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Let $R$ be a commutative ring and $M\neq 0$ an $R$-module. Show that if for every $R$-module $X\neq 0$, $\operatorname{Hom}_R(M,X)\neq 0$ then $M\otimes_R X\neq0$ for every $R$-module $X\neq 0$.

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I removed the "non-commutative algebra" tag as this is about commutative rings. –  Ted Dec 27 '12 at 21:58

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up vote 5 down vote accepted

If $X\ne0$, then $\operatorname{Hom}(X,X)\ne0$, hence by assumption $\operatorname{Hom}(M,\operatorname{Hom}(X,X))\ne 0$. But the latter is also $\operatorname{Hom}(M\otimes X,X)$.

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Perhaps you should say that you used the tensor - hom adjunction in your proof above. +1 anyway for your answer. –  user38268 Jan 1 '13 at 9:58

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