Endomorphisms of $p$-adic multiplication?

Question

I'm interested in characterizing endomorphisms of $p$-adic integer multiplication, when $p>2.$ The $p$-adic integers with multiplication do not form a group, they form a monoid, but we can still talk about endomorphisms in this setting.

That being said, I have (what I believe is) such a characterization. I will post it as an answer to this question. I invite others to post their solutions.

Answer 1

The endomorphisms $\phi$ are all of the form $$\phi(zp^k) = \mu^mb^k\exp(a\log(z/\mu)),$$ where

• $z$ is a unit,
• $\mu$ is the $p-1$ root of unity such that $\mu = z\operatorname{mod} p,$
• $b, a\in\mathbb Z_p,$
• $m \in \mathbb Z.$

Proof: Let $\phi:\mathbb Z_p\to\mathbb Z_p$ be an endomorphism. $\phi$ must send units to units, and nonunits can go anywhere. The nonunits coincide with the ideal generated by $p.$ Therefore, $\phi$ can be completely characterized by $\phi|_{U(\mathbb Z_p)}$ and $\phi(p),$ where $U(\mathbb Z_p)$ denotes the set of units. Conversely, any endormorphism over $U(\mathbb Z_p)$ and any choice $b=\phi(p)$ gives an endomorphism of $\mathbb Z_p.$ Therefore, it suffices to characterize the endomorphisms of $U(\mathbb Z_p).$

Now we can prove that $U(\mathbb Z_p)$ and $M_{p-1}\oplus\mathbb Z_p^+,$ where $M_{p-1}\subset U(\mathbb Z_p)$ denotes the subgroup of $p-1$ roots of unity and $\mathbb Z_p^+$ is considered an additive group, are isomorphic. (See this question.) The isomorphism is given by $z\mapsto (\mu, \log(z/\mu)/p),$ where once again $\mu\in M_{p-1}$ is chosen so that $\mu=z\operatorname{mod} p.$ The endomorphisms of $M_{p-1}$ are given by $\mu\mapsto\mu^m$ for any $m,$ since it is cyclic. Therefore, it suffices to characterize endomorphisms of $\mathbb Z_p^+.$

Let $\psi$ be an endomorphism of $\mathbb Z_p^+.$ Now $\psi(n) = na$ for all finite $n\in\mathbb Z_p$ and for some $a\in\mathbb Z_p$ (namely $a = \psi(1)$). This is an easy inductive proof, so I will omit it. For infinite $n\in\mathbb Z_p,$ denote $n = \cdots d_2d_1d_0$ and $S_k = d_kd_{k-1}\cdots d_0.$ Notice that

$$\lim_{k\to\infty}\left|\psi(n) - \psi\left(S_k\right)\right|_p = \lim_{k\to\infty} |\psi(n - S_k)|_p.$$ Since $n - S_k$ is divisible by $p^k,$ then the limit is 0, proving that $\psi(n) = an$ for infinite $n$ as well. Conversely, any choice for $a$ yields an endomorphism.

If you tie all this together, you will see that the characterization I gave above is the correct one.