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Let $M$ be a finitely generated module over a commutative ring and $N$ be a non zero proper submodule of $M$. Then is it always possible to have a non zero homomorphism $f$ from $M$ to $N$?

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Let $R$ be a commutative ring, and $M$ a non-zero finitely generated $R$-module. If $0\ne N\le M$, then $\operatorname{Hom}_R(M,N)\ne 0$.

It's clear that $\operatorname{Hom}_R(M,E(N))\ne 0$. Let $f\in \operatorname{Hom}_R(M,E(N))$, $f\ne 0$, and $\operatorname{Im}(f)=\langle y_1,\dots,y_n\rangle$ with $y_i\in E(N)$ not all zero. Since the extension $N^n\subset E(N)^n$ is essential, there is $a\in R$ such that $0\ne a(y_1,\dots,y_n)\in N^n$. Multiplication by $a$ defines a non-zero homomorphism from $\operatorname{Im}(f)$ to $N$, and thus we get a non-zero homomorphism from $M$ to $N$.

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  • $\begingroup$ Beautiful! For the first sentence: $\operatorname{Hom}_R(\_, E(N))$ is exact, so $0 \to N \to M$ induces $\operatorname{Hom}_R(M,E(N)) \to \operatorname{Hom}_R(N,E(N)) \to 0$ $\endgroup$ – zcn May 26 '14 at 22:07

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