Quadratic Gauss Sum¶

Example

Solve \(x^7-1 = 0\).
It's obvious that \(x^7-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)\). So we just need to solve \(x^6+x^5+x^4+x^3+x^2+x+1 = 0\). And that means we only need to solve \(x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3} = 0\). Let's set \(y = x + x^{-1}\). So the equation is converted into \(y^3+y^2-2y-1 = 0\).
We set \(\omega = e^{\frac{2\pi \mathrm{i}}{3}}\) and suppose that the solutions to the equation are \(y_1, y_2, y_3\). We set \(t(1) = y_1+y_2+y_3\), \(t(\omega) = y_1+y_2 \cdot \omega+y_3 \cdot \omega^2\), \(t(\omega^2) = y_1+y_2 \cdot \omega^2+y_3 \cdot \omega\). Then we have

\[ \begin{pmatrix} t(1) \\ t(\omega) \\ t(\omega^2) \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}. \]

\(t(\omega) \cdot t(\omega^2) = y_1^2+y_2^2+y_3^2+(y_1y_2+y_2y_3+y_3y_1)(\omega+\omega^2) = 7\) So \(t(\omega^2) = \frac{7}{t(\omega)}\).
Try \(t(\omega)^3 = y_1^3 + y_2^3 + y_3^3 + 6y_1y_2y_3 + 3\omega(y_1^2y_2+y_2^2y_3+y_3^2y_1) + 3\omega^2(y_1y_2^2+y_2y_3^2+y_3y_1^2)\). Set \(\alpha = y_1^2y_2+y_2^2y_3+y_3^2y_1, \beta = y_1y_2^2+y_2y_3^2+y_3y_1^2\). So \(t(\omega)^3 = 2 + 3\omega\alpha + 3\omega^2\beta\).
And we have \(\alpha + \beta = -1, \alpha \cdot \beta = -12\). So \(\{\alpha, \beta\} = \{-4, 3\}\).
Assume \(\alpha = 3, \beta = -4\). \(t(\omega)^3 = 14+21\omega\). Choose \(\eta\) such that \(\eta^3 = 14+21\omega\). \(t(\omega) = \eta\), \(t(\omega^2) = \frac{7}{\eta}\), \(t(1) = -1\).
So

\[ \begin{gather} y_1 = \frac{1}{3}(-1 + \eta + \frac{7}{\eta}) \\ y_2 = \frac{1}{3}(-1 + \omega^2\eta + \frac{7}{\omega^2\eta}) \\ y_3 = \frac{1}{3}(-1 + \omega\eta + \frac{7}{\omega\eta}) \end{gather} \]

Main Theorem

Define \(\mathfrak{G}_a = \sum_{t = 0}^{p-1}(\frac{t}{p}) \omega^{at}, \omega = e^{\frac{2\pi \mathrm{i}}{p}}\). Then \(\mathfrak{G}_1^2 = (-1)^{\frac{p-1}{2}}\cdot p = \begin{cases} p, p \equiv 1 \pmod 4 \\ -p, p \equiv -1 \pmod 4 \end{cases}\). And \(\mathfrak{G}_1 = \begin{cases} \sqrt{p}, p \equiv 1 \pmod 4 \\ \sqrt{-p}, p \equiv -1 \pmod 4 \end{cases}\).

Lemma

(i)

\[ \sum_{t = 0}^{p-1} \omega^{at} = \begin{cases} p, \ p \mid a \\ 0, \ p \not \mid a \end{cases} \]

(ii) Consider \(\chi: \mathbb{Z}/p\mathbb{Z} \rightarrow \mathbb{C} - \{0\}\).
(a) \(\chi(a + b) = \chi(a) \chi(b)\).
(b) \(\chi \neq 1 \Rightarrow \sum_{a = 0}^{p-1} \chi(a) = 0\).

(iii) \(\sum_{t = 1}^{p-1}\left(\frac{t}{p}\right) = 0\).

(iv) \(\mathfrak{G}_a = \left(\frac{a}{p}\right)\mathfrak{G}_1\).

Proof

(i) If \(p \mid a\), then \(\omega^{at} = 1\). So \(\sum_{t = 0}^{p-1} \omega^{at} = p\).
If \(p \not \mid a\), then \(\omega^{at} \neq 1\). So \(\sum_{t = 0}^{p-1} \omega^{at} = \dfrac{\omega^{ap} - 1}{\omega^a - 1} = 0\).

(ii) (b) \(\sum_{a = 0}^{p-1} \chi(a + b) = \chi(b) \sum_{a = 0}^{p-1} \chi(a)\), and \(\sum_{a = 0}^{p-1} \chi(a + b) = \sum_{a = 0}^{p-1} \chi(a)\). So \((\chi(b) - 1) \sum_{a = 0}^{p-1} \chi(a) = 0\). Let \(\chi(b) \neq 1\), we have \(\sum_{a = 0}^{p-1} \chi(a) = 0\).

(iii) Set \(\chi(a) = \left(\frac{a}{p}\right)\). So \(\sum_{t = 1}^{p-1}\left(\frac{at}{p}\right) = \chi(a) \sum_{t = 1}^{p-1}\left(\frac{t}{p}\right) = \sum_{t = 1}^{p-1}\left(\frac{t}{p}\right)\). \((a \not \mid p)\) Set \(\chi(a) \neq 1 \Rightarrow \sum_{t = 1}^{p-1}\left(\frac{t}{p}\right) = 0\).

(iv) (a) \(p \mid a\), \(\mathfrak{G}_a = \sum_{t = 0}^{p-1}(\frac{t}{p})\omega^{at} = 0 = \left(\frac{a}{p}\right)\mathfrak{G}_1\).
(b) \(p \not \mid a\), \(\mathfrak{G}_a = \sum_{t = 0}^{p-1}(\frac{t}{p})\omega^{at}\). So \(a^{-1}\) exists. Let \(t \mapsto a^{-1}t\), we have \(\mathfrak{G}_a = \sum_{t = 0}^{p-1}(\frac{a^{-1}t}{p})\omega^{t} = \left(\frac{a}{p}\right)\mathfrak{G}_1\). \((\left(\frac{a^{-1}}{p}\right) \cdot \left(\frac{a}{p}\right) = 1)\)

Proof of Main Theorem

(a) If \(p \not \mid a, \mathfrak{G}_a \cdot \mathfrak{G}_{-a} = \left(\frac{a}{p}\right)\mathfrak{G}_1 \cdot \left(\frac{-a}{p}\right)\mathfrak{G}_1 = (-1)^{\frac{p-1}{2}}\mathfrak{G}_1^2\).
(b) If \(p \mid a, \mathfrak{G}_a = 0\).
Sum them up, we have \(\sum_{a = 0}^{p-1} \mathfrak{G}_a \cdot \mathfrak{G}_{-a} = (-1)^{\frac{p-1}{2}}\cdot (p-1) \cdot \mathfrak{G}_1^2\).
Then we consider calculating \(\sum_{a = 0}^{p-1} \mathfrak{G}_a \cdot \mathfrak{G}_{-a}\). \(\mathfrak{G}_a \cdot \mathfrak{G}_{-a} = \sum_{x = 0}^{p-1} \sum_{y = 0}^{p-1} \left(\frac{x}{p}\right) \left(\frac{y}{p}\right) \omega^{a(x-y)}\).

\[\begin{align} \sum_{a = 0}^{p-1} \mathfrak{G}_a \cdot \mathfrak{G}_{-a} & = \sum_{x = 0}^{p-1} \sum_{y = 0}^{p-1} \left(\frac{x}{p}\right) \left(\frac{y}{p}\right) \sum_{a = 0}^{p-1} \omega^{a(x-y)} \\ &=\sum_{x, y} \left(\frac{x}{p}\right) \left(\frac{y}{p}\right) \sum_{a = 0}^{p - 1} \delta_{x, y} \\ &= \sum_{x = 0}^{p-1} p \cdot \left(\frac{x^2}{p}\right) \cdot 1 = p(p-1) \end{align}\]

Then \(p(p-1) = (-1)^{\frac{p-1}{2}}\cdot (p-1) \cdot \mathfrak{G}_1^2\). So \(\mathfrak{G}_1^2 = (-1)^{\frac{p-1}{2}}\cdot p\).

Collary

\(\pm \sqrt{\pm p} = \mathbb{Q}\)-linear combination of \(e^{\mathrm{i}\frac{2\pi}{p}}\).