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Finding the units of the ring Z[sqrt(-3)]

  1. Jun 11, 2009 #1
    1. The problem statement, all variables and given/known data

    In the ring [tex]Z[\sqrt{-3}][/tex] find all units and prove that 2 is irreducible but 7 is not.

    2. Relevant equations



    3. The attempt at a solution

    Well a unit is a non-zero element of the ring that when multiplied by some other non-zero element of the ring gives the unity of the ring.

    ie; [tex] ab = 1\;a,b \in Z[\sqrt{-3}] [/tex]

    in otherwords b is the multiplicative inverse of a.

    a is of the form [tex] a = x + y\sqrt{3} i [/tex] which just has the inverse of [tex]\frac{x - y\sqrt{3}i}{x^{2} + 3y^{2}}[/tex] (ie; b has this form)

    So everything with the form of b with integer coefficients is a unit and 0 is not a unit.

    I really don't understand what this question is asking, because I know what I've written down is completely stupid.

    How would one go about finding the units of this ring?
     
  2. jcsd
  3. Jun 11, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Which means that [itex]x^2+ 3y^2[/itex] is a factor of both x and y. For what x and y is that true?

    Another way to do this is to look at [itex](x+ y\sqrt{3})(a+ b\sqrt{3})[/itex][itex]= (ax+ 3by)+ (ay+bx)\sqrt{3}= 1[/itex]. What must x, y, a, and b be?

     
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