Is \sqrt{I} an Ideal of a Commutative Ring R?

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Homework Help Overview

The discussion revolves around the properties of the radical of an ideal in a commutative ring. The original poster is tasked with showing that the radical of an ideal, denoted as \(\sqrt{I}\), is itself an ideal of the ring \(R\). The context involves understanding the definitions and implications of elements belonging to the radical.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of an element being in the radical of an ideal and discuss how to demonstrate that the product of an element from the ring and an element from the radical remains in the radical. Questions arise about the addition of elements in the radical and how to show that their sum also belongs to the radical.

Discussion Status

Participants are actively engaging with the problem, confirming each other's reasoning and attempting to clarify the steps needed to show that \(\sqrt{I}\) is an ideal. Some guidance has been provided regarding the multiplication of elements and the need to show closure under addition, but no consensus has been reached on the complete solution.

Contextual Notes

There is an emphasis on the definitions and properties of ideals and radicals, with participants questioning their understanding of how these properties interact. The original poster expresses uncertainty about specific steps in the proof, indicating a need for further exploration of the concepts involved.

Juanriq
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Salutations!

Homework Statement

Let R be a commutative ring and let I \subseteq R be an ideal. Show \sqrt{I} is an ideal of R if \sqrt{I} is f \in R such that there exists an n \in \mathbb{N} \mbox{ such that } f^{n} \in I.



Homework Equations





The Attempt at a Solution

Pick an r \in R \mbox { and } x \in \sqrt{I}. We want to show that xr \in \sqrt{I}. Well, this would imply that (xr)^n = x^nr^n \in I, I think this means that x^n \in I, but I am a little befuddled on how to proceed. Thanks!
 
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Yes, that is correct.
So you still need to show that x,y\in \sqrt{I} implies that x+y is also in the radical.
There exists n and m such that x^n\in I and x^m\in I. Now try to show that (x+y)^{n+m}\in I.
 
Thanks micromass! So everything above is correct? I'm still not really sure how I see that xr \in \sqrt{I} though. Is it because both r^n \mbox{ and } x^n are in radical I? I know showing the summation will be fun... all the terms will be in the ideal I, I think and the union of a bunch of ideals is still an ideal
 
Well, since x^n\in I (by definition, since x\in \sqrt{I}), we got that r^nx^n\in I (since arbitrary multiplication preserves elements in I).
 
ohhhhh-gotcha! Thanks, I appreciate it. I'm sure I'll be back later after trying the second part. Thanks again!
 

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