Let $M$ be a $ \mathbb{Z}_{p}[[T]] $-module and $X=Hom(M,\mathbb{Q}_{p}/\mathbb{Z}_{p})$ be the dual of $M$. Let $X[p^n]$ denotes the $p^n$-torsion points of $X$. Is $X/X[p^n]$ the dual of $M[p^n]$ $?$ If not, then whose dual is $X/X[p^n]$ in terms of $M$ $?$
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$\begingroup$ More detail, please. What is $R$, and what notion of dual do you mean (is $X = Hom_R(X, R)$ or $Hom_{\text{ab.gps}}(X, \mathbb{Q} / \mathbb{Z})$ or what?) $\endgroup$– David LoefflerCommented Dec 19, 2013 at 10:40
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$\begingroup$ @DavidLoeffler I have edited my question. Hope it is fine now. $\endgroup$– AndrewCommented Dec 19, 2013 at 10:47
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$\begingroup$ What is the role of $[[T]]$? Is the $Hom$ over $\mathbb Z_p[[T]]$, and if so do you have a favorite way for $T$ to act pronilpotently on $\mathbb Q_p/\mathbb Z_p$? Or have I misunderstood the notation? $\endgroup$– Theo Johnson-FreydCommented Dec 24, 2013 at 4:51
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$\begingroup$ I might have misunderstood the question, in which case I apologize. But if $M = \mathbb Z_p$, then $X = \mathbb Q_p / \mathbb Z_p$, and $X[p^n] \neq X$, but $M[p^n] = 0$. So why would you think the answer to your first question would be "yes"? $\endgroup$– Theo Johnson-FreydCommented Dec 24, 2013 at 4:55
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2 Answers
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As @S.Carnahan shows. this is not true.
Perhaps the statement you want is that the dual of $M[p^n]$ is $X/p^nX$?
Take the exact sequence
$$0\to M[p^n] \to M \to M \to M/p^nM \to 0,$$
where the middle map is multiplication by $p^n$, and dualize it.
Edit: This also shows that $X/X[p^n]$ is the dual of $p^nM$, to answer the second question.
answered Dec 19, 2013 at 11:25
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Let $M = \mathbb{Z}_p$ with $T$ acting as $0$. Then (if I'm not missing anything) $X \cong \mathbb{Q}_p/\mathbb{Z}_p$, and $X/X[p^n] \cong X \neq 0$, while $M[p^n] = 0$.
answered Dec 19, 2013 at 11:18