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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?

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I must not be understanding what you mean. Isn't every element $e \in E(B)$ already an integral element, since it satisfies $e^2 - e = 0$? So that $E(B)$ is automatically integral over $E(A)$ even if $B$ isn't integral over $A$? –  Todd Trimble Mar 23 '11 at 9:38
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Besides, the sum of two idempotents does not have to be idempotent. Thus $E(A)$ is not a ring. –  Martin Brandenburg Mar 23 '11 at 10:17
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@Todd: You are absolutely right.. the sums just confused me. Consider this question answered @Martin: In the ring of idempotents the sum is defined by $e +' f = e(1-f) + f(1-e)$ where $+$ and $-$ are operation in the ring itself. Multiplication remains the same as the ring itself. So here each element is the additive inverse of itself. –  Jose Capco Mar 23 '11 at 10:28
    
Ah, thanks ! –  Martin Brandenburg Mar 23 '11 at 17:30
    
Jose, you could probably add the definition of $E(A)$ to the body of the question (as opposed to leaving it hidden in the comments) because it is not that standard, as far as I know. –  Mariano Suárez-Alvarez Mar 24 '11 at 7:42
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up vote 1 down vote accepted

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.)

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Nice, you pwned it! –  darij grinberg Mar 24 '11 at 11:38
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Take any $B$ such that $E(B)$ is infinite (e.g. $k^\mathbb{N}$, $k$ a field) and write it as a quotient of a domain $A$ (e.g. a polynomial ring over $k$). Then $E(A)=\mathbb{Z}/2\mathbb{Z}$, so $E(B)$ is not finite over $E(A)$.

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I think the OP is talking about "integral extensions" in the classical sense, where an extension really extends something (meaning, it is injective). –  darij grinberg Mar 24 '11 at 10:19
    
@darij grinberg: good point, but in my example $A$ injects into $A\times B$. –  Laurent Moret-Bailly Mar 24 '11 at 13:09
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