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Thread: composition functions

  1. #1
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    composition functions

    Hi again :

    I have an exercise a little complicated and extensive,




    Let $\displaystyle A= \mathbb{Z}\times{}\mathbb{Z}\times{}\mathbb{Z}.... . $ be the direct product of copies of $\displaystyle \mathbb{Z}$ 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 $\displaystyle \emptyset$ be the element of R defined by $\displaystyle \emptyset (a_1,a_2,a_3,......)=(a_2,a_3,........)$. Now , if $\displaystyle \Psi$ be the element of R defined by $\displaystyle \Psi(a_1,a_2,a_3......)=(0,a_1,a_2,a_3........)$ . Prove that :



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