Abstract Algebra Hello Experts: Proving Theorems About Ideals and Radicals

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Discussion Overview

The discussion revolves around proving theorems related to ideals and radicals within the context of abstract algebra, specifically in commutative rings. Participants seek to understand the proofs of specific properties of radicals of ideals.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant requests a detailed proof for two theorems regarding ideals I and J in a commutative ring R, specifically that the radical of I is contained in the radical of J, and that the radical of the radical of I equals the radical of I.
  • Another participant questions whether the problem is a homework question and asks what attempts have been made to solve it.
  • A participant clarifies that the problem is not from homework but expresses a desire to understand the reasoning behind the theorems.
  • A participant provides a proof for the first theorem, stating that if x is in the radical of I, then x raised to some power is in I, and since I is contained in J, x must also be in the radical of J.
  • The same participant offers a proof for the second theorem, explaining that if x is in the radical of the radical of I, then it can be shown that x is also in the radical of I.
  • Another participant thanks the previous contributor for the explanation and shifts the discussion to a different topic regarding a division ring D and its centralizer F, asking for clarification on the implications of an element x being in D but not in F while its square is in F.
  • A participant responds with an example involving quaternions to illustrate that it is possible for an element x in D to have its square in F while not being in F itself.
  • Another participant defines F as the center of D.

Areas of Agreement / Disagreement

Participants generally agree on the need for detailed proofs and explanations, but there is some contention regarding the classification of the original problem as homework. The discussion about the division ring and centralizer introduces additional complexity, with differing views on the implications of the properties of elements within these structures.

Contextual Notes

The discussion includes assumptions about the properties of ideals and radicals in commutative rings, as well as the characteristics of division rings and their centralizers. Some mathematical steps and definitions may be taken for granted, and the implications of certain properties remain unresolved.

DukeSteve
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Hello Experts,

I can't find the proof of this theorems please help me:

Given that there is a commutative ring R and 2 ideals I and J, also given that I is included in J

I need to prove
1) radical of I is in radical of J
2) radical of radical of ideal I = radical of ideal I.


Please give me a detailed proof, I want to understand the proof and your way.
 
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Is this a homework problem? If so, then this belongs in the homework forums...

As for the problem: what did you try already to solve the problem?
 
It's not a hw, I just want to understand why it's right. If you don't have the answer just don't answer, let others answer it.
 
I'm sure micromass knows the answer to this: he's a very clever guy. It's just against PF rules to give help on homework without seeing the poster's attempt.

1) This seems straightforward: Given [itex]x \in \sqrt{I}[/itex] then [itex]x^n \in I[/itex] for some positive integer n. Since [itex]I \subseteq J[/itex] then [itex]x^n \in J[/itex] as well.

2)Suppose [itex]x \in \sqrt{\sqrt{I}}[/itex]. Then there is some positive integer n s.t. [itex]x^n \in \sqrt{I}[/itex]. By the definition of [itex]\sqrt{I}[/itex], there is some pos. int. m s.t. [itex](x^n)^m \in I[/itex]. Then [itex]x^{nm} \in I[/itex], so [itex]x \in \sqrt{I}[/itex].

The converse is trivial: given [itex]x \in \sqrt{I}[/itex] then [itex]x^1 = x \in \sqrt{I}[/itex], so [itex]x \in \sqrt{\sqrt{I}}[/itex].
 
Thanks a lot for this great explanation. It's really not from HW.

Could you please do me a favor and explain why is this right:

If I have D ring with division but not a field and it's char !=2. If I know that F is centralizer of D (that means F is a field) and I have x in D such that x^2 is in F but x is not in F.

What does this mean that x is not in F? Why?
 
I'm not sure I understand the question. I'm pretty sure you can have x in D with x2 in F but still have [itex]x \notin F[/itex]. Take the quaternions for example (http://en.wikipedia.org/wiki/Quaternion). They form a division ring but the element j isn't in F since ij = k but ji = -k. However j2 = -1, which does commute with every element and thus is in F.
 
F = center(D).
 

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