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Math Help - Irreducibles

  1. #1
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    Irreducibles

    If we define Z\left[ {\sqrt { - 5} } \right] = \left\{ {a + b\sqrt { - 5} :a,b \in Z} \right\}, how would I show that 2, 3, 1 + \sqrt { - 5} and 1 - \sqrt { - 5} are all irreducible?

    I have shown that none of these are units of Z\left[ {\sqrt { - 5} } \right].

    For example, if I assume that 2 = xy, I need to show that at least one of x and y is a unit. Obviously if 2 = 2 x 1 = 1 x 2, then 1 is a unit, but how do I show that there are no other factorisations of 2 in Z\left[ {\sqrt { - 5} } \right]?

    Thanks for help!
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  2. #2
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    Quote Originally Posted by BeWareWereWolf View Post
    If we define Z\left[ {\sqrt { - 5} } \right] = \left\{ {a + b\sqrt { - 5} :a,b \in Z} \right\}, how would I show that 2, 3, 1 + \sqrt { - 5} and 1 - \sqrt { - 5} are all irreducible?

    I have shown that none of these are units of Z\left[ {\sqrt { - 5} } \right].

    For example, if I assume that 2 = xy, I need to show that at least one of x and y is a unit. Obviously if 2 = 2 x 1 = 1 x 2, then 1 is a unit, but how do I show that there are no other factorisations of 2 in Z\left[ {\sqrt { - 5} } \right]?

    Thanks for help!
    Let N be a multiplicative norm such that N(a + b\sqrt { - 5}) = a^2 + 5b^2 (verify this is an indeed multiplicative norm).

    We have N(\alpha \beta)=N(\alpha)N(\beta) for all \alpha, \beta in \mathbb{Z}\left[ {\sqrt { - 5} } \right], and N(\alpha)=1 if \alpha is a unit.

    Now consider 2 and suppose 2= \alpha \beta.
    Then 4=N(2)=N(\alpha)N(\beta). If 2 is not irreducible, N(\alpha) should be 2, but there is no integer satisfyng 2= a^2 + 5b^2 . Contradiction! Thus 2 is irreducible in \mathbb{Z}\left[ {\sqrt { - 5} } \right].

    Now consider 1 + \sqrt { - 5} and suppose 1 + \sqrt { - 5} = \alpha \beta.
    Then 6=N(1 + \sqrt { - 5})=N(\alpha)N(\beta). If 1 + \sqrt { - 5} is not irreducible, then N(\alpha) should be either 2 or 3. But there is no integer satisfyng a^2 + 5b^2=2 or 3. Contradiction! Thus 1 + \sqrt { - 5} is irreducible in \mathbb{Z}\left[ {\sqrt { - 5} } \right].

    I leave it to you to show the remainder elements are irreducible in \mathbb{Z}\left[ {\sqrt { - 5} } \right].
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