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Math Help - Quadratic integers

  1. #1
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    Quadratic integers

    1) Which of the following are quadratic integers in Q[sqrt(-5)]?
    a) 3/5
    b) (3+8sqrt(-5))/5
    c) i*sqrt(-5)

    Prove your result in each case.


    2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.
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  2. #2
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    [QUOTE=beta12;24319]1) Which of the following are quadratic integers in Q[sqrt(-5)]?
    a) 3/5
    b) (3+8sqrt(-5))/5
    c) i*sqrt(-5)

    Again, use what I said before, namely that,
    \{1,i\sqrt{5}\} forms a basis for the field \mathbb{Q}(\sqrt{-5}) viewed as a vector space over \mathbb{Q}.
    We note that,
    \mathbb{Q}\leq \mathbb{Q}(\sqrt{-5})
    Meaning, that the rationals are contained in this simple extension.

    a)Yes, as just explained all rationals are contained thin this field.

    b)Yes, it is \frac{3}{5}+\frac{8}{5}\sqrt{-5} so if you select a=3/5,b=8/5 \in \mathbb{Q} as a linear combination for a+b\sqrt{-5} then it works!

    c)No, unless you are asking for i\sqrt{5}.
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  3. #3
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    Hi Perfecthacker,

    What about (3+5sqrt(2))/2 ? It can be written as 3/2 + 5/2*sqrt(2) in the form of a +bsqrt(d). So is it a quadratic integer?
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  4. #4
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    Quote Originally Posted by beta12 View Post
    Hi Perfecthacker,

    What about (3+5sqrt(2))/2 ? It can be written as 3/2 + 5/2*sqrt(2) in the form of a +bsqrt(d). So is it a quadratic integer?
    Yes.

    But is it not a quadradic integer in \mathbb{Q}(\sqrt{-5})
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes.

    But is it not a quadradic integer in \mathbb{Q}(\sqrt{-5})
    Hi Perfecthacker,

    I see.

    Can you teach me how to solve the second question?
    2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.

    Thank you very much.
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  6. #6
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    Quote Originally Posted by beta12 View Post
    Can you teach me how to solve the second question?
    2) Find an element of Q[sqrt(-1)] tht is not a quadratic integer, and yet its norm in Q[sqrt(-1)] is an integer.
    Not sure about this one. See, I never studied algebraic number theory (noob ). However, I was able to answer those questions you had because I studied field theory. The following is going to be a guess, though I am highly certain about it.

    The field \mathbb{Q}(i) are the Gaussian integers (rationals is more proper). Hence, any can be expressed as a linear combination,
    a+bi\, \exists a,b\in \mathbb{Q}
    The number,
    \frac{ \sqrt{2} }{2}+i\frac{ \sqrt{2} }{2} is not Gaussian. Yet its norm is 1 which is Gaussian.
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  7. #7
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    I see. But you are still great at Quadratic field.
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