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

hmmm16

Member
Feb 25, 2012
31
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
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.