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Math Help - Comaximal Ideas in a Principal Ideal Domain

  1. #1
    Super Member Bernhard's Avatar
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    Comaximal Ideas in a Principal Ideal Domain

    Prove that in a prinicpal ideal domain, two ideals (a) and (b) are comaximal if and only if a greatest common divisor of a and b (in which case (a) and (b) are said to be coprine or realtively prime)

    Note: (1) Two ideals A and B of the ring R are said to be comaximal if A + B = R

    (2) Let I and J be two ideals of R
    The sum of I and J is defined as  I+J = \{ a+b | a \in I, b \in J \}
    Last edited by Bernhard; March 14th 2013 at 02:04 AM.
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    Re: Comaximal Ideas in a Principal Ideal Domain

    Prove that in a prinicpal ideal domain, two ideals (a) and (b) are comaximal if and only if a greatest common divisor of a and b is 1 (in which case (a) and (b) are said to be coprine or realtively prime)

    Let R be a principal ideal domain and suppose (a)+(b)=R. The following is a series of equivalences, let me know if you need clarification on any of the steps.

    (a)+(b)=R
    \iff (a)+(b)=(1)
    \iff (a)+(b)\ni 1
    \iff 1=ax+by for some x,y\in \mathbb{R}
    \iff gcd(a,b)=1
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    Super Member Bernhard's Avatar
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    Re: Comaximal Ideas in a Principal Ideal Domain

    Thanks for the help

    One immediate problem for me is that I cannot see why the following is true:

     (a) + (b) = R \Longleftrightarrow (a) + (b) = (1)

    Can you please show me explicitly why this is the case.

    Thanks

    Peter
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    MHF Contributor Siron's Avatar
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    Re: Comaximal Ideas in a Principal Ideal Domain

    Prove (1) \subseteq R and R \subseteq (1)
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    Super Member Bernhard's Avatar
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    Re: Comaximal Ideals in a Principal Ideal Domain

    Can clearly see that proving  (1) \subseteq R and  R \subseteq (1) is the way to go ... but ...

    How do we even know that R has a 1?

    Can someone help?

    Peter
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    Super Member Bernhard's Avatar
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    Re: Comaximal Ideals in a Principal Ideal Domain

    I think I just answered my most recent post.

    Just checked definitions of PID and Integral domain

    A PID is an integral domain in which every ideal is principal

    and

    An integral domain is a commutative ring with identity having no zero divisors

    So by definition a PID has a 1!

    Can someone confirm this is correct?

    Peter
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    Re: Comaximal Ideals in a Principal Ideal Domain

    1 just denotes the multiplicative identity on the ring.

    EDIT: Yes, you are right. A PID has to contain a multiplicative identity.
    Last edited by Gusbob; March 14th 2013 at 11:15 PM.
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