Why is $\text{Supp}(M)$ connected in the Zariski topology? 
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.)
 A: By taking a quotient, we can assume that the annihilator of $M$ is $0$. Then the support of $M$ is the full spectrum $X$. 
Suppose $X$ is not connected. By Corollary $2$ to Proposition $15$ Section II.$4.3$ of Bourbaki's Algèbre commutative, there are nontrivial ideals $\mathfrak a$ and $\mathfrak b$ satisfying 
$$
A=\mathfrak a+\mathfrak b,\quad\mathfrak a\cap\mathfrak b=0.
$$ 
By the Chinese Remainder Theorem, this implies 
$$
A\simeq\frac{A}{\mathfrak a}\times\frac{A}{\mathfrak b}\quad.
$$
Let $M=N\oplus P$ be the corresponding decomposition. 
As $\mathfrak b$ is contained in the annihilator of $N$, we have $N\neq M$, and similarly for $P$.
A: Yes, if a finitely generated module $M$ is indecomposable, its support is connected.  
Proof
The support of $M$ is $V(Ann(M))$.
Replacing  $R$ by $R/Ann(M)$, we may assume that $Ann(M)=0$ and that $Supp(M)=Spec(R)$.
If that space $SuppM=Spec(R)$ were non-connected, it is well-known  that there would exist two non-zero idempotents $e,f\in R$ with $ef=0$ and $e+f=1$.
Then we would have a non trivial decomposition $M=eM\oplus fM$. Contradiction.
Edit
The fact I mentioned on idempotents is Exercise 2.25 in Eisenbud's Commutative Algebra (the exercise is  practically solved by its copious hints). 
Second Edit
In order not to give an exercise as sole reference, here is a proof (different from Eisenbud's) that a ring with two non-zero idempotents $e,f\in R$ such that  $ef=0$ and $e+f=1$ is a non-trivial product of rings (and thus has non-connected spectrum). 
The trick is to realize that the ideal $eR$, although not a subring of $R$, is a ring of its own, with addition and multiplication inherited from $R$, but with unit element $e$. Similarly for $fR$.  
Then the morphism $R\to eR\times fR:r\mapsto (er,fr)$ is an isomorphism of rings.
(For surjectivity notice that $ (er_1,fr_2)$ is the image of $er_1+fr_2$.
Injectivity is the trivial observation that $er=fr=0$ implies $r=(e+f)r=0$)
