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My book says "it is easy to see that $A$ (a commutative ring) has an idempotent $e\neq 0,1$ if and only if it is a direct sum of rings $A=A_1\oplus A_2$ with $A_1=Ae$ and $A_2=A(1-e)$."

I know the $\implies$ implication, but if $A=Ae\oplus A(1-e)$, why is $e$ a nontrivial idempotent? I tried writing $e=ae+b(1-e)$ for $a,b\in A$ and squaring, but nothing cancels correctly. I also tried computing $e(1-e)$ with $1-e=1-ae-b(1-e)$, but again nothing cancels nicely.

Cye
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  • Dear Cye: I agree with you. Which book is it? (Welcome to MSE, and $+1$ for your nice question!) – Pierre-Yves Gaillard Mar 05 '12 at 09:28
  • @Pierre-YvesGaillard Thanks! It's the first few sentences of section 1.11 on page 28 of Miles Reid's Undergraduate Commutative Algebra. – Cye Mar 05 '12 at 09:32
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    Dear Cye: You're right. Here is a preview. I think it's just a typo. The correct statement is (I think) "iff it is a nontrivial direct sum of two ideals". ($A_1$ and $A_2$ are not subrings, but ideals. These ideals have a ring structure, but the inclusion is not a ring morphism.) (But it is the direct product of the two rings.) – Pierre-Yves Gaillard Mar 05 '12 at 10:01
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    @Pierre: If $e$ is an idempotent, then $Ae$ is a ring, and a subset. Whether it is a subring depends on whether you require rings to have unity (but subring isn't mentioned in the question). Of course, I really don't like the usage of "direct sum" in this fashion since it ought to imply a coproduct of sorts, but unfortunately this seems to be a common conventional usage too. :( –  Mar 05 '12 at 10:08

1 Answers1

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This is OK, I think. If $A = Ae \oplus A(1-e)$, then consider $e(1-e) = (1- e)e.$ This is in $Ae \cap A(1-e) = \{ 0\},$ so we must have $e(1-e) =0$ and hence $e = e^2.$ I think you should require both $Ae$ and $A(1-e)$ to be non-zero, though, strictly speaking to exclude $e \in \{0,1 \}.$

Geoff Robinson
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