Prove that P is prime ideal of R?

Question

$R$ is a commutative ring and $1\in R$. Let $I$ be an ideal of $R$ and $P$ be a prime ideal of $I$. Then show that $P$ is prime ideal of $R$.

I know how to prove that P is ideal of R. Suppose that $r\in R$ and $p \in R$ and $i\in R\backslash I$ so $i.(rp) = (ir).p$ and because I is ideal of R so $ir \in I$ and because P is prime ideal of I so $(ir).p \in P$ then ($P$ is prime ideal of $I$) $i\in P$ or $rp\in P$ because of choosing $i$, $rp\in P$.

How can I prove that $P$ is prime of $R$? Can anyone help me?

Answer 1

Take $6\Bbb Z\subset 3\Bbb Z\subset\Bbb Z$. Observe that

$$3n\cdot3m\in6\Bbb Z\iff n\;\text{or}\;m\;\text{are even}\;\iff 3n\in6\Bbb Z\;\text{or}\;3m\in6\Bbb Z\implies6\Bbb Z\text{ is prime in }3\Bbb Z.$$ Yet $6\Bbb Z$ is not prime in $\Bbb Z$, so the claim is false.