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Math Help - congruence and quadratic integers.

  1. #1
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    congruence and quadratic integers.

    Question 1)
    If alpha is a quadratic integers in Q[sqrt(-d)] , then define a notion of congruence (mod alpha).
    Furthermore, define +, -, and X for congruence classes , and show that this notion is well-defined.

    Question 2)
    Prove that ( the integers of ) Q[sqrt(-5)] do not form a UFD.
    Hint: Consider the quadratic integers that divide 6.

    Please teach me how to solve these two questions. Thank you very much.
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  2. #2
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    Quote Originally Posted by beta12 View Post
    Question 1)
    If alpha is a quadratic integers in Q[sqrt(-d)] , then define a notion of congruence (mod alpha).
    Furthermore, define +, -, and X for congruence classes , and show that this notion is well-defined.

    You can define two quadratic integers congruent when there corresponding parts are congruent modulo \alpha that means,
    x+y\sqrt{d}\equiv z+w\sqrt{d} whenever (actually if and only if, since all definitions are biconditional statements)
    x\equiv z\mbox{ and }y\equiv w modulo \alpha.

    Let the ordered pair (a,b) represent the quadratic integer a+b\sqrt{d}.

    We define (a,b)\to (c,d) when they are congruent (just like the integers).

    The relation \to forms an equivalence relation, this is easy to show because it is completely derived from modulo for integers (which does in fact from an equivalence relation).

    Now we define [a,b] as the set of all quadratic integers such that a congruent to (a,b).

    We define addition as [a,b]+[c,d]=[a+c,b+d] and show it is well-defined (invariant under the representatives). This is easy to show since again it is derived from the fact of well-defineness for integers modulo an integer.
    -----
    Again, I never studied quadratic fields, but this looks supprising. Does this have to represent the field of quotients of quadratic integers?
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  3. #3
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    Quote Originally Posted by beta12 View Post

    Question 2)
    Prove that ( the integers of ) Q[sqrt(-5)] do not form a UFD.
    Hint: Consider the quadratic integers that divide 6.
    .

    You can factor 6 proper and non-trivially as,
    6=(2)(3)
    or,
    6=(1+\sqrt{-5})(1-\sqrt{-5})
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  4. #4
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    Hi Perfecthacker,

    I got it. Thank you very much.
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