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Math Help - Common factors in integral domain Z[sqrt(-13)]

  1. #1
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    Common factors in integral domain Z[sqrt(-13)]

    Consider the integral domain \mathbb{Z}\[\sqrt{-13}\] = \{ x + y \sqrt{-13} | x,y \in \mathbb{Z} \}.

    Show that 2 and 3 + \sqrt{-13} have no common factors in \mathbb{Z}\[\sqrt{-13}\] except for 1 and -1.


    I can show that both 2 and 3 + \sqrt{-13} are irreducible. However, is there another way to prove that 1 and -1 are the only common factors?
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  2. #2
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    Re: Common factors in integral domain Z[sqrt(-13)]

    Showing irreducibility is not enough, you must also prove that the only units in your domain are 1 and -1.
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  3. #3
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    Re: Common factors in integral domain Z[sqrt(-13)]

    But should showing irreducibility be the first step though? Or is there a different solution that does not involve proving irreducibility?
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  4. #4
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    Re: Common factors in integral domain Z[sqrt(-13)]

    Suppose \alpha = a + \sqrt{-13} b is a common factor.

    Then 2 = \alpha x for some x = c + \sqrt{-13}d and N(2) = 2^2 = N(\alpha)N(x) = (a^2 + 13 b^2)(c^2 + 13d^2). It can be deduced that a = \pm 1 or a = \pm 2.

    We also have 3 + \sqrt{-13} = \alpha y for some y = e + \sqrt{-13}f and N(3 + \sqrt{-13}) = 22 = N(\alpha)N(y) = (a^2 + 13 b^2)(e^2 + 13f^2). It can be seen that a = \pm 1 is the only possible value.

    Hence \alpha = \pm 1 is the only common factor.
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