Jacobi sum

Remarks

Some books set the trivial character equal to $1$ at $0$; this changes $J$ only when $\chi$ or $\lambda$ is trivial. If $\chi$, $\lambda$ and $\chi\lambda$ are nontrivial, then $J(\chi, \lambda) = g(\chi)g(\lambda)/g(\chi\lambda)$, where $g$ is the Gauss sum with a fixed nontrivial additive character, and so $\abs{J(\chi, \lambda)} = \sqrt{q}$.

Example: let $q = 5$ and $\chi = \lambda$ the Legendre symbol modulo $5$. The terms with $a = 0$ and $a = 1$ vanish. For $a = 2, 3, 4$ the products $\chi(a)\chi(1-a)$ are $(-1)(1) = -1$, $(-1)(-1) = 1$ and $(1)(-1) = -1$. So $J(\chi, \chi) = -1$; here $\chi^2$ is trivial and the general rule gives $J(\chi, \chi^{-1}) = -\chi(-1) = -1$.

Non-example: the sum $\sum_{a} \chi(a) \lambda(a)$ is not a Jacobi sum, since the second argument is $a$ and not $1 - a$. By orthogonality it equals $q - 1$ if $\chi\lambda$ is trivial and $0$ otherwise.