Show that if $R$ is commutative, then $\mathrm{Ann}(n_1) + \mathrm{Ann}(n_2) = R$.

Question

Let $R$ be a commutative ring with $1$, and let $M$ be a cyclic $R$-module with generator $m$ (so that $M=Rm$). Suppose that $M = N_1 \bigoplus N_2$ for some submodules $N_1$ and $N_2$ of $M$. Let $n_1 \in N_1$ and $n_2 \in N_2$ such that $m = n_1 + n_2$. Show that if $R$ is commutative, then $\mathrm{Ann}(n_1) + \mathrm{Ann}(n_2) = R$.

Answer 1

$am=0\Rightarrow an_1+an_2=0\Rightarrow an_1=an_2=0$ (why?). So $\mathrm{Ann}(m)=\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)$. Moreover, $N_i=Rn_i$ for $i=1,2$. We have $R/\mathrm{Ann}(m)\simeq Rm$, $Rm=N_1\dotplus N_2$, and $N_1\dotplus N_2\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2)$, therefore $$R/\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2).$$

This shows that your question is in fact a converse to CRT, and for a proof see this answer.