Let m in Z, where m is not a square in Z
Define N: Z[sqrt(m)]---> N
Suppose that : a belongs to Z [sqrt(m)]. If N(a) is a prime in N . Show that a is an atom in Z [sqrt(m)]. Give an example of an m and a in Z [sqrt(m)] such that a is an atom but N(a) is not a prime
Can you give me some hints please?
Thank you
Sorry about that
N is the natural number and N is a norm.
Define N: Z[sqrt(m)]---> N be an Euclidean Domain with:
(i) If a,b belong to Z[sqrt(m)] and a | b then N(a) =<N(b)
(ii) If a in Z[sqrt(m)] then there exists q,r in Z[sqrt(m)] s.t b= qa + r where either r =0 or r is non zero and N(r) < N(a)
Anyway, I think I know how to do it now. Thank you so much for your time on this thread.
Cheers