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  • 2 Post By svatan
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February 3rd, 2012, 11:41 AM   #1
Joined: Feb 2012

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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!
Alex25 is offline  
October 2nd, 2014, 07:43 PM   #2
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ℤ[√±n] is not a UFD for non-square n > 2


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 non-prime.

Continuing the premise that non-square n >2, considering ℤ[√n], and using similar arguments for

n² - n = (n + √n)(n - √n)
ℤ[√n] cannot be UFD for non-square n > 2.
Thanks from Deveno and topsquark
svatan is offline  
October 3rd, 2014, 11:45 AM   #3
Joined: Oct 2014
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My previous post may be inaccurate

Alex's question is theorem 4 in

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 ...
svatan is offline  

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