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How to show this is a Hom.

  1. Feb 27, 2014 #1
    Define ##\Omega: \mathbb{Z_{p^n}} \rightarrow \mathbb{Z_{p^n}}## where ##p## is prime

    with ##\ \ \ \ \ \ \Omega(x) = x^{p}##


    I am trying to prove this is a ##Hom## under addition.

    any ideas?
     
  2. jcsd
  3. Feb 27, 2014 #2

    jgens

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    Essentially just apply the binomial theorem. Since the ring has characteristic p all the undesired terms vanish.
     
  4. Feb 27, 2014 #3
    The ring ##K## has order ##p^n## which makes it ##\cong \mathbb{Z_{p^n}}##
     
    Last edited: Feb 27, 2014
  5. Feb 27, 2014 #4
    You did notice that the ring I am working with has ##p^n## elements. It is an extension field of ##\mathbb{Z_{p^n}}##.
     
    Last edited: Feb 27, 2014
  6. Feb 27, 2014 #5

    jgens

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    Oh whoops you are right! Unless I am mistaken it turns out the result is actually false! Take p = n = 2 and consider the map Ω:Z4Z4 given by Ω(x) = x2. Then Ω(1+1) = Ω(2) = 22 = 0 but Ω(1)+Ω(1) = 12+12 = 2.
     
  7. Feb 27, 2014 #6
    yes, even in the general case ##\Omega(1+1) = 2^p \neq 2 = \Omega(1) + \Omega(1) \ \forall n, p \geq 2##
     
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