Idempotents in a local ring

Question

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 ?

Answer 1

Let $(R,\mathfrak m)$ 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 $Re = \mathfrak m (Re)$ and by Nakayama $Re = 0$ which implies $e = 0$.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

Remember that in a local ring $R$ the complement of the maximal ideal is the set of invertible elements, $R^{*}=R-m$.

The definition of maximal ideal implies the inclusion of $R^{*}\subset R-m$. If $x\in R-R^{*}$ then $(x)$ is different from $R$ and since $R$ has a unique maximal ideal $m$ we have $x\in (x) \subset m$. Therefore $R-R^{*}\subset m$. Thus, $R-m \subset R^{*}$.

Using this fact take $e$ an idempotent element, then $e^2=e$. Hence $e(e-1)=0 \in m$ implies that $e\in m$ or $e-1\in m$.

• If $e\in m$ then $e-1\in R-m=R^{*}$ and $e=0$.
• If $e-1\in m$ then $e\in R-m=R^{*}$ and $e=1$.

Answer 6

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.)