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$ \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\VP}{V.P.} %\DeclareMathOperator{\CCC}{C^1} \DeclareMathOperator{\contt}{C} \DeclareMathOperator{\PCC}{PC} \DeclareMathOperator{\re}{Re} \DeclareMathOperator{\im}{Im} %\DeclareMathOperator{\dd}{d} \newcommand{\ph}{\varphi} \newcommand{\inth}{\int_{-h}^0} \newcommand{\teta}{\theta} \newcommand{\R}{\mathbb{R}} \renewcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\<}{\leq} \renewcommand{\>}{\geq} \renewcommand{\d}{\mathrm{d}} \newcommand{\Int}{\int\limits} \newcommand{\Sum}{\sum\limits} \newcommand{\Max}{\max\limits} \newcommand{\Sup}{\sup\limits} \newcommand{\Lim}{\lim\limits} \newcommand{\lmin}{\lambda_{\min}} \newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}} \newcommand{\pc}[4]{\PCC_{#1}^{#2}\br{#3,#4}} \newcommand{\mh}{\mathfrak{h}} \newcommand{\bh}{\mathbf h} \newcommand{\bN}{\mathbf N} \newcommand{\eps}{\varepsilon} \newcommand{\dt}{\mathrm{d}t} \newcommand{\dd}{\mathrm{d}} \newcommand{\x}{\textup{x}} \newcommand{\y}{\textup{y}} \newcommand{\w}{\textup{w}} \newcommand{\al}{\alpha} \newcommand{\ti}{\times} \newcommand{\D}{\Delta} \newcommand{\del}{\delta} \newcommand{\til}[1]{\widetilde{#1}} %\newcommand{\CC}[2]{\CCC\br{#1,#2}} %\newcommand{\cc}[3]{\CCC_#1\br{#2,#3}} \renewcommand{\c}[4]{\contt_{#1}^{#2}\br{#3,#4}} \newcommand{\F}[1]{\mathbb{F}_{#1}}$ Consider finite field $\F{p},$ irreducible $q(x)\in\F{p}[x]$ with $\deg q=n\>1$, and mapping $$ \ph:~\F{p}[x]/(q(x))\to\F{p}[x]/(q(x)),~[f(x)]\mapsto [f(x^p)]\quad\forall f(x)\in\F{p}[x]. $$ I know that $\ph$ is well-defined mapping and is linear transformation between vector spaces over $\F{p}$. I'd like to find the minimal polynomial of $\ph.$

The solution to this problem claims that $g(x)=x^n-1\in\F{p}[x]$ is the answer. I have that:

  1. $\F{p}[x]/(q(x))\simeq\F{p^n}$;
  2. $y^{p^n}=y\quad\forall y\in\F{p^n}$;
  3. Degree of a polynomial over field is not less than number of its roots. Therefore $\deg f\> p^n$ if all elements of $\F{p^n}$ are roots of $f(x)\in\F{p}[x]$ with $\deg f\> 1.$
  4. $\Sum_{j=1}^m f_j(x)^{p}=\left(\Sum_{j=1}^m f_j(x)\right)^{p}\quad\forall \{f_j(x)\}_{j=1}^m\subset\F{p}[x],~\forall m\in\N\setminus\{0\}$;
  5. $f(x^{p})=f(x)^{p}\quad\forall f(x)\in\F{p}[x]$;
  6. $\ph([f(x)])=[f(x)]^p\quad\forall f(x)\in\F{p}[x]$;
  7. $\ph^n:~[f(x)]\mapsto [f(x)]\quad\forall f(x)\in\F{p}[x]$;
  8. Minimal polynomial divides characteristic one and any annihilating polynomial.
  9. Irreducible polynomial has a root iff it is linear. Therefore $g(x)$ is reducible (since it has root $1$). But why factors of $g(x)$ are not annihilating at $\ph$?

Seventh point implies that $g(x)=x^n-1$ is annihilating polynomial of $\ph.$ Why the third one gives here that $g(x)$ is minimal?

Dmitry
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  • You need the polynomial $q(x)$ to be irreducible for otherwise the quotient ring will not be a field. – Jyrki Lahtonen Mar 30 '22 at 19:34
  • If $g(x)$ were not the minimal polynomial, then the minimal polynomial $m(x)$ would have degree $<n$. But in that case $m(\phi)$ would be a polynomial of degree $<p^n$. Yet, it will have $p^n$ zeros in the field $\Bbb{F}_{p^n}$. This is a contradiction and also explains why we needed $q(x)$ to be irreducible. – Jyrki Lahtonen Mar 30 '22 at 19:37
  • I explained this as a key component of this old answer of mine, so perhaps I should just comment this time. – Jyrki Lahtonen Mar 30 '22 at 19:40
  • @JyrkiLahtonen My bad, forgot to mention that $q$ is irreducible. Thank you for your comments! Now it is evident that I should have considered $m(\phi(\cdot))$ as polynomial over $\mathbb{F}_{p^n}.$ – Dmitry Apr 06 '22 at 10:51

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