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Math Help - integral domain w/ complex #'s

  1. #1
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    integral domain w/ complex #'s

    Let C={a+bi|a,b in Z}. Show that C is an integral domain.

    Please show details of this proof. I'm very confused and this is a practice problem for my final exam.

    Thank you!
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by mpryal View Post
    Let C={a+bi|a,b in Z}. Show that C is an integral domain.

    Please show details of this proof. I'm very confused and this is a practice problem for my final exam.

    Thank you!
    Hi mpryal.

    It is enough to show that C is a subring of the complex field \mathbb C=\{a+bi:a,b\in\mathbb R\} containing 1. Any subring of a field containing the multiplicative identity is automatically an integral domain.
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  3. #3
    Super Member Gamma's Avatar
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    More directly.
    (a+bi)(c+di)=ac-bd + (ad+bc)i=0

    1)
    ac=bd

    2)
    ad=-bc

    multiply 1) by d on both sides

    acd=bd^2
    Apply 2
    -bc^2=bd^2
    b(c^2+d^2)=0

    Case1) b=0 then your first one a+bi is just an integer which is definitely not a zero divisor, I think you can check that easily enough

    Case2) c^2+d^2=0 but this means both c and d must be 0 because integers squared are nonnegative. But that is exactly the condition that the second gaussian integer was 0.
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  4. #4
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    Quote Originally Posted by mpryal View Post
    Let C={a+bi|a,b in Z}. Show that C is an integral domain.

    Please show details of this proof. I'm very confused and this is a practice problem for my final exam.

    Thank you!
    Here is another way. You need to know that |z_1z_2| = |z_1||z_2| where z_1,z_2\in \mathbb{C}.

    If (a+bi)(c+di) = 0 \implies |(a+bi)(c+di)| = 0 \implies |a+bi|\cdot |b+di| = 0 \implies  (a^2+b^2)(c^2+d^2)=0.

    Thus, a^2+b^2=c^2+d^2 = 0 \implies a=b=c=d=0.

    EDIT: It should be "a=b=0 OR c=d=0" not "a=b=0 AND c=d=0".
    Last edited by ThePerfectHacker; May 1st 2009 at 02:30 PM.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If (a+bi)(c+di) = 0 \implies |(a+bi)(c+di)| = 0 \implies |a+bi|\cdot |b+di| = 0 \implies  (a^2+b^2)(c^2+d^2)=0.

    Thus, a^2+b^2=c^2+d^2 = 0 \implies a=b=c=d=0.
    Hi ThePerfectHacker.

    Thatís not the right conclusion. What it implies is that either a^2+b^2=0 (so a+ib=0) or c^2+d^2=0 (so c+id=0).
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  6. #6
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    ThePerfectHacker
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