Factor principal ideals in a field with class number 2

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Homework Help Overview

The discussion revolves around the properties of ideals in a number field, specifically focusing on the factorization of an irreducible element's ideal when it is not prime. The context involves understanding the implications of the class number being 2 and how it relates to the structure of ideals in the ring of integers of the field.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the class number and the degree of the extension K/Q, questioning their relevance to the problem. There is a discussion about the nature of irreducible elements and their implications for the factorization of ideals, particularly whether they can be principal.

Discussion Status

The discussion has progressed with participants offering clarifications and insights into the problem. Some guidance has been provided regarding the factorization of ideals and the implications of the class group structure. There is an acknowledgment of the original poster's confusion, which has been addressed through dialogue.

Contextual Notes

There is a specific focus on the definition of irreducible elements within the ring of integers of K, and the implications of the class number being 2 are under consideration. The original poster expresses uncertainty about the relationship between these concepts.

R.P.F.
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Homework Statement



Let K be a number field. Let a be an irreducible element of K. Prove that if <a>(the ideal generated by a) is not a prime ideal, then it is a product of two prime ideals.


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The Attempt at a Solution



I'm actually quite clueless...:frown: I have been thinking maybe it is possible for me to obtain some information about the extension K:Q from the condition that the class number is 2, like the degree of the extension, etc..?
Any help is appreciated!
 
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The class number of K and the degree of the extension K/Q have nothing to do with each other. (There are quadratic extensions of Q with arbitrarily large class number.)

In your problem I assume by an "irreducible element of K" you really mean an irreducible element in the ring of integers of K. Anyway, just start with what you're given: (a) is not prime, so it factors into a product of prime ideals. Some of these will be principal, some not. Try to use the fact that the class group is Z/2Z to simplify your factorization.
 
Thanks for your reply!
morphism said:
In your problem I assume by an "irreducible element of K" you really mean an irreducible element in the ring of integers of K.
Yes. This is what I meant.
morphism said:
Some of these will be principal, some not.

Could they really be principal? Would that imply that a has a proper factor, which is a contradiction?
 
R.P.F. said:
Could they really be principal? Would that imply that a has a proper factor, which is a contradiction?
Good! Can you use this observation to finish off the solution?
 
Figured it out! Finally can take a break after 8 hours in the library. :approve:

Thanks so much. :biggrin:
 
Well done!
 

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