A note on Jacobi sums. III
The notation and convention remain the same as in part II (cf. preceding review). In the present paper, a new proof is given to the fact that \(\mathbb{Q} (J({\mathfrak p}))= \mathbb{Q}({\mathfrak p})\) if \(f\) is odd. For \(\omega\in \mathbb{Z} [{\mathcal G}]\), define the quotient space \(H({\mathfrak p}^ \omega)= {\mathcal G} ({\mathfrak p}^ \omega)/ {\mathcal G} ({\mathfrak p})\). Now we identify \({\mathcal G}\) with the cyclic group \(\mathbb{F}^*_ \ell\). Then for any \(\omega= \sum_{t\in \mathbb{F}^*_ \ell} a(t) \sigma_ t\in \mathbb{Z}[{\mathcal G}]\), \(H({\mathfrak p}^ \omega)\) is simply given by \[ H ({\mathfrak p}^ \omega)= \Bigl\{ s\in \mathbb{F}\mid \chi(s)=1 \text{ for all } \chi\in \widehat {\mathbb{F}} \text{ such that } \sum_{t\in \mathbb{F}^*_ \ell} a(t) \overline{\chi} (t) \neq 0\Bigr\}. \] Here \(\mathbb{F}\) stands for \(\mathbb{F}^*_ \ell/ (\mathbb{F}^*_ \ell )^ g\). The proof of the statement then boils down to show that \(H({\mathfrak p}^ \omega) =1\) if \(f\) is odd. The new proof is somewhat inspired by Stickelberger's theorem.
scientific article; zbMATH DE number 549030
scientific article