Let $V$ be an A-module over a commutative ring $A$.
(a) Let $x,y \in V$. Then $x=y$ $\Leftrightarrow \frac {x}{1}=\frac{y}{1}$ in $V_M$ for all $M\in Spm A$.
If I have $x=y$ then for each $m\in M$ I will have $m(x-y)=0$. So, I am done.
Conversly, if $\frac {x}{1}=\frac{y}{1} $ in $V_M$ then I have $m\in M $ such that $m (x-y)=0 $ for all $M\in Spm A$. But I am not sure how exactly I should proceed with the proof.
(b) Prove that following statements are equivalent.
(i) $V=0$
(ii) $V_P = 0 $ for all $P\in Spec A$
(iii) $V_M =0$ for all $M\in Spec A$.
(iii) $\Rightarrow$(ii) holds as every maximal ideal is a prime ideal.
(i) $\Rightarrow$(iii) is clear.
(ii) $\Rightarrow$ (i) I am not getting any intuition on how exactly should I approach this part and would appreciate hints.
Kindly help.
Hints. For (a), you can always assume $y=0$. So you have to prove that $x=0$ if and only if $x/1=0/1$ for all maximal ideals $M$.
Assume that $x\neq 0$. Show that $Ann(x)$ is an ideal different from $A$. Hence it is contained in some maximal ideal $M_0$. Then prove that $x/1\neq 0/1$ in $V_{M_0}$ by contradiction.
For $(b)$, just apply $(a)$ with $y=0$.