Let $A \not= \{0 \}$ be a commutative ring and let $M$ be an $A$-module. Define $$\text{Supp} (M) = \{ P \in \operatorname{Spec} A : M_P \not= 0 \}$$
My first question is if $0$ is the element $0/1$?
Also can I have a example of a ring $A$ and a module $M$ such that $M_p=0$ for some prime ideal and where $M_p \not= 0$ for some prime ideal? I am having troubles to get a picture in my head of $\text{Supp}(M)$.
$0$ means the trivial module (whose underlying additive group is the trivial group).
As an example pick $A = \Bbb Z$, $M = \Bbb Z/2\Bbb Z$; then $M_{(2)} = M$ but $M_{\mathfrak p}$ is trivial for all other primes $\mathfrak p$.
To see this last statement, recall the definition of $S^{-1}M$; it's equivalence classes of fractions $a/b$ ($a \in M, b \in S$), where $a/b$ is equivalent to $c/d$ if there's an $s \in S$ with $s(ad-bc)=0$. We want to show that $a/b=0/1$ for all $a,b$. Now $2 \in \Bbb Z \setminus (p)$ for all $p \neq 2$; and $2(a)=0\in M$, as desired.