Prelude (Disclaimer: This paragraph has no relevance to the actual question, but tries to illustrate the motivation behind my actual elementary question.) During the last days I wondered about the construction of certain Automorphisms of the Galois field extension $\overline{\mathbb{F}}_p\vert \mathbb{F}_p$, where $\mathbb{F}_p$ denotes the field of prime order $p$ and $\overline{\mathbb{F}}_p$ its algebraic closure. In particular I tried to construct a counterexample to the Fundamental Theorem of Galois Theory for non-finite Galois field extensions. For this I showed that the fixed point set of the Automorphism subgroup $\langle\mathrm{Fr}\rangle$ (generated by the Frobenius Automorphism $\mathrm{Fr}\colon x \mapsto x^p$) is only base field $\mathbb{F}_p$. If one now shows that this is a proper subgroup of the Galois group, we have a counterexample to the classical Fundamental Theorem of Galois Theory. Using some field theory one can show that $\overline{\mathbb{F}}_p \cong \bigcup_{d\in\mathbb{N}}\mathbb{F}_{p^d}$ and I reduced the problem to the construction of a Automorphism $$ \psi\colon \overline{\mathbb{F}}_p\to\overline{\mathbb{F}}_p, \psi\vert_{\mathbb{F}_{p^d}} = \mathrm{Fr}^{m(d)} $$ where $m\colon \mathbb{N} \to \mathbb{Z}$ is a function such that $\psi$ is well-defined and $\psi \notin\langle\mathrm{Fr}\rangle$. This question can be reduced onto a simple artihmetic problem which forms my actual question.
Question Is there a function $m\colon \mathbb{N} \to \mathbb{Z}$ such that
- $d\vert m(d\cdot k) - m(d)$
- $d \mapsto (m(d)\mod d)$ takes at least two different values?
Remark I have already resolved my original problem, namely the construction of some Automorphism as above which is no constant multiple of the Frobenius Automorphism, but by slightly different methods. Nevertheless, I am curious if a function $m$ as given above exists.
Unless I'm missing something, there are lots of easy ways to satisfy the conditions.
As a simple example, if $m:\mathbb{N}\to \mathbb{Z}$ is given by $$m(n) = n+1$$ then, for all positive integers $d,k$, we have $$m(kd) - m(d) = (kd+1) - (d+1) = (k-1)d$$ which is a multiple of $d$.
Thus, condition $(1)$ is satisfied.
Then, letting $r:\mathbb{N}\to \mathbb{Z}$ be given by $$r(d) = (m(d)\;\text{mod}\;d)$$ we have \begin{align*} r(1) &= (2\;\text{mod}\;1)=0\\[4pt] r(2) &= (3\;\text{mod}\;2)=1\\[4pt] \end{align*} so $r$ is not constant.
Thus, condition $(2)$ is satisfied.