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Math Help - composition functions

  1. #1
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    Brasilia
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    composition functions

    Hi again :

    I have an exercise a little complicated and extensive,




    Let  A= \mathbb{Z}\times{}\mathbb{Z}\times{}\mathbb{Z}....  . be the direct product of copies of \mathbb{Z} indexed with the positive integers (so A is a ring under componentwise addition and multiplication) and let R be the ring of all group homomorphism from A to itself with addition defined as pointwise addition of functions . If \emptyset be the element of R defined by \emptyset (a_1,a_2,a_3,......)=(a_2,a_3,........). Now , if \Psi be the element of R defined by \Psi(a_1,a_2,a_3......)=(0,a_1,a_2,a_3........) . Prove that :



    a)  \emptyset\Psi(composition functions) is the identity of R ,but \emptyset\Psi is not the identity of R (i.e \Psi is a right inverse for \emptyset , but not a left inverse.
    b)Exhibit infinitely many right inverses for \emptyset
    c)Find a nonzero element  \Pi \in{} R such that \emptyset\Pi=0 , but  \Pi\emptyset\neq0
    d)Prove that is no nonzero element p \in{} R such that  p\emptyset=0 (i,e
     \emptyset is a left zero divisor but not a right zero divisor)



    Any help will be well welcome
    Thanks
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  2. #2
    MHF Contributor

    Joined
    Mar 2011
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    Tejas
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    Re: composition functions

    a) you have this written down wrong.

    you want to show that θψ = 1R but ψθ ≠ 1R

    b) what happens if you replace 0 in the definition of ψ with some other integer k?

    c) how about π(a1,a2,a3,....) = (1,0,0,0,......) ?

    d) suppose ρθ = 0, that is: ρθ(a1,a2,a3,....) = (0,0,0,....).

    this means that ρ(a2,a3,....) = (0,0,0,......), no matter how we choose a2,a3,....

    so ρ = 0.
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