I am working with a presheaf of rings and I'm having problems to show that it is in fact a sheaf. Specifically: Let $A$ be a commutative ring with identity. Let $E(A)$ be it's ring of idempotents with sum given by $a \oplus b= a+b+ab$ ; $a,b\in E(A)$ and the product is the same as in the ring $A$. We consider $Spec E(A)$ and we see that the basic open sets for the Zariski topology are in this case clopens. They have the following form:
$$ \Omega_e=\{I\in SpecE(A):e\in I\} $$
For $e\in E(A)$. And then we associate with every basic open set $\Omega_e$, the ring $A/(e)$. Well, all of this is just motivation for the question.
My problem is the following: Suppose that $\Omega_e=\bigcup_{i\in I}\Omega_{e_i} $ and that $\{r_i+(e_i)\}_{i\in I}$ is a collection of element such that $r_i+(e_i)\in A/(e_i)$ for all $i\in I$. Suppose further that we have for all $i,j\in I$
$$r_i-r_j\in(e_i+e_j+e_ie_j)$$
Then, my question is, ¿how I paste this elements in $A/(e)$?; formally, I can't show the existence of an element $r\in A$ such that $r+(e_i)=r_i+(e_i)$ for all $i\in I$.
This is what I tried: We have equations
$$r_i-r_j=(e_i+e_j+e_ie_j)s_{ij}$$
For all $i,j\in I$. Now, let $k,g,p\in I$, then we have:
$$r_p=(e_p+e_k+e_pe_k)s_{pk}-r_k$$ $$r_p=(e_p+e_g+e_pe_g)s_{pg}-r_g$$
So, $0=e_p(s_{pk}-s_{pg})+e_ks_{pk}+e_pe_ks_{pk}+r_k-e_gs_{pg}-e_pe_gS_{pg}-r_g$, thus
$$e_ks_{pk}+r_k-e_gs_{pg}-r_g=e_p(s_{pg}-s_{pk}-e_ks_{pk}+e_{g}s_{pg})\in(e_p)$$
This says that $e_ks_{pk}+r_k+(e_p)=e_gs_{pg}+r_g+(e_p)$ for all $h,g,p\in I$. Finally, if we take $p=k$ and $s_{kk}=0$ we obtain that $r_k+(e_k)=e_gs_{kg}+r_g+(e_k)$ for all $k,g\in I$. If I can show that there is a $g$ such that $s_{kg}=c$ for all $k$, we are done, because the element will be $e_gc+r_g$. But I don't see how to figure it out.
I don't know if there is a more easy way to do this, this is just what I attempted so far. I will be very grateful if someone can help me with this, I am really stuck with this. Thanks in advance.