Proposition

A local ring contains no idempotent $\ne 0, 1$.

Solution

Let $A$ be a local ring that contains no idempotent $\ne 0, 1$.

Let $x$ be an idempotent $\ne 0, 1$. If no such element exists, we are done.

Suppose it is a unit and let $y$ be the inverse of $x$. Then $x^2 = x$ implies $y(x^2) = yx$, so $x = 1$. This is a contradiction, so $x$ is not a unit.

Let $m$ be the maximal ideal of $A$. Since $m$ is the only maximal ideal, the Jacobson radical $\mathfrak{R} = m$.

By Corollary 1.5[Atiyah], $x$ is contained in a maximal ideal since $x$ is not a unit. Since there is only one maximal ideal, $x \in m$. Therefore, $x \in \mathfrak{R}$.

By Proposition 1.9, $1 - xy$ must be a unit in $A$ for all $y \in A$. Let $y = x$. Then $(1 - x^2)^2 = (1 - x)^2 = 1 - 2x + x^2 = 1 - x^2$. Therefore, $1 - x^2 = 1 - x$ is an idempotent. Since $x \ne 0, 1$, $1 - x \ne 0, 1$. Using the same argument as above, we conclude that $1 - x$ is not a unit. Therefore, this is a contradiction, so there cannot be an idempotent $\ne 0, 1$.