# 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

Question
Commutative Algebra
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 $$\displaystyle{P}:{I}={P}$$

2020-11-28
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.

### Relevant Questions

Suppose that R and S are commutative rings with unites, Let PSJphiZSK be a ring homomorphism from R onto S and let A be an ideal of S.
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b. If A is maximal in S, show that $$\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}$$ is maximal $$\displaystyle\in{R}$$.
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$$\displaystyle{P}{\left({A}\right)}={\left\lbrace{x}{\mid}{x}\subseteq{A}\right\rbrace}$$
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(where $$\displaystyle{y}^{{c}}$$ is the complement of y) the following statement istrue:
If A and B are ideals of a commutative ring R with unity and A+B=R show that $$\displaystyle{A}\cap{B}={A}{B}$$
Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in $$\displaystyle{R}\oplus{S}$$ if and only if a or b is a zero-divisor or exactly one of a or b is 0.
Let R be a commutative ring with unity and a in R. Then $$\displaystyle{\left\langle{a}\right\rangle}={\left\lbrace{r}{a}:{r}\in{R}\right\rbrace}={R}{a}={a}{R}$$