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Let $K \supset \mathbb{Q}_p$ be the $p$-adic field and let $O_K$ be its ring of integers and $M_K$ be the maximal ideal with integral closure $\bar{M}_K$. A power series is invertible if its lowest degree coefficient is unit, otherwise it is non-invertible. Consider a non-invertible power series $f(x) \in O_K[[x]]$ without constant term of weierstrass degree $\text{wideg}(f)=p^r$, $r$ is a natural number. Now consider the following two sets (for fixed $n$): \begin{align*} A_n&=\{x \in \bar{M}_K ~|~ f^n(x)=0 \ \text{and} \ f^{n-1}(x) \neq 0, \ n \geq 1 \}, \\ B_{n,m}&=\{x \in \bar{M}_K ~|~ f^n(x)=f^m(x), \ \text{where} \ n>m \geq 0 \},\\ C_n&=\{x \in \bar{M}_K ~|~ f^n(x)=f^{n-1}(x) \}. \end{align*} Here $f^n=f \circ f \circ \cdots \circ f$ ($n$ times) is the iteration of $f$.

In fact, $A_n$ is the set of Torsion points of $f$ of exactly $n$-th iteration while $B_{n,m}$ and $C_n$ contains preperiodic points of $f$.

Question:

When is there a bijective correspondence between the sets $A_n$ and $B_{n,m}$ on some condition of $m$? Is there a bijective correspondence between the sets $A_n$ and $C_n$ ?

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    $\begingroup$ Do you really just want a bijective correspondence between those sets? Aren't they all infinite countable sets, so they have the same cardinality. Also, instead of putting "arithmetic dynamics" in the title, it would be better as a tag. (And the set-theory tag probably isn't needed, if you're limited to 4 tags.) $\endgroup$ Commented Oct 22, 2020 at 17:36
  • $\begingroup$ @JoeSilverman, Thank you for your comment. Sorry I forgot to assume weierstrass degree of $f$. Here $\text{wideg}(f)=p^r$. Further, I have fixed $n$ whenever I am saying $A_n$, $B_{n,m}$ or $P_n$. There can be only finitely many torsion points of each itearates of $f$ in the maximal ideal $\bar{M}_K$. Probably, you are talking about the union of $A_n$ or of $B_{n,m}$, which are ofcourse infinitely countable. $\endgroup$
    – MAS
    Commented Oct 22, 2020 at 18:47

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