Show that $R/[R,R]$ is commutative.
Before I show what I did, I wan't to find out if I understood how such an ideal is defined. I know that for a sequence that ends somewhere the ideal is defined as follows. $$ (x_1, \cdots ,x_n) \quad = \quad \{ \ r_1x_1 + \cdots r_nx_n \ : \ r_1, \cdots + r_n \in R \ \} $$
For an infinite set, I guess that just finitely many elements are allowed to be non-zero but I still doubt about it a bit.
I understand that I need to pick $x,y \in R,$ and than I need two find some $r \in [R,R]$ such that $xy = ryx$. That would mean that $r=xyx^{-1}y^{-1}$, a commutator as I them from group theory. Can you help me to find the right element?
Adjustment I messed up something. I should find an $r$ such that $xy + r = yx$, instead of finding an $r$ that satisfies $xy =ryx$. Everything is easier now.
If you mod out by $[R,R]$, then $\overline{xy-yx}=\overline{xy}-\overline{yx}=\overline{x}\overline{y}-\overline{y}\overline{x}=\overline{0}$. Whence $\overline{x}\overline{y}=\overline{y}\overline{x}$.