If $J$ is the ideal generated by all idempotents in a prime ideal, then $R/J$ has only trivial idempotents

Question

Let $R$ be a commutative ring with identity, $P$ be a prime ideal in $R$ and define $$X := \lbrace t \in P \mid t^2=t \rbrace. $$ Also let $J$ denote the smallest ideal of $R$ that contains $X$.

Prove that if $$a^2+J=a+J$$ in $R/J$, then $$a+J=J$$ or $$a+J=1+J$$

I am trying to solve this but I am new to algebra (aprox. 3 months), so I generally don't know how to start such questions. So any help is appreciated.

Answer 1

First note that $J$ is the ideal generated by all idempotents in $P$, so $J\subset P$.
Since $a^2-a\in J$ we have $a^2-a=a_1t_1+\cdots+a_nt_n$ with $t_i\in P$, $t_i^2=t_i$. Then $(1-t_1)\cdots(1-t_n)(a^2-a)=0$, so $(1-t)(a^2-a)=0$ for some $t\in P$, $t^2=t$. (In particular, $t\in J$.)

Since $a^2-a\in P$ we have two possibilities:
$(i)$ $a\in P$; from $(1-t)(a^2-a)=0$ we get $a-ta=(a-ta)^2$, so $a-ta\in J$, and thus $a\in J$;
$(ii)$ $1-a\in P$; similarly $(1-a)-t(1-a)$ is idempotent, so $1-a\in J$.