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Math Help - Please help with abstract algebra

  1. #1
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    Please help with abstract algebra

    Im studying for my final next week and I need a little help.

    "+ denotes an external direct product"

    1. Find the number of elements of order 4 in Z_60+Z_12+Z_12.
    2. In the ring Z+Z show: A = {(3x,y)|x,y in Z} is a maximal ideal.

    For number 1 the LCM(60,12)=60 then is that the number of elements of order 4?

    For number 2, I need some serious help starting.

    Thanks Guys
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  2. #2
    Super Member Gamma's Avatar
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    easier way to look at 1

    Consider the fact that \mathbb{Z}_{60}\cong\mathbb{Z}_{5}\oplus\mathbb{Z}  _{12} and \mathbb{Z}_{12}\cong\mathbb{Z}_3\oplus\mathbb{Z}_4.

    Applying these relations we have the following:
    \mathbb{Z}_{60}\oplus\mathbb{Z}_{12}\oplus\mathbb{  Z}_{12}\cong\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\m  athbb{Z}_3\oplus\mathbb{Z}_4\oplus\mathbb{Z}_4\opl  us\mathbb{Z}_4\oplus\mathbb{Z}_5

    It is clear that since 4 does not divide 3 or 5 there will be no elements of order 4 in \mathbb{Z}_3 or \mathbb{Z}_5 respectively (Lagrange). Furthermore, it is obvious that 1 and 3 have order 4 in \mathbb{Z}_4. Thus any combination of these elements in the 3 copies of \mathbb{Z}_4 will work so long as you choose 0 in the rest of the places. Note obviously not ALL of the componants can be 0 or else it is the identity and has order 1. I claim that there will be 3*3*3-1 elements, you oughta be able to prove why there are no more and no less.
    Last edited by Gamma; December 5th 2008 at 12:36 AM.
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  3. #3
    Super Member Gamma's Avatar
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    Problem 2

    Certainly \mathbb{Z} is a Euclidean Domain which can be shown to be a Principal Ideal Domain. In a PID every nonzero prime ideal is maximal. It is pretty clear that the second coordinate is all of \mathbb{Z}, so really this just amounts to showing that the first coordinate is maximal (I think anyway). The first coordinate is simply the principal ideal generated by (3) which is a prime ideal, so it is also maximal. Thus any ideal containing A would have to contain all of \mathbb{Z} in the second coordinate, and would have to contain (3) in the first coordinate, but since (3) is maximal in \mathbb{Z} this means any ideal satisfying this is necessarily all of \mathbb{Z} or else (3) again, and so if B were an ideal containing A then either B=(3)\oplus\mathbb{Z} or B=\mathbb{Z}\oplus\mathbb{Z} and so A is maximal.

    Not sure where exactly you got hung up on this one, but it is pretty easy to show both that E.D. implies PID, and then in a PID that prime ideals are maximal, just follow definitions, nothing too tricky about that, plus almost any algebra book would have proofs of these facts I should think.
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