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For a commutative ring $R$, and ideal $I$, I am trying to prove that $Rad(I)= \{ r \in R : r^n \in I$, for some $n\in \mathbb{N} \} $ from the definition which was presented to me of the radical as the intersection of prime ideals which contain the ideal $I$. I am following the proof from the first page of https://math.mit.edu/~fgotti/docs/Courses/Ideal%20Theory/3.%20Radical%20and%20Primary%20Ideals/Radical%20and%20Primary%20Ideals.pdf. I am following everything apart from the statement "M is a multiplicative subset of R that is disjoint from I. Therefore I is contained in a prime ideal P that is disjoint from M" where $M := \{r^n+a: n\in \mathbb{N}, a \in I\}.$ The first sentence makes sense, but why must this imply there exists a prime ideal containing $I$ disjoint from $M$?

Thanks

letsgetthismsc
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  • This is one of the first things proven in any text introducing prime ideals for commutative rings. In fact, it is Theorem 1, in Chapter 1, on page 1 of Kaplansky's book "Commutative rings." – rschwieb Aug 19 '24 at 17:29

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