Question

Let R be a commutative ring. If I and P are idelas of R with P prime such that \(\displaystyle{I}!\subseteq{P}\), prove that the ideal \(\displaystyle{P}:{I}={P}\)

Answer 1

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.