The ideal quotient of \(\displaystyle{P}{\quad\text{and}\quad}{I}{i}{s}{P}:{I}={\left\lbrace{x}\in{R}:{x}{I}\subset{P}\right\rbrace}\) 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 \(\displaystyle{x}{I}\subset{P}\) since P is an ideal.

Therefore, \(\displaystyle{P}\subset{P}:{I}\)

Conversely, let \(\displaystyle{x}\in{P}:{I}\)

Then \(\displaystyle{x}{I}\supset{P}\). That is \(\displaystyle{x}{y}\in{P}\) , for every \(\displaystyle{y}\in{I}.\)

Since \(\displaystyle{I}!\supset{P}\), there exists y in IZSK such that \(\displaystyle{y}\notin{I}.\) But \(\displaystyle{x}{y}\in{P}.\)

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.

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 \(\displaystyle{x}{I}\subset{P}\) since P is an ideal.

Therefore, \(\displaystyle{P}\subset{P}:{I}\)

Conversely, let \(\displaystyle{x}\in{P}:{I}\)

Then \(\displaystyle{x}{I}\supset{P}\). That is \(\displaystyle{x}{y}\in{P}\) , for every \(\displaystyle{y}\in{I}.\)

Since \(\displaystyle{I}!\supset{P}\), there exists y in IZSK such that \(\displaystyle{y}\notin{I}.\) But \(\displaystyle{x}{y}\in{P}.\)

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.