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 $R'$.

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03GF. Beware of the difference between the letter 'O' and the digit '0'.