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

[hmmmmm]

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]

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.