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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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5 Answers 5

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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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

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 $e$ 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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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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