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Thread: 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 $\displaystyle \alpha$ that means,
    $\displaystyle x+y\sqrt{d}\equiv z+w\sqrt{d}$ whenever (actually if and only if, since all definitions are biconditional statements)
    $\displaystyle x\equiv z\mbox{ and }y\equiv w$ modulo $\displaystyle \alpha$.

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

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

    The relation $\displaystyle \to$ 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 $\displaystyle [a,b]$ as the set of all quadratic integers such that a congruent to $\displaystyle (a,b)$.

    We define addition as $\displaystyle [a,b]+[c,d]=[a+c,b+d]$ 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,
    $\displaystyle 6=(2)(3)$
    or,
    $\displaystyle 6=(1+\sqrt{-5})(1-\sqrt{-5})$
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  4. #4
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    Hi Perfecthacker,

    I got it. Thank you very much.
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