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

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?

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 $$f\equiv 0$$.
If $$aN = 0$$ and $$am = 0$$ implies $$m = 0$$ then
$$0 = f(an) = af(n) \implies f(n) = 0$$
for all $$n$$.