It's well known there exists only one finite field with $p^n$ elements, up to isomorphism, where $p$ is a prime and $n \geq 1$. Let $n = m t$, where $m, t &gt; 1$. Then $F = Z_p[X]/(f(X))$ is a field with $p^n$ elements, provided $f$ is some irreducible polynomial of degree $n$ in $Z_p$. By the same token, $G = Z_p[X]/(g(X))$ is a field with $m$ elements ($g$ irreducible of degree $m$). Then $G$ has $p^m$ elements. Finally let: $H = G[X]/(h(X))$ where $h$ is an irreducible polynomial of degree $t$ with coefficients in $G$. Now, in my understanding $F$ and $H$ have both $p^n$ elements. So my question is: Are $F$ and $H$ isomorphic?
Fraleigh(7ed) Theorem33.12. Let $p$ be a prime and let $n\in\mathbb{Z}^+$. If $E$ and $E&#39;$ are fields of order $p^n$, then $E \simeq E&#39;$.   Proof in the text: Both $E$ and $E&#39;$ have $\mathbb{Z}_p$ as prime field, up to isomorphism. By Corollary 33.6(A finite extension $E$ of a finite field $F$ is a simple extension of $F$), $E$ is a simple extension of $\mathbb{Z}_p$ of degree $n$, so there exists an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Z}_p[x]$ such that $E\simeq \mathbb{Z}_p[x]/ \langle f(x) \rangle$. Because the elements of $E$ are zeros of $x^{p^n}-x$, *we see that $f(x)$ is a factor of $x^{p^n}-x$ in $\mathbb{Z}_p[x]$*. Because $E&#39;$ also consists of zeros of  $x^{p^n}-x$, we see that $E&#39;$ also contains zeros of irreducible $f(x)$ in $\mathbb{Z}_p[x]$. Thus, because $E&#39;$ also contains exactly $p^n$ elements, $E&#39;$ is also isomorphic to $E\simeq \mathbb{Z}_p[x]/\langle f(x) \rangle$.  I don't know why $f(x)$ divides $x^{p^n}-x$. $E$ has a zero $\alpha$ of $f(x)$, but it doesn't need to have all the zeros of $f(x)$. So $f(x)=(x-\alpha)g(x)$ in $E[x]$ and $g(x)$ need not be splitted into linear factors. How can $f(x)$ divide $x^{p^n}-x$?
\nameref breaks for star versions of section/paragraph when titlesec is loaded \nameref breaks for the star versions of section, paragraph when titlesec is loaded. I have a document that numbers paragraphs by default, but I have a few I do not want to number but would like to refer to. These are marked with \paragraph*, however, if the titlesec package is loaded at the same time, it causes nameref to display an empty string (the link box of hyperref is still visible).  MWE:  \documentclass[12pt]{article} \usepackage[sf]{titlesec} %comment this out and it works fine. \usepackage{hyperref} \begin{document} \section{Sections} \label{sec} This is \nameref{sec} \paragraph{A named Paragraph} \label{par} This is \nameref{par} \paragraph*{A starred named Paragraph} \label{par:star} This is \nameref{par:star} \end{document}   I suspect titlesec causes this defect for ALL star variants of sectioning commands but have not tested this. 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