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Thread: generators and relations

  1. #1
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    generators and relations

    Here is the problem I'm working on, from an old qualifying exam:

    Let M be the $\displaystyle \mathbb{Z}$-module generated by a, b, c with the relations $\displaystyle 4a+3b+3c=2a-b+3c=0$. Express M as a direct sum of cyclic modules. What are the orders of these modules?

    There's only two relations, so I'm guessing that one of the modules is going to be $\displaystyle \mathbb{Z}$. I managed to find a vector that is perpendicular to (4,3,3) and (2,-1,3), which is (6, -3, -5). I'm not sure what to do with it though, or if I'm even going in the right direction.

    Thanks,
    Hollywood
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  2. #2
    MHF Contributor
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    Re: generators and relations

    I think I've made some progress on this problem.

    $\displaystyle 4a+3b+3c=0$
    $\displaystyle 2a-b+3c=0$

    Subtracting twice the second from the first gives:

    $\displaystyle 2a-b+3c=0$
    $\displaystyle 5b-3c=0$

    And now subtracting the second from the first gives:

    $\displaystyle 2a-6b+6c=0$
    $\displaystyle 5b-3c=0$

    Let $\displaystyle e=a-3b+3c$, so $\displaystyle a=e+3b-3c$ gives:

    $\displaystyle 2e=0$
    $\displaystyle 5b-3c=0$

    The first equation gives $\displaystyle \mathbb{Z}/2\mathbb{Z}$, and for the second, my reasoning is that $\displaystyle \mathbb{Z}\times\mathbb{Z}$ gets divided into cosets of $\displaystyle \{(3x,5x):x\in\mathbb{Z}\}$, and so that gives a factor of $\displaystyle \mathbb{Z}$. So the answer is $\displaystyle M\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}$.

    Is that correct?

    - Hollywood
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