Def: A ring $R$ is called prime ring if $I_{1}I_{2}\subseteq \{0\}$ for ideals $I_{i}$ then $I_{1}$ or $I_{2}$ is $\{0\}$.
Def: A left $R$-module $M$ is prime if $Ann_{R}(N)=Ann_{R}(M)$ for every nonzero zero left $R$-submodule $N$.
It is clear that $R$ is prime ring iff $aRb=\{0\}$ impies $a$ or $b$ is zero iff the annhilator of every nonzero left ideal is $\{0\}$ iff the annhilator of every nonzero right ideal is $\{0\}$.
If $R$ is a prime left $R\otimes_{\mathbb{Z}}R^{op}$ module then $R$ is a prime ring(we can consider elements $a\otimes 1$ to prove this property).
But I can not find any example that $R$ is a prime ring but not a prime left $R\otimes_{\mathbb{Z}}R^{op}$-module. The most difficult part of this question is that we need to find a nonzero ideal $I$ and $\sum\limits_{i=1}^{n}a\otimes b$ such that $\sum\limits_{i=1}^{n}atb=0$ for every $t\in I$ and $\sum\limits_{i=1}^{n}arb\neq 0$ for some $r \in R$.
Is every prime ring a prime left $R\otimes_{\mathbb{Z}}R^{op}$-module?