# How are Gauss sums the analogue of the gamma function for finite fields?

I've seen this statement on the internet in a few places and I don't really get the connection. Would anyone mind fleshing out the details? Thanks.

Paraphrasing from Wikipedia:

The Gamma function is the integral of the additive character $e^{-x}$ against the multiplicative character $x^z$ with respect to the Haar measure $\frac{dx}{x}$ on the Lie group $\mathbb{R}^+$.

Similarly, a Gauss sum on a finite commutative ring $R$ is given by: $$\sum_{r\in R^\times} \chi(r) \psi(r)$$

Where $\chi:R^\times\to\mathbb{C}$ is a multiplicative character, and $\psi:R\to\mathbb{C}$ is an additive character.

This is an integral over the discrete Lie group $R^\times$ with the counting measure.

Note that Haar measure is unique up to a constant, so the appearance of $\frac{dx}{x}$ on $\mathbb{R}^+$ is less arbitrary than it might appear.

• Thank you. Would you mind explaining what it means to be a character with respect to a measure, or directing me towards some material that explains it? – Vik78 May 3 '16 at 0:57
• @Vik78 The character is not with respect to a measure, the integral is with respect to a measure. The characters are just (continuous) group homomorphisms $(R,+)\to (\mathbb{C},\cdot)$ and $(R^\times,\cdot) \to (\mathbb{C},\cdot)$. – Slade May 3 '16 at 1:59