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


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  2. #2
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    Re: Interception is lcm

    What do $\displaystyle \left(m\right)$ and $\displaystyle \left(n\right)$ 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 $\displaystyle R$ is a subset $\displaystyle I$ of $\displaystyle R$ such that


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

    - Definition

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

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

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

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

    since $\displaystyle m$ is a multiple of $\displaystyle a$ and a multiple of $\displaystyle b$

    $\displaystyle m\in (a)\cap (b)$

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

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

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

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