If $R$ is an associative ring then $[R,R]$ is the subgroup generated by the elements $[r,s]= rs-sr,$ for $r,s\in R$. Show that $Trace : M_n(R)\longrightarrow R/[R,R]$ induces an isomorphism $$ M_n(R)/[M_n(R),M_n(R)] \longrightarrow R/[R,R]$$
What I have tried so far is this: I have a map $trace : M_n(R) \longrightarrow R$ the usual trace which is surjective Then $j : R\longrightarrow R/[R,R]$ which is also surjective.
So I have a surjection $Trace$ ($(j\circ trace))$ form $ M_n(R)\longrightarrow R/[R,R]$
Now I need to show that Kernel(Trace) is my $[M_n(R),M_n(R)] $
Since $trace [AB-BA] \in [R,R]$ I have one way containment which is: $$ [M_n(R),M_n(R)] \subset Kernel(Trace)$$
I am having trouble in proving $$Kernel(Trace)\subset [M_n(R),M_n(R)]$$
What I mistakenly proved is that for any $r\in [R,R]$ there exists $A,B \in M_n(R)$ such that $trace(AB-BA) = r$. Which doesn't help me.
So if you guys could help me out I will be delighted. Thank you.
View both $M_n(R)$ as a free left $R$-module with basis $(E_{ij})_{1\le i,j\le n}$. This trace map is obviously a $R$-module homomorphism (for all $n\ge 0$). It is obviously surjective for $n\ge 1$.
It remains to prove injectivity ($n\ge 0$). This means one has to prove that if some element $M\in M_n(R)$ has its trace in $[R,R]$ then it belongs to $[M_n(R),M_n(R)]$. Note that this is the case for $M=E_{ij}=[E_{ii},E_{ij}]$, and for $M=E_{ii}-E_{jj}=[E_{ij},E_{ji}]$. This allows to reduce to the case when $M=rE_{11}$, with $r\in [R,R]$, say $r=\sum [a_i,b_i]$: then $M$ equals $\sum [a_{i}E_{11},b_iE_{11}]$.