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Math Help - Prime and irreducable U.F.D

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    Prime and irreducable U.F.D

    This question is from Dummit and Foote Abstract Algebra pg 293 #5.

    Let R=\mathbb{Z}[\sqrt{-n}] where n is a square free integeter greater than 3.

    (a) PRove that 2, \sqrt{-n},1+\sqrt{-n} are irreducable in R.

    (b) Prove that R is not a U.F.D Conclude that the quadratic inter ring is not a U.F.D for n \equiv 2,3 \mod 4, n < -3

    Show that \sqrt{-n} \mbox{ or } (1+\sqrt{-n}) is not prime.

    (c) Give an explicit ideal in R that is not principal. Hint: Use (b) and cosider a maximal ideal containing the non prime ideal \sqrt{-n} \mbox{ or } 1+\sqrt{-n}

    For part (a) I have shown that none of the of them are units i.e they don't have a norm of 1 and they are non zero. so the only thing I have left to do is show if they do factor one of the factors is a unit

    so for 2

    suppose that there is a factorization then there exits a,b \in \mathbb{Z}\sqrt{-n} such that

    a\cdot b =2 Since the norm is a multiplicitive function I get

    N(a)\cdot N(b)=N(2) \iff N(a)\cdot N(b)=4

    Since a and b are not units and n(a) and N(b) are integers greater than 1

    N(a)=N(b)=2

    so let a=\alpha +\beta \sqrt{-n}

    N(a)=\alpha^2+\beta^2n=2

    but since n > 3 \beta = 0 and there does not exista an integer \alpha such that \alpha^2=2

    There for 2 is irreducable.

    This is where I am stuck.

    I don't see how to generalize this argument.

    Thanks TES
    Last edited by TheEmptySet; April 8th 2009 at 08:12 AM.
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    Quote Originally Posted by TheEmptySet View Post
    (a) PRove that 2, \sqrt{-n},1+\sqrt{-n} are irreducable in R.
    You proved that 2 is irreducible. You can show that the other two are irreducible by a similar argument. If \sqrt{-n} = \alpha \beta for non-units, then by taking norms n = N(\alpha)N(\beta). Notice that 1 < N(\alpha) < n. Let \alpha = a + b\sqrt{-n} then N(\alpha) = a^2 + nb^2. For this to be less than n it is necessary for b=0. Therefore, \alpha must be a real number. By a similar argument \beta must be a real number. However, it is impossible to have \alpha \beta to be an imaginary number. Thus, it is not possible to write \sqrt{-n} = \alpha \beta for non-units.

    Say that 1+\sqrt{-n} = \alpha \beta for non-units. By taking norms, 1 + n = N(\alpha)N(\beta). Thus, we require that 1 < N(\alpha) < n+1. If \alpha = a + b\sqrt{-n} then 1< a^2 + nb^2 < n+1. We cannot have |b|\geq 2 so b^2=0,1. Likewise for \beta = c + d\sqrt{-n}. If b^2=1 then 1 < a^2 + n < 1 \implies 1 < a^2 < 1 which is a contradiction. Therefore, b=0, and now repeat the same argument as above.
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    Quote Originally Posted by ThePerfectHacker View Post
    You proved that 2 is irreducible. You can show that the other two are irreducible by a similar argument. If \sqrt{-n} = \alpha \beta for non-units, then by taking norms n = N(\alpha)N(\beta). Notice that 1 < N(\alpha) < n. Let \alpha = a + b\sqrt{-n} then N(\alpha) = a^2 + nb^2. For this to be less than n it is necessary for b=0. Therefore, \alpha must be a real number. By a similar argument \beta must be a real number. However, it is impossible to have \alpha \beta to be an imaginary number. Thus, it is not possible to write \sqrt{-n} = \alpha \beta for non-units.

    Say that 1+\sqrt{-n} = \alpha \beta for non-units. By taking norms, 1 + n = N(\alpha)N(\beta). Thus, we require that 1 < N(\alpha) < n+1. If \alpha = a + b\sqrt{-n} then 1< a^2 + nb^2 < n+1. We cannot have |b|\geq 2 so b^2=0,1. Likewise for \beta = c + d\sqrt{-n}. If b^2=1 then 1 < a^2 + n < 1 \implies 1 < a^2 < 1 which is a contradiction. Therefore, b=0, and now repeat the same argument as above.
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    Quote Originally Posted by TheEmptySet View Post
    (b) Prove that R is not a U.F.D Conclude that the quadratic inter ring is not a U.F.D for n \equiv 2,3 \mod 4, n < -3
    Remember a result about irreducible elements in a UFD. If \pi \in R is irreducible and R is a UFD then \pi is a prime element. Above we have shown that 2 is a prime element. Assume that R is a UFD. Notice that 2|(n+1) (in the case when n is odd) and that n+1 = (1 + \sqrt{-n})(1-\sqrt{-n}). Therefore, 2|(1+\sqrt{-n}) or 2|(1-\sqrt{-n}) so 2(a+b\sqrt{-n}) = 1 \pm \sqrt{-n} for some a,b\in \mathbb{Z}. This would imply 2a = 1 which is impossible.

    In the even case then 2|-n however -n = \sqrt{-n}^2 would would imply 2|\sqrt{-n}, which is again impossible.
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