Let R be a commutative ring. If I and P are idelas of R with P prime such that I !sube P, prove that the ideal P:I=P

The ideal quotient of ${\displaystyle P\text{and}IisP:I=\{x\in R:xI\subset P\}}$ which is again an ideal of R.
Given that P is a prime ideal.
To prove that ${\displaystyle P:I=P}$
We prove this set equivalence by proving that one set is the subset of other.
For, let ${\displaystyle x\in P}$
Then ${\displaystyle xI\subset P}$ since P is an ideal.
Therefore, ${\displaystyle P\subset P:I}$
Conversely, let ${\displaystyle x\in P:I}$
Then ${\displaystyle xI\supset P}$. That is ${\displaystyle xy\in P}$ , for every ${\displaystyle y\in I.}$
Since ${\displaystyle I!\supset P}$, there exists $y\in I$ such that ${\displaystyle y\notin I.}$ But ${\displaystyle xy\in P.}$
Also since P is a prime ideal, x must be ${\displaystyle \in P.}$
Hence ${\displaystyle x\in P}$ which implies that ${\displaystyle P\supset P:I}$
Hence proved.