If p is a prime ideal of P then Id( R/P ) = { P , P+1 }

Question

Please can you help me how to show that :

If p is a prime ideal of R then Id( R/P ) = { P , P+1 }

Id is the set of all idempotent elements in R/P and R is commutative ring with identity

Answer 1

Assuming $R$ to be a commutative unital ring:

$P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain; this is common knowledge.

If in any integral domain $D$ we have an idempotent $u \in D$, i.e.,

$u^2 = u, \tag 2$

then

$u(u - 1) = u^2 - u = 0; \tag 3$

if $u \ne 0$, then since $D$ has no zero divisors, we must have

$u - 1 = 0, \tag 4$

or

$u = 1; \tag 5$

we thus see that the only idempotents in an integral domain such as $D$ are $0$ and $1$.

Applying the above to the integral domain $R/P$ immediately shows that the only idempotents in $R/P$ are $0 + P = P$ and $1 + P$. And that does it!