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February 3rd, 2012, 11:41 AM  #1 
Newbie Joined: Feb 2012 Posts: 1 Thanks: 0  Why is Z[??n] not a UFD for n>2?
Hi guys, Could anyone help me with the following problem For n>2 show that is NOT a Unique Factorisation Domain(UFD) I have managed to show 2 is irreducible in this ring for all n>2. Now If we firstly consider the case for n odd, this means n+1 is even and so . But and 2 does not divide or so 2 is NOT prime. However in a UFD all primes and irreducibles are the same so cannot be a UFD for when n is odd and bigger than 2. For n even though I don't know how to prove it? I have tried to construct a similar argument as that for n odd but it does not seem to work? I have also tried to factor n in two different ways but again am getting nowhere with it. Any help would be very much appreciated! 
October 2nd, 2014, 07:43 PM  #2 
Newbie Joined: Oct 2014 From: USA Posts: 2 Thanks: 2  ℤ[√±n] is not a UFD for nonsquare n > 2
@Alex25, Let p ∈ ℕ be a prime factor of n(n + 1) = n² + n. E.g. p can be taken as 2. p is irreducible in ℤ[√n], using Norm properties n² + n = (n + √n)(n  √n)  (1) p∣ left side of (1) but ∤ either factor on the right, so it is not a prime. ℤ[√n] cannot be UFD, since it has an irreducible which is a nonprime. Continuing the premise that nonsquare n >2, considering ℤ[√n], and using similar arguments for n²  n = (n + √n)(n  √n) ℤ[√n] cannot be UFD for nonsquare n > 2. 
October 3rd, 2014, 11:45 AM  #3 
Newbie Joined: Oct 2014 From: USA Posts: 2 Thanks: 2  My previous post may be inaccurate
Alex's question is theorem 4 in http://www.ias.ac.in/resonance/Volum.../00720079.pdf It's an exercise in Michael Artin's book, Algebra When the radicand is positive, the answer is more complicated Per quali d l’anello Z[√d] è UFD, PID, ED?  Carlo Mazza I do not know how to edit a previous article, being new to writing math articles, using unicode, etc. ...Will continue to research on this intriguing question ... 

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