Vanishing set of $\text{Ann} (M)$, where $M$ is a finitely generated $A$ module Let $M$ be a finitely generated $A$ module, generated by say $x_1, ..., x_n$.
Let $V(S)$ denote the set of primes of $A$ containing $S$.
I am guessing that 
$$
V(\text{Ann}(M)) = \cup_{1 \leq i \leq n} V(\text{Ann}(x_i)),
$$
where $\text{Ann}$ denotes annihilator, holds but I am having trouble proving it.
I would appreciate any hint! Thanks!
 A: Suppose $M$ if a finitely generated $A$-module: $M = \langle x_1, \ldots, x_n \rangle$.

Claim: $\displaystyle \text{ann}(M) = \bigcap_{k = 1}^n \text{ann}(x_k)$.
Proof: Let $r \in \text{ann}(M)$, then since $x_k \in M$ for $k = 1, 2, \ldots, n$, then $rx_k = 0$ which means $r \in \text{ann}(x_k)$. Conclude that $\displaystyle r \in \bigcap_{k = 1}^n \text{ann}(x_k)$. For the reverse containment, let $\displaystyle r \in \bigcap_{k = 1}^n \text{ann}(x_k)$, then $rx_k = 0$ for $k = 1, 2, \ldots, n$. Let $m \in M$ be arbitrary, since $M$ is finitely generated, we can write it as $m = a_1x_1 + \cdots + a_nx_n$ where $a_k \in A$ for $k = 1, 2, \ldots, n$. Observe that $$rm = r(a_1x_1 + \cdots + a_nx_n) = a_1(rx_1) + \cdots + a_n(rx_n) = a_1 \cdot 0 + \cdots + a_n \cdot 0 = 0$$ and conclude that $r \in \text{ann}(M)$.

Claim: $\displaystyle V\bigg( \bigcap_{k = 1}^n \mathfrak a_k \bigg) = \bigcup_{k = 1}^n V(\mathfrak a_k)$.
Proof: Let $\displaystyle \mathfrak p \in V\bigg( \bigcap_{k = 1}^n \mathfrak a_k \bigg)$, then $\displaystyle \bigcap_{k = 1}^n \mathfrak a_k \subseteq \mathfrak p$. since $\mathfrak p$ is prime, it can be shown (with a bit of work) that $\mathfrak a_k \subseteq \mathfrak p$ for some $k = 1, 2, \ldots, n$, in which case we have $\mathfrak p \in V(\mathfrak a_k)$. Hence we have that $\displaystyle \mathfrak p \in \bigcup_{k = 1}^n V(\mathfrak a_k)$.
For the reverse containment, let $\displaystyle \mathfrak p \in \bigcup_{k = 1}^n V(\mathfrak a_k)$, then $\mathfrak p \in V(\mathfrak a_k)$ for some $k = 1, 2, \ldots, n$. Suppose, without loss of generality, that $\mathfrak p \in V(\mathfrak a_1)$, then $\mathfrak a_1 \subseteq \mathfrak p$. But notice that $\displaystyle \bigcap_{k = 1}^n \mathfrak a_k \subseteq \mathfrak a_1 \subseteq \mathfrak p$ and hence $\displaystyle \mathfrak p \in V\bigg( \bigcap_{k = 1}^n \mathfrak a_k \bigg)$.

Finally, we have $$\displaystyle V \big(\text{ann}(M)\big) = V\bigg(\bigcap_{k = 1}^n \text{ann}(x_k)\bigg) = \bigcup_{k = 1}^n V \big(\text{ann}(x_k) \big).$$
