I would like to point out that this question has a satisfactory answer here. I am just trying a slightly different way.
Let $R$ be a local ring with maximal ideal $P$. Let $M$ and $N$ be two $R$-modules. Let $k=R/P$. Let $M\otimes_R N=0$
We have $$(M\otimes_R k)\otimes_R(N\otimes_R k)\cong M\otimes_R(k\otimes_R(N\otimes_R k))$$ $$\cong M\otimes_R(k\otimes_R(k\otimes_R N))$$ $$\cong M\otimes_R((k\otimes_Rk)\otimes_R N))$$ $$\cong M\otimes_R( N\otimes_R(k\otimes_Rk))$$ $$\cong M\otimes_R( N\otimes_R(k\otimes_Rk))$$ $$\cong (M\otimes_RN)\otimes_R(k\otimes_Rk))$$ $$\cong (M\otimes_RN)\otimes_R(R/P\otimes_RR/P))$$ $$\cong (M\otimes_RN)\otimes_R(R/P)$$
For the last step we are using the fact that for ideals $I$ and $J$ of $R$ we have $$R/I\otimes_R R/J\cong R/(I+J)$$
Since $M\otimes_R N=0$ we conclude that $$(M\otimes_R k)\otimes_R(N\otimes_R k)=0$$
My questioin:
Can we conclude from $(M\otimes_R k)\otimes_R(N\otimes_R k)=0$ that either $(M\otimes_R k)=0$ or $(N\otimes_R k)=0$?