The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Example 10.143.3. Let $p$ be a prime number. The ring extension

\[ R = \mathbf{Z}[1/p] \subset R' = \mathbf{Z}[1/p][x]/(x^{p - 1} + \ldots + x + 1) \]

has the following property: For $d < p$ there exist elements $\alpha _0, \ldots , \alpha _{d - 1} \in R'$ such that

\[ \prod \nolimits _{0 \leq i < j < d} (\alpha _ i - \alpha _ j) \]

is a unit in $R'$. Namely, take $\alpha _ i$ equal to the class of $x^ i$ in $R'$ for $i = 0, \ldots , p - 1$. Then we have

\[ T^ p - 1 = \prod \nolimits _{i = 0, \ldots , p - 1} (T - \alpha _ i) \]

in $R'[T]$. Namely, the ring $\mathbf{Q}[x]/(x^{p - 1} + \ldots + x + 1)$ is a field because the cyclotomic polynomial $x^{p - 1} + \ldots + x + 1$ is irreducible over $\mathbf{Q}$ and the $\alpha _ i$ are pairwise distinct roots of $T^ p - 1$, whence the equality. Taking derivatives on both sides and substituting $T = \alpha _ i$ we obtain

\[ p \alpha _ i^{p - 1} = (\alpha _ i - \alpha _1) \ldots \widehat{(\alpha _ i - \alpha _ i)} \ldots (\alpha _ i - \alpha _1) \]

and we see this is invertible in $R'$.


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