Suppose $M$ is a indecomposable module, so that it cannot be written as $M_1\oplus M_2$ for $M_1\neq M$ and $M_2\neq M$, which is finitely generated over a commutative ring $R$. Why is $\text{supp}(M)$ connected in the Zariski topology?
Now I know $\text{supp}(M)=\{\mathfrak{p}\in\text{Spec}(R): M_\mathfrak{p}\neq 0\}$. Since $M$ is finitely generated over $R$, I was able to derive that $\text{supp}(M)=Z(\text{ann}(M))$, with $Z(\mathfrak{a})=\{\mathfrak{p}\in\text{Spec}(R):\mathfrak{p}\supset\mathfrak{a}\}$. For suppose $\mathfrak{p}\in\text{Spec}(R)$ such that $M_\mathfrak{p}\neq 0$. If $aM=0$, $aM_\mathfrak{p}=0$, so $a\notin\mathfrak{p}$ then $M_\mathfrak{p}=0$, and thus $\text{ann}(M)\subset\mathfrak{p}$.
In the other direction, take $\text{ann}(M)\subset\mathfrak{p}$, and suppose $m_1,\dots,m_r$ are generators for $M$ over $R$. If $M_\mathfrak{p}=0$, then there is some $a_i\notin\mathfrak{p}$ such that $a_im_i=0\in M$ for all $i$. Then setting $a=a_1\cdots a_r$, $aM=0$, and $a\in\mathfrak{p}$, contradiction.
So I suppose $\text{supp}(M)=Z(\text{ann}(M))=A\cup B$ where $A$ and $B$ are closed sets in the Zariski topology. To see $\text{supp}(M)$ is connected, I'm trying to conclude one of $A$ or $B$ is empty. Since $M$ is indecomposable, I know it can't be written as $M=M_1\oplus M_2$ for $M_1\neq M$ and $M_2\neq M$. I'm also aware that $\text{supp}(M_1\oplus M_2)=\text{supp}(M_1)\cup\text{supp}(M_2)$.
I thought maybe if it isn't connected, I could possibly show that $M$ is actually decomposable. Can someone please explain the correct idea here? Thanks.
(If possible, I'd appreciate an explanation that uses mostly commutative algebra, if one exists, since I'm unfamiliar with ideas from algebraic geometry.)