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

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

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