Given a commutative ring $R$ and two isomorphic free $(R,R)$-bimodules $M$ and $N$, is it true that the tensor product $M\otimes_R M$ is isomorphic to the tensor product $N \otimes_R N$?
Here's my idea thus far: Given an $R$-module isomorphism $\varphi:M\rightarrow N$, we know there is a unique $R$-module homomorphism $\Phi=(\varphi\otimes\varphi):M\otimes_R M\rightarrow N\otimes_R N$ given by $\Phi(m_1\otimes m_2)=\varphi(m_1)\otimes\varphi(m_2)$. Clearly this map is surjective, given the surjectivity of $\varphi$, but I'm struggling to determine if it's injective. I know the kernel of $\Phi$ is given by the elements $m_1\otimes m_2 \in M\otimes_R M$ such that $\varphi(m_1)\otimes\varphi(m_2)=0$, but is it possible to then determine that $m_1\otimes m_2=0$?