Exercise 3.2.25* ($\genfrac{(}{)}{}{}{a}{p} = \prod_{h \in \mathscr{H}} \frac{\sin 2\pi ah/p}{\sin 2\pi h/p}$)

Proof.

Let $p$ be an odd prime. Note first that, as the name suggests, a one-half set has $(p-1)∕2$ elements: since $h\in \mathscr{H}⟺-h\notin \mathscr{H}$, the map

is a bijection, thus $|\mathscr{H}|=\left|{\text{∁}}_{{𝔽}_p^{\text{∗}}}\mathscr{H}\right|$, therefore $|\mathscr{H}|=(p-1)∕2$.

(a)

The first property is a generalization of Gauss’ lemma (Theorem 3.2). We mimic the proof of Theorem 3.2.

Suppose that the integer $a$ is prime to $p$ (here we will write $\mathit{ah}=\bar{a}h$ for $h\in \mathbb{Z}∕\mathit{p\mathbb{Z}}$). Let $h_1,h_2,\dots ,h_{\nu }$ be the distinct elements of $\mathscr{H}$ satisfying $r_i=a{h}_i\in 𝒦$, $(1\le i\le \nu )$, and let $l_1,l_2,\dots ,l_n$ denote the other elements of $\mathscr{H}$, so that $s_j=a{l}_j\in \mathscr{H},(1\le j\le n)$. Then

$$\begin{array}{lll}\hfill \mathscr{H}=\{h_1,h_2,\dots ,h_{\nu },l_1,l_2,\dots ,l_n\}.& & \hfill \text{(1)}\end{array}$$

Since $\mathrm{Card}\mathscr{H}=(p-1)∕2$, $\nu +n=(p-1)∕2$.

The map $\phi :{𝔽}_p^{\text{∗}}\to {𝔽}_p^{\text{∗}}$ defined by $\phi (x)=\mathit{ax}$ is injective (one-to-one), since $p\nmid a$ , so the elements $r_1,r_2,\dots ,r_{\nu }$ are distinct. Similarly $s_1,s_2,\dots ,s_n$ are distinct. Also no $s_j$ is equal to $-r_i$, otherwise $a{h}_i=-a{l}_j$, where $a\ne 0$, so $h_i=-l_j$, but $h_i\in \mathscr{H}$, and $-l_j\notin \mathscr{H}$ by definition of a one-half set. This contradiction shows that $-r_i\ne s_j$ for all $i,j$, $1\le i\le \nu ,1\le j\le n$. Therefore the set $\{-r_1,-r_2,\dots ,-r_{\nu },s_1,s_2,\dots ,s_n\}\subseteq \mathscr{H}$ contains $(p-1)∕2$ distinct elements, where $\mathrm{Card}\mathscr{H}=(p-1)∕2$, hence

$$\begin{array}{lll}\hfill \mathscr{H}=\{-r_1,-r_2,\dots ,-r_{\nu },s_1,s_2,\dots ,s_n\}=\{-a{h}_1,-a{h}_2,\dots ,-a{h}_{\nu },a{l}_1,\dots ,a{l}_n\}.& & \hfill \text{(2)}\end{array}$$

Put $P={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{h\in \mathscr{H}}h$, the product of all elements in $\mathscr{H}$. The equalities (1) and (2) give

$$P=\underset{h\in \mathscr{H}}{\prod }h={(-1)}^{\nu }{a}^{(p-1)∕2}\underset{i=1}{\overset{\nu }{\prod }}{h}_i\underset{j=1}{\overset{n}{\prod }}{l}_j={(-1)}^{\nu }{a}^{(p-1)∕2}P.$$

Simplifying by $P\ne 0$, we obtain ${(-1)}^{\nu }=a^{(p-1)∕2}$, thus ${(-1)}^{\nu }\equiv a^{(p-1)∕2}\equiv \left(\frac{a}{p}\right)(\mathit{mod}p)$, where ${(-1)}^{\nu }=\pm 1,\left(\frac{a}{p}\right)=\pm 1$, and $p$ an odd prime, hence

$$\left(\frac{a}{p}\right)={(-1)}^{\nu }.$$

(b)

Now we show that $\mathit{a\mathscr{H}}$ and $\mathit{a𝒦}$ are complementary one-half sets. Since $\phi :x\mapsto \mathit{ax}$ is bijective, $\phi (\mathscr{H})=\mathit{a\mathscr{H}}$ and $\phi (𝒦)=\mathit{a𝒦}$ are complementary sets.

If $k\in \mathit{a\mathscr{H}}$, then $k=\mathit{ah}$, where $h\in \mathscr{H}$. Since $\mathscr{H}$ is a one-half set, $-h\notin \mathscr{H}$, thus $-h\in 𝒦$, and $-\mathit{ah}\in \mathit{a𝒦}$, so $-k=-\mathit{ah}\notin \mathscr{H}$. This shows that for all $k\in {𝔽}_p^{\text{∗}}$, $k\in \mathit{a\mathscr{H}}\Rightarrow -k\notin \mathit{a\mathscr{H}}$. Similarly, $k\in \mathit{a𝒦}\Rightarrow -k\notin \mathit{a𝒦}$. Since $\mathit{a\mathscr{H}}$ and $\mathit{a𝒦}$ are complementary sets, this gives $k\notin \mathit{a\mathscr{H}}\Rightarrow -k\in \mathit{a\mathscr{H}}$. So, for all $k\in {𝔽}_p^{\text{∗}}$,

$$k\in \mathit{a\mathscr{H}}⟺-k\notin \mathit{a\mathscr{H}}.$$

This shows that $\mathscr{H}$ is a one-half set, and so is $𝒦$. We have proved that $\mathit{a\mathscr{H}}$ and $\mathit{a𝒦}$ are complementary one-half sets.

(c)

Put $\zeta =e^{2\mathit{i\pi }∕p}$. Since ${\zeta }^p=1$, ${\zeta }^{a+\mathit{kp}}={\zeta }^a$, thus ${\zeta }^a$ doesn’t depend of the choice of the representative $a\in \alpha$ where $\alpha$ is an element of $\mathbb{Z}∕\mathit{p\mathbb{Z}}$, so we can define without ambiguity ${\zeta }^{\alpha }$ for $\alpha \in \mathbb{Z}∕\mathit{p\mathbb{Z}}$ (and similarly $\sin <!--\; FUNCTION\; APPLICATION\; -->2\mathit{\pi a\alpha }∕p$).

Put

$$\mathrm{\Pi }=\underset{h\in \mathscr{H}}{\prod }\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{2\mathit{\pi ah}}{p}\right)}{\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{2\mathit{\pi h}}{p}\right)}.$$

Using $\sin <!--\; FUNCTION\; APPLICATION\; -->x=\frac{e^{\mathit{ix}}-e^{-\mathit{ix}}}{2i}$, we obtain

$$\mathrm{\Pi }=\underset{h\in \mathscr{H}}{\prod }\frac{{\zeta }^{\mathit{ah}}-{\zeta }^{-\mathit{ah}}}{{\zeta }^h-{\zeta }^{-h}}.$$

Morover, using equality (1),

$$\begin{array}{llll}\hfill \underset{h\in \mathscr{H}}{\prod }({\zeta }^{\mathit{ah}}-{\zeta }^{-\mathit{ah}})& =\underset{i=1}{\overset{\nu }{\prod }}({\zeta }^{a{h}_i}-{\zeta }^{-a{h}_i})\cdot \underset{j=1}{\overset{n}{\prod }}({\zeta }^{a{l}_i}-{\zeta }^{-a{l}_i})& \hfill & \\ \hfill & ={(-1)}^{\nu }\underset{i=1}{\overset{\nu }{\prod }}({\zeta }^{-a{h}_i}-{\zeta }^{a{h}_i})\cdot \underset{j=1}{\overset{n}{\prod }}({\zeta }^{a{l}_i}-{\zeta }^{-a{l}_i}).& \hfill & \end{array}$$

By equality (2), $\mathscr{H}=\{-a{h}_1,-a{h}_2,\dots ,-a{h}_{\nu },a{l}_1,\dots ,a{l}_n\}$, therefore

$$\begin{array}{llll}\hfill \underset{h\in \mathscr{H}}{\prod }({\zeta }^{\mathit{ah}}-{\zeta }^{-\mathit{ah}})& ={(-1)}^{\nu }\underset{h\in \mathscr{H}}{\prod }({\zeta }^{\mathit{ah}}-{\zeta }^{-\mathit{ah}}).& \hfill & \end{array}$$

This gives

$$\mathrm{\Pi }={(-1)}^{\nu },$$

so, using part (a),

$$\underset{h\in \mathscr{H}}{\prod }\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{2\mathit{\pi ah}}{p}\right)}{\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{2\mathit{\pi h}}{p}\right)}=\left(\frac{a}{p}\right).$$