Use Nakayama's Lemma to show that $I$ is principal, generated by an idempotent. [duplicate]
Solution 1:
Well, you know that by Nakayama's that there is an $x$ such that $1-x \in I, xI = 0$. For any element $y$ in $I$ we have that $$(1-x)y = y-xy = y$$ Which means that any element $z \in I$ can be written in the form $$(1-x)z\in R$$ which gives the inclusion $I \subseteq (1-x)$ and the inverse inclusion comes from the fact that $1-x \in I$. We also have idempotence because $$(1-x)(1-x) = 1-x$$ By our first claim. Therefore $I = (1-x)$ and satisfies all listed properties.