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Math Help - Help on Intergral Domain

  1. #1
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    Help on Intergral Domain

    Consider the integral domain Z(\sqrt10)={ a+b\sqrt10|a,b \in Z}

    (a) Using the fact that the norm N(a+b\sqrt10)=a^2 - 10b^2 is multaplicative, describle the units of Z(\sqrt10).

    (b) Show that all four numbers: 2,3,4+\sqrt10, 4-\sqrt10 are irreducible. Are any of these numbers prime?

    (c) Deduce that Z(\sqrt10) is not unique factorization domain.

    My lecturer go though this topic too fast, so I have difficulty on above question. Please help me.
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  2. #2
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    Quote Originally Posted by kleenex View Post
    Consider the integral domain Z(\sqrt10)={ a+b\sqrt10|a,b \in Z}

    (a) Using the fact that the norm N(a+b\sqrt10)=a^2 - 10b^2 is multaplicative, describle the units of Z(\sqrt10).
    Put an absolute value on it, then N(a+b\sqrt{10}) = |a^2 - 10b^2| will satisfy N(\alpha \beta) = N(\alpha)N(\beta).
    If u is a unit then there is v so that uv=1 but then N(uv) = N(1) = 1 thus N(u)N(v) = 1 which means N(u)=1. Thus, the units must have norm equal to 1. Conversely, if N(a+b\sqrt{10}) = 1 then \tfrac{a}{a^2 - 10b^2}  - \tfrac{b}{a^2 - 10b^2} \sqrt{10} (which is in \mathbb{Z}(\sqrt{10})) is inverse of a+b\sqrt{10}. To find the units we need to solve |a^2 - 10b^2| = 1 \implies a^2 - 10b^2 = \pm 1. This equation is Pellian and got infinitely many solution. I assume all you need to do is find a test to determine if a number is a unit or not, that is how to do it.

    (b) Show that all four numbers: 2,3,4+\sqrt10, 4-\sqrt10 are irreducible. Are any of these numbers prime?
    A non-zero non-unit element \alpha is called irreducible iff \alpha = \beta \gamma is an improper trivial factorization, i.e. one of them, say \beta, is associate to \alpha and the other one \gamma is a unit. Hence it cannot be reduced further. A non-zero non-unit element \alpha is called prime iff \alpha | \beta \gamma \implies \alpha | \beta \mbox{ or }\alpha | \gamma. A simple consequence is that prime elements are irreducible, while not necessarily irreducible elements are prime.

    (c) Deduce that Z(\sqrt10) is not unique factorization domain.
    In a unique factorization domain the irreducible elements are also prime. Thus, if you can find an irreducible non-prime element (which is probably from the above list) then it would mean this is not a unique factorization domain.
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