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Let $R$ be a commutative ring with identity and let $I$ be an ideal of $R$. Define $\operatorname{Rad}(I)=\{a\in R:\exists n\in\mathbb N, a^n\in I\}$. Show that $Rad(I)$ is the intersection of all prime ideals $P$ of $R$ containing $I$.

I could prove that $$\operatorname{Rad}(I)\subset \bigcap_{I\subset P \text{ prime}}P$$

However I am not able to prove the other direction. I found the same question has been asked before but I do not follow the arguments placed there. I know nothing about localization or that of maximal ideal disjoint from a subring, etc. I would be obliged if someone helps me to prove this in the simplest terms possible.

Here's the link to the "identical" question: $\operatorname{rad}(I)=\bigcap_{I\subset P,~P\text{ prime}}P$

I don't actually get any of the answers, from the first line themselves.

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Landon Carter
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  • If you do not understand the arguments at the question, then your question here should be about the parts you don't understand there. If you just ask the same question as the other one, you may attract close votes. – rschwieb Oct 06 '15 at 13:58
  • I don't understand anything that's written there. – Landon Carter Oct 06 '15 at 16:01
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    Localization is the easiest path for this. If you are unfamiliar with it, do you know why there are prime ideals in a ring like yours? You can then mimic that argument and avoid localization (or at least mask it). – Mohan Oct 06 '15 at 16:09
  • @LandonCarter If you can't even express the first single thing you don't understand about the post, I don't think many people will be excited about reexplaining the entire thing to you. We don't even know what post you are talking about, so someone might post a solution here that is nearly identical that you still "don't understand anything" about. You really have to throw us a bone, here. – rschwieb Oct 06 '15 at 17:40
  • @LandonCarter On the other hand, I am certain people would react positively to explaining concrete statements that you're having a hard time grasping. Just read it and pick the statements you don't get. And maybe link the solution you're looking at. – rschwieb Oct 06 '15 at 17:41
  • Sure, let me present the link to the "identical" question asked. Please see the edit after 1 minute. – Landon Carter Oct 06 '15 at 17:42
  • @LandonCarter Also, this is one of those situations where avoiding the popular solution is probably making things harder on yourself. Writing around localization will probably be longer and a lot more confusing than actually biting the bullet and trying to understand the popular proof. – rschwieb Oct 06 '15 at 17:43
  • Edited. I understand that. However, I was getting daunted by such terms and was looking for a more elementary proof. – Landon Carter Oct 06 '15 at 17:46

1 Answers1

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Let $x\notin Rad(I)$. By definition of $Rad(I)$ you know $S=\{1,x,x^2,x^3,\ldots\}$ is a multiplicatively closed set disjoint from $I$.

At this point, I highly recommend you get your head around the idea in item 1 of this post. It is a standard bit of commutative algebra that you should not avoid. It is literally Theorem 1 on page 1 of Kaplansky's Commutative Rings, so you should not be afraid of learning it. You don't have to go clear into localization, but it is well worth your time to get this bit down.

After you understand why

  • an ideal maximal with respect to avoiding a multiplicative set $S$ is prime,

it is not hard to adapt the proof to prove that

  • given an ideal $I$ not intersecting multiplicative set $S$, there is an ideal maximal with respect to avoiding $S$ which also contains $I$, and is necessarily prime for the same reasons as in the last point.

After that is established and you find your prime $P$, you'll have to see why $x\notin P$. At that point, you will have established that $x\notin Rad(I)$ implies $x\notin \cap \{P\mid P\text{ prime and } I\subseteq P\}$, and that's the contrapositive of the containment that you are seeking.

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