I encountered following proposition:
Let $p$ be a prime, $R=\mathbb{Z}/p^2\mathbb{Z}$, $M=\mathbb{Z}/p\mathbb{Z}$ as $R$-module. Then we have the map $M^*\otimes_RM\rightarrow$Hom$_R(M,M)$ given by $\varphi\otimes m\mapsto \varphi(\cdot)m$ identically $0$.
However, I cannot see why it is identically $0$.
if we define $\varphi(1)=a$, then $\varphi(n)=an$, then $\varphi\otimes m$ will map to $f(n)=\varphi(n)m=amn$, it is no reason to be zero. right?