Proving Z[√(-3)] is not a euclidean domain

  • Context: MHB 
  • Thread starter Thread starter hmmmmm
  • Start date Start date
  • Tags Tags
    Domain Euclidean
Click For Summary
SUMMARY

The discussion focuses on proving that the ring $\mathbb{Z}[\sqrt{-3}]$ is not a Euclidean domain by identifying an irreducible element that is not prime. The user proposes the element 2 as irreducible, as its norm N(2) equals 4, which leads to the conclusion that any factorization involving 2 does not yield prime factors. Specifically, 2 divides 4, which can be expressed as the product of the factors (1 + √(-3))(1 - √(-3)), yet does not divide either factor, confirming its status as irreducible but not prime.

PREREQUISITES
  • Understanding of Euclidean domains and their properties
  • Familiarity with the concept of irreducibility in ring theory
  • Knowledge of norms in algebraic number theory
  • Basic experience with algebraic integers, specifically in the context of $\mathbb{Z}[\sqrt{-3}]$
NEXT STEPS
  • Study the properties of Euclidean domains and their implications in algebraic structures
  • Explore the concept of irreducibility versus primality in more depth
  • Learn about norms in algebraic number fields and their applications
  • Investigate other examples of rings that are not Euclidean domains
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in algebraic number theory, as well as students and researchers interested in the properties of rings and domains.

hmmmmm
Messages
27
Reaction score
0
I am trying to show that $\mathbb{Z}[\sqrt{-3}]$ is not a euclidean domain, now I know that in every euclidean domain we have that an element is prime iff it is irreducible so I need to find an irreducible element of $\mathbb{Z}[\sqrt{-3}]$ that is not prime, I can't seem to think of one though, is there a general method for finding one?

Thanks for any help
 
Physics news on Phys.org
Re: Proving $\mathbb{Z}[\sqrt{-3}]$ is not a euclidean domain

how about 2? 2 is irreducible, since N(ab) = N(2) = 4 implies N(a) = 1,2 or 4. if N(a) = 1, then a = 1 or -1, which are both units. there are no solutions to N(a) = 2, and if N(a) = 4, then b is a unit.

but 2 divides 4 = (1+√(-3))(1-√(-3)), and 2 does not divide either factor, so 2 is not prime.
 

Similar threads

Replies
48
Views
5K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
  • · Replies 84 ·
3
Replies
84
Views
11K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K