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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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4 Answers 4

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.

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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

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