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Math Help - UFD Question

  1. #1
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    UFD Question

    Hello all,

    I am finishing up this chapter in my book with regards to UFD's, and it presented an interesting question that ties in with unique prime factorization. Here is what it says:

    It can be shown directly that Z[\sqrt{-3}] is not a UFD by finding an integer which factors into primes in more than one way. These two factorizations are related to the unique prime factorization in the quadratic integers in Q[\sqrt{-3}]

    Can anyone show me this directly and explain how those factorizations are related?

    Thank you!
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  2. #2
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    Quote Originally Posted by Samson View Post
    Hello all,

    I am finishing up this chapter in my book with regards to UFD's, and it presented an interesting question that ties in with unique prime factorization. Here is what it says:

    It can be shown directly that Z[\sqrt{-3}] is not a UFD by finding an integer which factors into primes in more than one way. These two factorizations are related to the unique prime factorization in the quadratic integers in Q[\sqrt{-3}]

    Can anyone show me this directly and explain how those factorizations are related?
    Look at the factorisations 4 = 2\times2 = (1 + \sqrt{-3})(1 - \sqrt{-3}).
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    Thank you Opalg! How would you describe how those factorizations are related though? Is there some property or law that describes that relationship?
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    Quote Originally Posted by Samson View Post
    How would you describe how those factorizations are related though? Is there some property or law that describes that relationship?
    The ring \mathbb{Z}[\sqrt{-3}] has a norm associated with it, namely a multiplicative function N from the ring to the positive integers. In this case, N(a+b\sqrt{-3}) = a^2+3b^2. If you want to factorise a number in \mathbb{Z}[\sqrt{-3}] then you need to factorise its norm. For example, N(4) = 16. So if a+b\sqrt{-3} is a factor of 4 then N(a+b\sqrt{-3}) must be a factor of 16. The only factor of 16 that can be expressed in the form a^2+3b^2 is 4, which is equal to both 2^2+3\cdot0^2 and 1^2+3\cdot1^2. So the only factors of 4 in \mathbb{Z}[\sqrt{-3}] are \pm2 and \pm1\pm\sqrt{-3}]. These numbers all have norm 4, and there are no proper factors of 4 that can be expressed in the form N(a+b\sqrt{-3}). So 2 and 1\pm\sqrt{-3} are all irreducible in the ring \mathbb{Z}[\sqrt{-3}]. Thus 4 has two essentially different factorisations in \mathbb{Z}[\sqrt{-3}], which shows that this ring is not a UFD.
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    Thank you!
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