Why is a commutative ring with an infinite number of idempotent elements unstable? In a book of model theory I found the following statement:

A commutative ring with an infinite number of idempotent elements unstable.

I haven't manage to prove it yet. As stability in the model theoretic sense is somewhat difficult, I would be pleased if somebody could help answering the following, success bringing question:
In a commutative ring with an infinite set $I$ of idempotent elements one can define a partial order on $I$ by $a\leq b $ iff $a\cdot b=a$. Is it possible though to extract an infinite totally ordered chain out of $I$? 
I hope the answer to be positive, cause that would lead to the desired instability... 
thanks for reading and thinking!
 A: Let 
$$
J=\{\text{ monotone chains of projections }\},
$$
ordered by inclusion. If we take a chain of chains, then it has an upper bound (namely, the union); so Zorn's Lemma applies and there is a maximal chain $\{a_\beta\}_{\beta\in B}$. 
To finish, we need to argue that this maximal chain has to be infinite. Assume on the contrary that $B$ is finite. So our maximal chain is of the form $a_1\geq\cdots \geq a_n$. If $a_1$ is not a maximal projection, then there exists a projection $a_{0}$ with $a_{0}\geq a_1$, $a_{0}\ne a_1$, and this contradicts the maximality of the chain. With a similar reasoning we show that $a_n$ has to be a minimal projection. 
Now let 
$$
p_1=a_1-a_2, \ p_2=a_2-a_3,\ldots,\ p_{n-1}=a_{n-1}-a_n, \ \ p_n=a_n.
$$
Each of these is an idempotent, and $p_kp_j=0$ if $k\ne j$. So we can see that each chain of idempotents is in correspondence with a "partition of the unity" like that $\{p_j\}$.
The maximality of the chain $\{a_k\}_{k=1}^n$ is equivalent to the fact that all $p_k$  are minimal: if $b\leq p_k$ for an idempotent $b$, then
$
p_k=b+(p_k-b),
$ a sum of two idempotents with product zero, and so 
$$
p_1,\ldots,p_{k-1},b,p_k-b,p_{k+1},\ldots,p_n
$$
would be a longer chain of minimal idempotents, a contradiction. 
Note also that $\sum_{k=1}^np_k=a_1$, and $a_1$ is a maximal idempotent. But it turns out that a maximal idempotent has to be maximum: because if $c,d$ are idempotents, then so is $e=c+d-cd$, and we have $c\leq e$, $d\leq e$, and then if both $c$ and $d$ are maximal, $c=d=e$. Then $a_1$ is the maximum, and $ba_1=b$ for all idempotents $b$.  So
$$
b=\sum_{k=1}^nbp_k,
$$
and the minimality of each $p_k$ tells us that $bp_k$ is either $0$ or $p_k$. This implies that there are only finitely many idempotents in the ring, a contradiction. 
Thus our maximal chain of idempotents is infinite. 
A: I don't know if it is of some interes but for those who don't believe in the axiom of choise, there is a more direct way to proof the claim above:
For two arbitrary idempotents $a$ and $b$ let $inf(a,b):=ab$ and $sup(a,b):=a+b-ab$. Knowing that, it is quite simple to show, that every finite chain can be extended by another idempotent.
Let $a_n>\cdots >a_0$ be any chain and $c\notin D:=\{a_0,\cdots, a_n\}\cup\bigcup_{i=1\cdots n-1}\{a_k-a_i+a_{i-1}\mid n\geq k>i\}$ another idempotent (which exists since we have infinitly many of them). 
Then you can increase the length of the chain in the following way: 
case 1: $ca_n=c$, then we are done by $a_0>\cdots>a_n>c$.
case 2: $ca_n\notin \{c,a_n\}$. Then $ca_n<a_n$ and we're done.
case 3: $ca_n=a_n$. Then let $k:=max_{0\leq i\leq n}\{ca_i=a_i\}$.
case 3.1: If $k=n$ we're done. 
case 3.2.1: If $k<n$ and $a_{k+1}b\neq a_k$, we are also done by $a_n>\cdots>a_{k+1}>ba_{k+1}>a_k>\cdots >a_0$ 
case 3.2.2: So assume that $a_{k+1}b=a_k$. Then $sup(a_{k+1},b)=:b'$ still is in $D$ and $b'>a_{k+1}$. Now we can start all over and by finiteness of the chain we will receive another idempotent enlarging the size of the chain. 
Some remarks/consequences: 
1.) $sup$ and $inf$ in boolean rings are exactly conjunction and disjunction.
2.) Let $(X,\tau)$ be a topological space where $\tau$ consists of clopen sets. Then if  $X$ is compact so $|\tau |<\aleph_0$.
So far, s.
