Are all (q) prime ideals in Z(\rho)?

  • Context: Graduate 
  • Thread starter Thread starter learningphysics
  • Start date Start date
  • Tags Tags
    Primes Rings
Click For Summary
SUMMARY

The discussion centers on the conditions under which the ideal (q) is a prime ideal in the Kummer ring Z(ρ), where ρ = e^(2πi/p) and p and q are rational primes. It is established that (q) is not a prime ideal if the polynomial x^(p-1) + x^(p-2) + ... + 1 is reducible over Z_q, the ring of integers modulo q. The conversation highlights the complexity of this topic, particularly for p = 3, and emphasizes the need for specific cases of q to draw definitive conclusions.

PREREQUISITES
  • Understanding of prime ideals in ring theory
  • Familiarity with Kummer rings and their properties
  • Knowledge of irreducibility criteria for polynomials
  • Basic concepts of modular arithmetic and integer rings
NEXT STEPS
  • Study the irreducibility of the polynomial x^(p-1) + x^(p-2) + ... + 1 over various fields
  • Explore the structure of Kummer rings in greater detail
  • Investigate specific cases of prime ideals in Z(ρ) for different values of p and q
  • Review the properties of integral domains and their relation to prime ideals
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra, ring theory, and number theory, as well as students studying advanced algebraic structures.

learningphysics
Homework Helper
Messages
4,097
Reaction score
7
This is the last question in Elements of Abstract Algebra by Allan Clark.

When is (q) a prime ideal in Z(\rho) (the Kummer ring) where \rho = e^{2\pi i /p}, where p and q are rational primes.

This seems to be a difficult question to answer in general... since considerable effort goes into answering for p =3 alone in the book.

I think we need x^{p-1} + x^{p-2} + x^{p-3} + ... + 1 to be irreducible over Zq - ring of integers mod q (they show this result specifically for Z(w) the kummer ring for p=3 which I'm generalizing here for all primes... hope I'm correct)...
is there any more we can say immediately without going into specific cases of q ?

Appreciate any help or hints. Thanks.
 
Last edited:
Physics news on Phys.org
Consider the case in which ##\mathbb{Z}[\rho]/q\cdot \mathbb{Z}[\rho]## is no integral domain, i.e. ##(q)## no prime ideal. Then there have to be elements ##a=a_0+a_1\rho+\ldots+a_{p-1}\rho^{p-1}\, , \,b=b_0+b_1\rho+\ldots+b_{p-1}\rho^{p-1}## such that ##a\cdot b \in (q)##, i.e. ##q\,|\,a\cdot b##.

All other ##(q)## are then prime ideals.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
Replies
48
Views
5K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
988
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K