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Lecture 15 - 2025-03-10
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@@ -1061,4 +1061,239 @@ We get unique factorization, i.e., prime iff irreducible. \[
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\] so \[
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g^2 = 5
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\]
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\end{example}
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\end{example}
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\begin{remark}
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So far, for an odd prime \( p \), we have
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\begin{itemize}
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\item \( \left( \frac{-1}{p} \right) = 1 \iff p \equiv 1 \pmod{4} \)
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\item \( p = x^2 + y^2 \iff p \equiv 1 \pmod{4} \)
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\item \( \left( \frac{2}{p} \right) = 1 \iff p \equiv \pm 1 \pmod{8} \)
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\end{itemize}
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\end{remark}
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\begin{example}
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What is \( \left( \frac{-2}{p} \right) \)? \[
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\left( \frac{-2}{p} \right) = \left( \frac{2}{p} \right) \left( \frac{1}{p} \right)
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\]
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\begin{table}
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\centering
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\begin{tabular}{c|c|c|c}
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\( p \mod 8 \)
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& \( \left( \frac{-1}{p} \right) \)
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& \( \left( \frac{2}{p} \right) \)
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& \( \left( \frac{-2}{p} \right) \) \\
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\hline
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1 & 1 & 1 & 1 \\
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3 & -1 & -1 & 1 \\
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5 & 1 & -1 & -1 \\
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7 & -1 & 1 & -1
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\end{tabular}
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\end{table}
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\end{example}
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\begin{definition}[Gauss Sum]\index{Gauss Sum}
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For \( a \in \F_p \), \( x \in \Z// = \F_p \), define \[
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\psi_a(x) = e^{\frac{2 \pi i ax}{p}}
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\]
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This is non-trivial / injective if and only if \( a \not\equiv 0 \pmod{p} \).
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\end{definition}
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\begin{definition}
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\[
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g_2(a; p) = \sum_{x \in \F_p} \left( \frac{x}{p} \right) e^{\frac{2 \pi i ax}{p}} = \sum_{x \in \F_p} \left( \frac{x}{p} \right) \zeta_p^{ax}
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\]
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\end{definition}
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\begin{theorem}
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\[
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{g_2}^2(a; p) = \left( \frac{-1}{p} \right) p
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\]
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\end{theorem}
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\begin{remark}
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This theorem / identity is purely algebraic. It is true for any field in which \( \zeta_p = e^{\frac{2 \pi i}{p}} \) is defined.
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\end{remark}
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\begin{example}
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Can we figure out \( \left( \frac{5}{p} \right) \)?
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Say \( p \neq 2, 5 \) and \( \zeta_5 \in \F_q \supset \F_p \)
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Then in \( \F_q \), \[
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\sqrt{5} = g_2(5; p) = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4
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\]
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We have \( \sqrt{5} \in \F_p \) if and only if \( (\sqrt{5})^p = \sqrt{5} \).
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We have, \[
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(\sqrt{5})^p = {\zeta_5}^p - {\zeta_5}^{2p} - {\zeta_5}^{3p} + {\zeta_5}^{4p}
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\] and we know that \( \left( \frac{5}{p} \right) \) depends only on \( p \pmod{5} \).
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\begin{itemize}
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\item If \( p \equiv 1 \pmod{5} \), we have \[
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{g_2}^p = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4 = g_2 \implies \left( \frac{5}{p} \right) = 1
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\]
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\item If \( p \equiv 2 \pmod{5} \), we have \[
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{g_2}^p = \zeta_5^2 - \zeta_5^4 - \zeta_5^6 + \zeta_5^8 = - \left( \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4 \right) = -g_2 \implies \left( \frac{5}{p} \right) = -1
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\]
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\item If \( p \equiv 3 \pmod{5} \), we have \[
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\left( \frac{5}{p} \right) = -1
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\]
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\item If \( p \equiv 4 \pmod{5} \), we have \[
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\left( \frac{5}{p} \right) = 1
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\]
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\end{itemize}
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\end{example}
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\begin{lemma}
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\[
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\sum_{x \in \F_p} \psi_a(x) = \begin{cases}
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p & \text{if } a = 0 . \\
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0 & \text{otherwise}
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\end{cases}
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\]
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\end{lemma}
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\begin{proof}
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If \( a = 0 \), then \[
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\psi_a(x) = e^{\frac{2 \pi i 0 x}{p}} = 1 \implies \sum_{x \in \F_p} \psi_a(x) = p
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\]
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If \( a \neq 0 \), then define \[
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S_a = \sum_{x \pmod{p}} \psi_a(x)
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\] so \[
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\psi_a(1) S_1 = \sum_{x \pmod{p}} \psi_a(x) \psi_a(1) = \sum_{x \pmod{p}} \psi_a(x + 1) = \sum_{y \pmod{p}} \psi_a(y) = S_a
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\] so \[
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\psi_a(1) S_a = S_a
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\] and \[
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a \neq 0 \implies \psi_a(1) = \zeta_a \neq 1
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\] so \[
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S_a = 0
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\]
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\end{proof}
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\begin{lemma}
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For \( a \neq 0 \pmod{p} \), \[
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g_2(a; p) = \left( \frac{a}{p} \right) g_2 (1; p)
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\]
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\end{lemma}
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\begin{proof}
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Take \( y = ax \),
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\[
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g_2(a; p) = \sum_{x \pmod{p}} \left( \frac{x}{p} \right) \psi_1(ax) = \sum_{y \pmod{p}} \left( \frac{a^{-1} y}{p} \right) \psi_1(y)
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\] and \( \left( \frac{a^{-1}y}{p} \right) \) is a multiplicative character so \[
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\left( \frac{a^{-1}y}{p} \right) = \left( \frac{a^{-1}}{p} \right) \left( \frac{y}{p} \right)
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\] and we have the expression \[
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\left( \frac{a^{-1}}{p} \right) \sum_{y \pmod{p}} \psi_1(y) = \left( \frac{a}{p} \right) g_2(1; p)
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\]
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\end{proof}
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\begin{remark}
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A well known fact in algebra is that \[
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I = \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}
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\]
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This is an analogue of the Gauss sum.
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\end{remark}
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\begin{lemma}
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\[
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g_2(a; p) = \sum_{x \pmod{p}} e^{\frac{2 \pi i ax^2}{p}}
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\]
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\end{lemma}
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\begin{proof}[Proof (of Theorem)]
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WTS \[
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{g_2}^2(a; p) = \left( \frac{-1}{p} \right) p
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\]
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By lemme, we assume \( a = 1 \) without loss of generality.
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Then,
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\begin{align*}
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[ g_2(1; p) ]^2
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& = \left( \sum_{x \pmod{p}} \left( \frac{x}{p} \right) \right)^2
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\\
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& = \sum_{x, y \in \F_p} \left( \left( \frac{x}{p} \right) \psi(x) \right) \left( \left( \frac{y}{p} \right) \psi(y) \right)
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\\
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& = \sum_{x, y \in {F_p}^\times} \left( \frac{xy}{p} \right) \psi(x + y)
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\\
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& = \sum_{x, z \in {\F_p}^\times} \left( \frac{x^2z}{p} \right) \psi(x + xz)
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& \text{take } y = xz
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\\
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& = \sum_{z \in {\F_p}^\times} \left[ \left( \frac{z}{p} \right) \sum_{x \in \F_p} \psi((1 + z)x) \right]
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\end{align*} and for the inner sum we have \begin{align*}
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\sum_{x \in \F_p} \psi((1 + z)x)
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& = \sum_{x \in \F_p} \psi_{z+1}(x) - 1
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\\
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& = \begin{cases}
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p - 1 & \text{if } z = -1 \\
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-1 & \text{otherwise}
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\end{cases}
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\end{align*} so \begin{align*}
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[ g_2(1; p) ]^2
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& = \left( \frac{-1}{p} \right) (p - 1) + \sum_{z \neq -1} \left( \frac{z}{p} \right) (-1)
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\\
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& = \left( \frac{-1}{p} \right) p - \sum_{z \in {\F_p}^\times} \left( \frac{z}{p} \right)
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\\
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& = \left( \frac{-1}{p} \right) p
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& \text{second term is 0 by lemma}
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\end{align*}
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\end{proof}
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\begin{proof}[Proof (of Quadratic Reciprocity)]
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WTS \[
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\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}
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\]
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Computing \( \left( \frac{p}{q} \right) \), define \[
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\sqrt{p^\ast} = \sqrt{\left( \frac{-1}{p} \right) p} = \sum_{a \pmod{p}} \left( \frac{a}{p} \right) \zeta_p^a
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\] in an extension of \( \F_q \). \[
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\sqrt{p^*} \in \F_q \iff \left( \sqrt{p^*} \right)^q = \sqrt{p^*}
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\] and \[
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\left( \sqrt{p^*} \right)^q = \sum_{a \pmod{p}} \left( \frac{a}{p} \right) \zeta_p^{aq}
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\] in an extension of \( \F_p \).
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This is precisely a Gauss sum \[
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g_1(q;p) = \left( \frac{q}{p} \right) g_2(1; p)
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\] by Lemma so \[
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\left( p^* \right)^{q/2} = \left( \frac{q}{p} \right) (p^*)^{1/2} \implies
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\left( \frac{p^*}{q} \right) \sqrt{p^*} = \left( \frac{q}{p} \right) \sqrt{p^*}
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\] and so \[
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\left( \frac{p^*}{q} \right) = \left( \frac{q}{p} \right)
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\]
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\end{proof}
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\begin{note}
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The following 2 statements are equivalent for \( p, q \) instinct odd primes.
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\begin{enumerate}
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\item \( \displaystyle \left( \frac{p^*}{q} \right) = \left( \frac{q}{p} \right) \)
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\item \( \displaystyle \left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}} \)
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\end{enumerate}
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We can complete the proof via case work of whether \( p \equiv 1 \pmod{4} \) or \( p \equiv 3 \pmod{4} \).
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\end{note}
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\begin{remark}
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How can we compute \( \left( \frac{a}{p} \right) \) for \( a \in \F_p \) effectively?
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\begin{definition}[Jacobi Symbol]
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For any \( a \in \Z \), \( n \in \N_{\geq 1} \) odd, define \[
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\left( \frac{a}{n} \right) = \left( \frac{a}{p_1} \right)^{e_1} \left( \frac{a}{p_2} \right)_{e_2} \cdots \left( \frac{a}{p_k} \right)^{e_k}
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\] where \[
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n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}
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\] is the prime factorization of \( n \).
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\end{definition}
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There exists quadratic reciprocity for the Jacobi symbol, \[
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\left( \frac{m}{n} \right) \left( \frac{n}{m} \right) = (-1)^{\frac{m-1}{2} \frac{n-1}{2}}
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\]
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Suppose we want to compute \( \left( \frac{7}{101} \right) \). We have \[ \left( \frac{7}{101} \right) = \left( \frac{101}{7} \right) = \left( \frac{3}{7} \right) = - \left( \frac{7}{3} \right) = - \left( \frac{1}{3} \right) = -1 \]
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\end{remark}
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