# Value of the quadratic Gauss sum viewed in $\mathbb{C}_p$

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?

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 $$q = p$$).

• 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. Mar 8 at 22:25
• 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.) Mar 8 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 $$\sgn(-1)q$$.