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Let $R$ be a ring, and $M,N$ are $R$-modules, and $I=Ann(N)$. If $I$ contains an $M$-regular element, then $\text{Hom}_{R}(N,M)=0$.

The above statement is from Proposition 1.2.3 of Bruns and Herzog's book "Cohen-Macaulay Rings." The author says that it is evident; but not for me. Could you explain why this holds?

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2 Answers 2

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Let $x_1\in I $ be a nonzerodivisor on $M $ and let $n\in N $. Notice that $x_1n=0$. For a linear map $f:N\to M $, we have $$f (x_1n)=0\implies x_1f (n)=0.$$ Thus $f (n)=0$ and hence $f\equiv 0$.

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If $aN = 0$ and $am = 0$ implies $m = 0$ then

$$0 = f(an) = af(n) \implies f(n) = 0$$

for all $n$.

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