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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?

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Since $M\cong R/pR$ as an $R$-module, an element of $M^*$ must map every element in $M$ to something in $R$ whose product with $p$ is zero. The only elements of $R$ which have this property are the multiples of $p$.

Combine these two facts.

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    Oh I see.. I need to satisfy structure on $R$, thx!.2012-11-01