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Thread: Interception is lcm

  1. #1
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    Interception is lcm

    Hi!

    I have problems with this demostration (I do not know how to start)

    If a,b \in{\mathbb{Z}} , prove that (a) \cap{(b)} = m where m = lcm(a,b)


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  2. #2
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    Re: Interception is lcm

    What do \left(m\right) and \left(n\right) mean? Please give the definition.
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  3. #3
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    Re: Interception is lcm

    Ok

    - Definition

    An ideal in a commutative ring R is a subset I of R such that


    a) 0\in{I}
    b) a,b \in{I} \Rightarrow{a+b \in{I}}
    c) If a\in{I}, r\in{A} then ra \in{I}

    - Definition

    If b_1,b_2,...,b_n lie in a commutative ring R , then the set of all linear combinations, denoted by

    (b_1,b_2,...,b_n) is an ideal in R, called the ideal generated by b_1,b_2,...,b_n . In particular if n=1, then

    (b) = \{rb:r \in{R}\}

    consists of all the mulpiples of b ; it is called the principal generated by b
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  4. #4
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    Re: Interception is lcm

    since m is a multiple of a and a multiple of b

    m\in (a)\cap (b)

    Therefore (m)\subseteq (a)\cap (b)

    If x is a multiple of a and a multiple of b then x is a multiple of m and therefore an element of (m)

    Therefore (a)\cap (b)\subseteq (m)
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  5. #5
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    Re: Interception is lcm

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