Let $p > 2$ be a prime and $q = p^r$ for some $r \in \mathbb{Z}^+$. I will assume that all roots of unity lie in $\mathbb{C}_p^{\times}$. Let $\zeta$ a primitive $p$-th root of unity. Let $Tr : \mathbb{F}_q \to \mathbb{F}_p$ be the trace. Also, denote $\pi$ to be the maximal prime in $\mathbb{Z}_p[\zeta]$ such that $\pi^{p-1} = -p$.

For a multiplicative character $\psi: \mathbb{F}_q \to \mathbb{C}_p^{\times}$ ($\psi(0) = 0$), the Gauss sum for $\psi$ is defined to be \begin{align*} G(\psi) = \sum_{c \in \mathbb{F}_q} \psi(c) \zeta^{Tr_{\mathbb{F}_q/\mathbb{F}_p}(c)} \end{align*}

I feel that the following should be true:

If $\psi$ is order 2 and $r =1$ then $G(\psi) = \pi^{(p-1)/2}$ or $-i\pi^{(p-1)/2}$, where $i \in \mathbb{C}_p$ is a solution to $X^2 + 1 = 0$.

My thought is that $\pi^{(p-1)/2}$ is playing the role of $i\sqrt{p}$ in the traditional value for the Gauss sum (where we view everything taking place in $\mathbb{C}$). Is this correct? Or is there something more subtle going on that I'm missing?


2 Answers 2


That $G(\psi,\zeta)^2 = \psi(-1)p$ is pure algebra, so it holds in $\mathbf C$ or $\mathbf C_p$ or any other field not of characteristic $2$ that contains a nontrivial $p$th root of unity.

You could write down a $p$-adic formula for your quadratic Gauss sum using the Gross–Koblitz formula.

First let's normalize the link between your nontrivial $p$th root of unity and your choice of $\pi$ such that $\pi^{p-1} = -p$. To each $\pi$ there is a unique nontrivial $p$th root of unity $\zeta$ such that $\zeta \equiv 1 + \pi \bmod \pi^2$, where the congruence means $\lvert\zeta - (1 + \pi)\rvert_p \leq \lvert\pi\rvert_p^2$, or equivalently $\lvert\zeta - (1 + \pi)\rvert_p < \lvert\pi\rvert_p$ since $\mathbf Q_p(\pi) = \mathbf Q_p(\zeta)$. Write the $\zeta$ fitting that congruence mod $\pi^2$ as $\zeta_{\pi}$.

Every character of $\mathbf F_q^\times$ with values in $\mathbf C_p$ is a power of the Teichmüller character $\omega_q$ (interpret $\mathbf F_q$ as $\mathbf Z_p[\zeta_{q-1}]/(p)$). For the Gross–Koblitz formula it is convenient to write characters of $\mathbf F_q^\times$ as powers of $\omega_q^{-1}$, say as $\omega_q^{-k}$ for $0 \leq k < q-1$. The quadratic character $\psi$ of $\mathbf F_q^\times$ is $\omega_q^{(q-1)/2} = \omega_q^{-(q-1)/2}$, so $k = (q-1)/2$. Let the base $p$ expansion of $k$ be $d_0 + d_1p + \cdots + d_{f-1}p^{f-1}$. When $k = (q-1)/2 = (p^f-1)/2$, all of its base $p$ digits are $(p-1)/2$, so the sum of the base $p$ digits is $f(p-1)/2$.

The Gross–Koblitz formula for the quadratic character $\psi$ says $$ -G(\psi,\zeta_\pi) = \pi^{f(p-1)/2}\Gamma_p\left(\frac{(p-1)/2}{q-1}\right)^f, $$ where $\Gamma_p$ is Morita's $p$-adic Gamma-function. Note the minus sign on the left side: normalizing Gauss sums with an overall minus sign is reasonable for various purposes, like here and in the Hasse–Davenport relation. On the right side of the formula above, $\pi^{f(p-1)/2}$ is a square root of $\pi^{f(p-1)} = (-p)^f = (-1)^fq$.

$\newcommand\sgn{\genfrac(){}{}}$In the special case $q = p$ (so $f = 1$), you're working with the classical quadratic Gauss sum for $\mathbf F_p$ and the Legendre symbol. In this case $$ -G\left(\sgn\cdot p,\zeta_\pi\right) = \pi^{(p-1)/2}\Gamma_p\left(\frac{1}{2}\right), $$ where $\pi^{(p-1)/2}$ is a square root of $\pi^{p-1} = -p$. For $p > 2$ it is known that $\Gamma_p(1/2)^2 = -\sgn{-1}p$, so if you square the right side above then you get $\pi^{p-1}\Gamma_p(1/2)^2 = -p(-\sgn{-1}p) = \sgn{-1}p p$, which is the formula for the square of the mod $p$ quadratic Gauss sum that I mentioned at the start of this answer (when $q = p$).

  • $\begingroup$ Thank you for your answer. This question came up because I am trying to learn about a certain value of $\Gamma_p$ (via Gross-Koblitz) $\Gamma_p(1- 1/4)^2 = \pi^{(p-1)/2}p^{-2}G(\omega^{(p^2-1)/4} \mod p^2 (p \equiv 1 \mod 4$), which involves a quartic gauss sum. I use Hasse-Davenport to get that $G(\omega^{(p^2 -1)/4}) = G^2(\omega^{(p -1)/4}) = G(\omega^{(p -1)/2})J(\omega^{(p -1)/4})$, which is now a quadratic Gauss sum, and a Jacobi sum that I have successfully calculated mod $p^2$. Things aren't matching up when I check my answer in sage though, so this is a part of my trouble shooting. $\endgroup$ Mar 8, 2022 at 22:25
  • $\begingroup$ TeX note: there is a baroque syntax for 'generalised fractions' using the 6-argument command \genfrac. To get the Legendre symbol $\genfrac(){}{}a b$, you can use \genfrac(){}{}a b, and TeX will take care of things like sizing the delimeters for you. I edited accordingly. (To get some tiny flavour of what is possible with \genfrac, binomial coefficients can be typeset using $\genfrac(){0pt}{}a b$ \genfrac(){0pt}{}a b, although here of course we have the pleasant shorthand \binom a b.) $\endgroup$
    – LSpice
    Mar 8, 2022 at 23:53

$\DeclareMathOperator\sgn{sgn}$Just as in the complex case, we have that $G(\sgn, \chi)$ lies in $\mu_4(\mathbb C_p)\sqrt q$.

We have that $G(\sgn, \chi)$ equals $\sum_{c \in \mathbb F_q} \chi(c^2)$, where $\chi : \mathbb F_q \to \mathbb C_p^\times$ is the character you wrote down. We may define more generally $\mathfrak G(\theta, \chi) = \sum_{v \in V} \chi(\theta(v))$, where $\chi$ is any additive character and $(V, \theta)$ is a quadratic space over $\mathbb F_q$. (I'd prefer to use $q$ for a quadratic form, but it's taken!) Then $\mathfrak G(\theta, \chi)$ is multiplicative for orthogonal sums, and $$ \mathfrak G(\theta, \chi) = \sum_{(x, y) \in \mathbb F_q^2} \chi(x y) = \sum_{y \in \mathbb F_q^2} 1 + \sum_{x \ne 0} \sum_{y \in \mathbb F_q} \chi(x y) $$ equals $q$ for $(\mathbb F_q^2, \theta)$ a hyperbolic plane and any non-trivial character $\chi$. Since $(\mathbb F_q, \sigma) \oplus (\mathbb F_q, -\sigma)$ is a hyperbolic plane (where $\sigma : c \mapsto c^2$ is the squaring form), we have that $\mathfrak G(\sigma, \chi)\mathfrak G(-\sigma, \chi)$ equals $q$. Since $\mathfrak G(-\sigma, \chi) = G(\sgn, \chi_{-1})$ clearly equals $\sgn(-1)G(\sgn, \chi)$, it follows that $G(\sgn, \chi)^2$ equals $\sgn(-1)q$.


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