I’m considering the ring Z[√−n], where n≥3 and square free. I want to see why it’s not a UFD.

I defined a norm for the ring by |a+b√−n|=a2+nb2. Using this I was able to show that 2, √−n and 1+√−n are all irreducible. Is there someway to conclude that Z[√−n] is not a UFD based on this? Thanks.

**Answer**

If n is even, then 2 divides √−n2=−n but does not divide √−n, so 2 is a nonprime irreducible. In a UFD, all irreducibles are prime, so this shows Z[√−n] is not a UFD.

Similarly, if n is odd, then 2 divides (1+√−n)(1−√−n)=1+n without dividing either of the factors, so again 2 is a nonprime irreducible.

This argument works equally well for n=3, but fails for n=1,2, and in fact Z[√−1] and Z[√−2] are UFDs.

**Attribution***Source : Link , Question Author : Danielle Intal , Answer Author : Chris Eagle*