# Tag Info

The issue here is that you need to know what you mean by "unique". You can show that such an object $L$ exists ; just take it to be the submodule of $M$ of elements mapped to $0$ under $f$, and check the two conditions. For "unicity", if you have two modules $L_1,L_2$ that satisfy property (i) and (ii), you cannot hope to show that $L_1 = L_2$, but you can ...
$am=0\Rightarrow an_1+an_2=0\Rightarrow an_1=an_2=0$ (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 ...