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Is it true that a local ring, i.e., a commutative ring with a unique maximal ideal, doesn't contain idempotent elements $\neq 0, 1$ ? Why ?

Any hint ?

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If $e$ is an idempotent which is not 0,1, then $e(e-1)=e^2-e=0$ shows that both $e$ and $e-1$ are zero divisors and in particular not invertible. Hence they must be in the maximal ideal, but then $1=e-(e-1)$ is also in the maximal ideal - contradiction.

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Let $(R,m)$ be local ring. Suppose that $e \in R$ is such that $e^2 = e$. If $e$ is a unit, then $e = 1$. If $e$ is not a unit, then $e \in m$ and by idempotency $(Re) = e (Re)$. Hence $Re = m (Re)$ and by Nakayama $Re = 0$ which implies $e = 0$.

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Let $e$ be a nontrivial idempotent. Then $eR\oplus (1-e)R=R$ is a nontrivial splitting of $R$ into two proper ideals. But both $eR$ and $(1-e)R$ are contained in the maximal ideal: how could they add up to $R$? This shows no such $e$ can exist.

This works even for noncommutative local rings.

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I very much like this. However, is it even correct to state the equality in terms of a direct sum? I don't see an immediate reason to assume uniqueness of summands. Either way, a direct sum isn't needed - your argument would still hold. – polynomial_donut Feb 15 at 23:02
@polynomial_donut It is indeed a direct sum. What cause do you have to doubt it? I do not understand your comment about uniqueness of summands since there is nothing like that in the definition, nor does it occur often. – rschwieb Feb 16 at 0:10
Well, I don't know which definition you are talking about, but here… and in probably any book on algebra, such as for example Atiyah/McDonald's Commutative Algebra to name a standard one, you will discover that direct sums are understood as either unique sum representations or as isomorphic to cartesian products of rings/modules, from which that unique sum representation would follow. There are also generic sums of submodules/rings, where uniqueness is not required and one would use ordinary plus symbols... – polynomial_donut Feb 17 at 5:50
@polynomial_donut Don't worry, I'u using the standard definition. It looks like you just misexpressed yourself earlier by talking about "uniqueness of summands". That is totally different from uniqueness of sum representations. It's elementary to see that $R=eR+(1-e)R$ and $eR\cap (1-e)R=\{0\}$, and that is the notion of direct sum I am working with which appears in all your books and links. – rschwieb Feb 17 at 10:40
@polynomial_donut If $ex=(1-e)y$, multiply on the left by $e$ to see $ex=0$. So the intersection is trivial. If you are talking about a direct sum decomposition $A\oplus B$, the summands are $A$ and $B$, not sums of pairs of elements. – rschwieb Feb 19 at 9:10

For every idempotent $e\notin \{0,1\} $ , $e\in Jac(R) $ so $1-e$ is invertible. Then from $e(1-e)=0$ we conclude that $e=0$, a contradiction.

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Welcome to math.stackexchange. You may want to give some idea of what the Jacobian is, and why being in the Jacobian implies that $1-e$ is invertible. – Joe Tait Jul 3 '14 at 12:53
Jacobson radical of$R$ denoted by $Jac(R)$. there is atheorem in commutative and noncommutative ring theory. – sajad Jul 5 '14 at 21:48

Recall that there is also an elementary characterization of local rings, not needing the concept of (maximal) ideals: A ring $A$ is a local ring if and only if $1 \neq 0$ in $A$ and, for any $x, y \in A$ such that $x+y$ is invertible in $A$, $x$ or $y$ is invertible in $A$ (inclusive or). (If you don't fear the empty set, you can phrase this more succinctly as: A ring is a local ring if and only if, for any finite sum which happens to be invertible, at least one summand is invertible.

With this characterization, the proof of your statement is easy: Assume $e^2 = e$. Then $e (1-e) = 0$. Since $e + (1-e)$ is invertible, $e$ is invertible or $1-e$ is invertible. In the first case, it follows that $1-e = 0$. In the second case, we have $e = 0$.

(The elementary characterization is crucial in constructive mathematics, where maximal ideals don't behave as well as in classical mathematics.)

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