Example 10.147.3. 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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