Let $R$ be a non-commutative ring, and, consider the tensor product $$R \otimes_R R $$ where we consider the 'first $R$' as a right module over $R$ and the 'second $R$' as a left module over itself.
Now, I've seen that $R \otimes_R R \cong R$, but I'm not sure why? How would I prove this? I've tried:
Define an $R$-middle-linear ($R$-balanced) map $m(a,b) = ab$ from $R \times R \rightarrow R$, but I'm not too sure where to go from here?
That gives you one direction, namely $R\otimes_RR\to R$.
If $R$ has an identity element, then its inverse will be $r\mapsto 1\otimes r$.
Note that, by the tensor property, we have $$ab\otimes 1=a\otimes (b\cdot 1) =a\otimes b=(1\cdot a)\otimes b=1\otimes ab$$