For any commutative ring $A$, the set of idempotents of $A$ will be denoted as $E(A)$. This set has a (canonical) ring structure. With addition defined by:
$$e+'f=e(1−f)+f(1−e)$$
where $+$ and $−$ are operation in the ring itself. The multiplication operation is the same as the ring itself.


Suppose now I have a commutative unital ring $A$ and let $B$ be another commutative ring that is an integral extension of $A$. Then clearly $E(B)$ is an over-ring of $E(A)$, but is it clear that $E(B)$ is an integral extension of $E(A)$. Are there easy counter examples for this? Would it help if I assumed that $B$ is a finite integral extension of $A$?


**Edit:** The question above had a trivial answer as Todd pointed out. Now Im curious what happens if I really want $E(B)$ to be a finite integral extension of $E(A)$. Does $B$ being a finite integral extension of $A$ gauarantee that? 



ring of idempotents of the integral extension of a ring
Every element $e$ of $E(B)$ is already integral since it satisfies $e^2 - e = 0$. (This was effectively in a comment above, but perhaps it's better placed as an answer.) 


