Why is Z[√−n],n≥3\mathbb{Z}[\sqrt{-n}], n\ge 3 not a UFD?

I’m considering the ring Z[n], where n3 and square free. I want to see why it’s not a UFD.

I defined a norm for the ring by |a+bn|=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)(1n)=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

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