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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

  1. $d\vert m(d\cdot k) - m(d)$
  2. $d \mapsto (m(d)\mod d)$ 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.

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  • $\begingroup$ By $(m(d) \mod d)$ I mean the remainder of division after dividing $m(d)$ by $d$. This condition rules out "non-triviality" of the function $m$. In particular $m(d) = d$ does not fulfill this condition since $(m(d)\mod d)$ would be constant zero. $\endgroup$
    – Peter Wildemann
    Commented Mar 2, 2018 at 23:52
  • $\begingroup$ Not necessarily. $\endgroup$
    – Peter Wildemann
    Commented Mar 3, 2018 at 0:00
  • $\begingroup$ I think you misunderstand something. We map $d$ onto $(m(d)\mod d)$ and consider the image under all $d \in \mathbb{N}$. This image should contain at least two elements. $\endgroup$
    – Peter Wildemann
    Commented Mar 3, 2018 at 0:04
  • $\begingroup$ Then isn't $(2)$ equivalent to the simpler statement: $d\not\mid m(d)$ for some $d\in\mathbb{N}$? $\endgroup$
    – quasi
    Commented Mar 3, 2018 at 0:11
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    $\begingroup$ First fix a prime $\ell$. Select an $\ell$-adic integer $m_\ell=\sum_{i=0}^\infty a_{i,\ell}\ell^i$. For each exponent $t$ set $m(\ell^t)=m_\ell$ (may want to reduce it modulo $\ell^t$ because it makes no difference). If $d=\prod_j \ell_j^{t_j}$ is the factorization of $d$, let $m(d)$ be the unique solution (modulo $d$) to the congruences $m(d)\equiv m(\ell_j^{t_j})\pmod {\ell_j^{t_j}}$ for all $j$. This way your condition $(1)$ is automatic. The resulting automorphism is not a power of Frobenius if for at least one prime $\ell$ there are infinitely many non-zero coefficients in $m_\ell$. $\endgroup$
    – Jyrki Lahtonen
    Commented Mar 3, 2018 at 5:22

1 Answer 1

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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.

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