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Is $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \mathbf{Z}_p$?

My proof: $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \operatorname{Hom}(\mathbf{Z}_p, \varprojlim\mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \mathbf{Z}_p$.

Is this correct?

  • 0
    I take it from the working $Z_p$ is the p-adic integers?2011-09-27
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    Yes, $\mathbf{Z}_p = \varprojlim \mathbf{Z}/p^n$.2011-09-27
  • 1
    Did you mean to write $\varprojlim \mathbf{Z}/p^n$ before your last equals sign?2011-09-27
  • 2
    What category is this Hom in?2011-09-27
  • 0
    In the category of groups (without topology).2011-09-27
  • 4
    @user5262: You need to show/argue that $\mathrm{Hom}(\mathbf{Z}_p,\mathbf{Z}/p^n\mathbf{Z})\cong\mathbf{Z}/p^n\mathbf{Z}$; otherwise fine. But why do you have the same expression in the penultimate and antepenultimate terms of your equality chain?2011-09-27
  • 0
    It was a copy/paste mistake. Thank you!2011-09-27
  • 0
    Big ups for use of penultimate and antepenultimate.2011-11-15
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    I sometimes forget these are not, apparently, terms in common use in English (at least in the US). We used them all the time when I was growing up (whenever somebody wanted to go last and 'called it' by yelling "Last!", the next person would yell "Penultimate!", quickly followed by "Antepenultimate!", for those too slow to get the coveted last and next-to-last spots...)2011-11-15

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