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$R$ commutative ring with unity. $I$ $R$-ideal. Then $\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$. That is, the radical of $I$ is the intersection of all prime ideals containing $I$.

There is a proof of this in my textbook, but I do not understand a certain piece of it. Here is the entire proof:

Clearly, $\operatorname{rad}(I) \subset \bigcap P$. Conversely, if $f\notin \operatorname{rad}(I)$, then any ideal maximal among those containing $I$ and disjoint from $\{f^n\mid n\geq1\}$ is prime, so $f\notin \bigcap P$.

I understand the first containment, but I do not know how to show that any ideal maximal among those containing $I$ and disjoint from $\{f^n\mid n\geq1\}$ is prime.

Could someone please give me a push in the right direction?

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  • $\begingroup$ I stated it the way it is stated in my textbook. $\endgroup$
    – Sam
    Sep 29, 2014 at 14:00

2 Answers 2

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In general, if $S$ is a subset of $R$, closed under multiplication, then an ideal which is maximal among those disjoint from $S$ is a prime ideal.

It is easiest to show this if you know about localizations, because this amounts to saying that a maximal ideal of $S^{-1} R$ pulls back to a prime ideal of $R$. But we can also do the heavy lifting ourselves:

Suppose that $I$ is maximal among ideals disjoint from $S$, and take two elements $x,y\notin I$. Then the ideal $(I,x)$, which is strictly larger than $I$, must contain an element of $S$. Similarly, $(I,y)$ also contains an element of $S$. Use this to show that $(I,xy)$ contains an element of $S$, so $xy\notin I$.

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  • $\begingroup$ Thank you! I appreciate it! $\endgroup$
    – Sam
    Sep 30, 2014 at 3:00
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The following theorem, or some equivalent, probably appears in your book.

In a ring $R$, given an ideal $I$ and a multiplicative subset $S$ such that $S \cap I = \emptyset$, there exists a maximal ideal among those containing $I$ and disjoint from $S$; any such ideal is prime.

To prove this, note that such ideals are in one-to-one order-preserving correspondence with proper ideals of $S^{-1}R$ containing $S^{-1}I$. Thus it is enough to take the inverse image of a maximal ideal containing $S^{-1}I$ under the canonical homomorphism $R \to S^{-1}R$. The inverse image of a maximal ideal is always prime.

In the present situation, apply the theorem to $S = \{1 \} \cup \{f^n|n \geq 1\}$.

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