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Thread: Help on Intergral Domain

  1. #1
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    Help on Intergral Domain

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

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

    (b) Show that all four numbers: $\displaystyle 2,3,4+\sqrt10, 4-\sqrt10$ 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 $\displaystyle Z(\sqrt10)=${$\displaystyle a+b\sqrt10|a,b \in Z$}

    (a) Using the fact that the norm $\displaystyle N(a+b\sqrt10)=a^2 - 10b^2$ is multaplicative, describle the units of $\displaystyle Z(\sqrt10)$.
    Put an absolute value on it, then $\displaystyle N(a+b\sqrt{10}) = |a^2 - 10b^2|$ will satisfy $\displaystyle N(\alpha \beta) = N(\alpha)N(\beta)$.
    If $\displaystyle u$ is a unit then there is $\displaystyle v$ so that $\displaystyle uv=1$ but then $\displaystyle N(uv) = N(1) = 1$ thus $\displaystyle N(u)N(v) = 1$ which means $\displaystyle N(u)=1$. Thus, the units must have norm equal to $\displaystyle 1$. Conversely, if $\displaystyle N(a+b\sqrt{10}) = 1$ then $\displaystyle \tfrac{a}{a^2 - 10b^2} - \tfrac{b}{a^2 - 10b^2} \sqrt{10}$ (which is in $\displaystyle \mathbb{Z}(\sqrt{10})$) is inverse of $\displaystyle a+b\sqrt{10}$. To find the units we need to solve $\displaystyle |a^2 - 10b^2| = 1 \implies a^2 - 10b^2 = \pm 1$. 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: $\displaystyle 2,3,4+\sqrt10, 4-\sqrt10$ are irreducible. Are any of these numbers prime?
    A non-zero non-unit element $\displaystyle \alpha$ is called irreducible iff $\displaystyle \alpha = \beta \gamma$ is an improper trivial factorization, i.e. one of them, say $\displaystyle \beta$, is associate to $\displaystyle \alpha$ and the other one $\displaystyle \gamma$ is a unit. Hence it cannot be reduced further. A non-zero non-unit element $\displaystyle \alpha$ is called prime iff $\displaystyle \alpha | \beta \gamma \implies \alpha | \beta \mbox{ or }\alpha | \gamma$. 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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