I saw the following exercise in the Kaplansky's book that is due to D. Lizard. Where can i find the main text for the proof of this exercise?
Let $P$ be a prime ideal of $R$, $I$ the ideal generated by all the idempotent in $P$. Prove that $R/I$ has no non trivial idempotent.(page 8 of kaplansky'book, exercise 6, commutative ring theory)
I think this is a proof if $R$ has $1$. Suppose $e + I$ is an idempotent in $R/I$. This means $e^2 \equiv e \pmod{I}$, so $e(1-e) \in I \subseteq P$. Since $P$ is prime, this implies that either $e \in P$ or $1 - e \in P$. Let's say $e \in P$. Since $e(1-e) \in I$, we may write $e(1-e) = \sum_{i=1}^n r_i e_i$, where each $e_i$ is an idempotent that is in $P$. Now, let $f = e \cdot \prod_{i=1}^n (1 - e_i)$. Then, since each $1 - e_i$ is an idempotent, we have $$f^2 = e^2 \cdot \prod_{i=1}^n (1 - e_i) = \left( e - \sum_{i=1}^n r_i e_i\right)\cdot \prod_{i=1}^n (1 - e_i) = e\cdot \prod_{i=1}^n (1 - e_i)= f, $$ so $f$ is an idempotent (and $f \in P$, because $e \in P$), so $f \in I$. Since $f \equiv e \pmod{I}$, this implies $e \in I$. So $e + I$ is zero in $R/I$.
