Each prime ideal contains an idempotent element

Question

An element of ring $e$ is called idempotent iff $e^2=e$.

Let $R$ be a commutative ring that contains the identity element and a non-trivial idempotent element.

I want to show that each of its prime ideals contain also an idempotent element.

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We have that $P$ is a prime ideal of $R$ iff $\forall a,b\in R$ : $a\cdot b\in P\Rightarrow a\in P \text{ or } b\in P$.

We have that $1\in R$ and $e^2=e\in R$.

So, $1\cdot e^2\in P \Rightarrow e^2\in P$, since $1\notin P$.

Since $e^2=e$, we also have that $e\in P$.

Does this imply that $e^2=e\in P$ ?

Answer 1

You can't assume $e^{2}$ $\in$ $P$. However $0$ $\in P$ and since $e^{2}-e=0$ $\in$ $P$, we have either $e$ $\in$ $P$ or $e-1$ $\in$ $P$. In either case we are done (why?).

Answer 2

Hint: $e^2=e$ implies $e(e-1)=0$.

Answer 3

No. In $1\cdot e²\in P$ you assumed that $e²=e\in P$, but this is the thesis.

Since $e²=e$, then $e²-e=e(e-1)=0$. Therefore $\overline{e(e-1)}=\overline{0}$ in $R/P$. If $P$ is prime, $R/P$ is a domain; so $\overline{e}=\overline{0}$ or $\overline{e-1}=\overline{0}$. If $\overline{e-1}=\overline{0}$, we have $e-1\in P$ and $$(1-e)²=(e-1)²=e²-2e+1=e-2e+1=1-e\in P.$$ If $e\in P$, done. We prove that every prime ideal in $R$ contain an idempotent element.