Additive character

Remarks

Then $\psi(0) = 1$ and $\psi(-x) = \psi(x)^{-1}$. If $R$ is finite, every value $\psi(x)$ is a root of unity, so $\psi$ is a character of the abelian group $(R, +)$ with values in the unit circle. For topological rings, such as $\R$ or the $p$-adic numbers, one asks $\psi$ to be continuous with values in the unit circle.

Example: for a prime $p$, the map $\psi(x) = e^{2\pi i x / p}$ on $\Z/p\Z$ is well defined, since changing $x$ by a multiple of $p$ does not change the value. It is an additive character, because $e^{2\pi i (x + y)/p} = e^{2\pi i x/p} e^{2\pi i y/p}$.

Example: for the finite field $\F_q$ with $q = p^m$, the map $\psi(x) = e^{2\pi i \operatorname{Tr}(x) / p}$, where $\operatorname{Tr} \colon \F_q \to \F_p$ is the field trace, is a nontrivial additive character, the canonical one. Every additive character of $\F_q$ is $x \mapsto \psi(ax)$ for exactly one $a \in \F_q$.

Non-example: for an odd prime $p$, the map $x \mapsto e^{2\pi i x^2/p}$ on $\Z/p\Z$ is not an additive character. At $x = y = 1$ it gives $e^{8\pi i/p}$ for $x + y$, but $e^{2\pi i/p} \cdot e^{2\pi i/p} = e^{4\pi i/p}$, and these differ because $2/p$ is not an integer.