Prove Prime Ideal Problem: I/J ⊆ P

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Discussion Overview

The discussion centers on the Prime Ideal Problem in ring theory, specifically examining the relationship between ideals I, J, and a prime ideal P under the condition that the intersection of I and J is a subset of P. Participants explore the implications of this condition and aim to prove that either I or J must be contained in P.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that if P is a prime ideal and I intersect J is a subset of P, then it must be shown that either I is a subset of P or J is a subset of P.
  • Another participant argues that even if an element ab is in I intersect J, it does not necessarily imply that both a and b are in I intersect J, which raises questions about the assumptions made.
  • A different participant points out that if neither I nor J is contained in P, it leads to a contradiction since there exist elements a in I and b in J that are not in P, yet their product ab is in P due to the properties of prime ideals.
  • One participant elaborates that to negate the containment of I or J in P, one must find elements a in I and b in J that are not in P, but this leads to a contradiction, reinforcing the original claim.
  • Another participant concurs, stating that if an element a can be found in I not in P, then it follows that no element b can be found in J not in P, leading to the conclusion that if I is not contained in P, then J must be contained in P, and vice versa.

Areas of Agreement / Disagreement

Participants generally agree on the implications of the prime ideal condition and the contradictions that arise when assuming neither ideal is contained in P. However, there are nuances in the reasoning presented, indicating that while some points are accepted, the discussion remains somewhat contested regarding the details of the proof.

Contextual Notes

The discussion does not resolve the proof for an arbitrary finite number of ideals, as the follow-up request suggests further exploration is needed in that area.

mathmajor2013
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Let R be a ring with ideals I, J, and P. Prove that if P is a prime ideal and I intersect J is a subset of P, then I is a subset of P or J is a subset of P.
 
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If i like ab in I intersect J, then ab is in P. Therefore a in P or b in P since P is prime. Neither a or b need be in I intersect J though.
 
If neither I nor J is contained in P, some element a in I and b in J are neither in P, as you say. But this is a contradiction, since ab is in P and P is prime, hence one of the ideals is contained in P.

Follow-up: Show this for an arbitrary finite number of ideals.
 
Last edited:
obviously if either I or J is a subset of P, there is nothing to prove. so to negate that, we need some a in I with a NOT in P, AND b in J with b NOT in P.

but since I is an ideal, ab is in I. since J is an ideal ab is in J. therefore ab is in I∩J, and thus in P. since P is prime, either a is in P, contradicting our choice of a, or b is in P, contradicting our choice of b.

the only conclusion is that we cannot pick such a in I AND b in J outside of P, either I or J must lie within P.
 
Exactly. So if we can pick an a in I not in P, we cannot pick a b in J not in P. Hence if I is not contained in P, J must be contained in P. Oppositely, if J is not contained in P, I must be contained in P.
 

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