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Lecture 14 - 2025-03-05

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Lance Li
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@@ -876,11 +876,11 @@ We get unique factorization, i.e., prime iff irreducible. \[
We have the following cases
\begin{itemize}
\item \( a^2 + b^2 = p \), and \( c^2 + d^2 = p \)
\item \( a^2 + b^2 = 1 \), and \( c^2 + d^2 = p^2 \)
\item \( a^2 + b^2 = p \), and \( c^2 + d^2 = p \)
This is impossible. \( a^2 + b^2 = 1 \implies a + ib \in \{ \pm 1, \pm i \} \) so \( \gcd(p, u + i) = 1 \) which is a contradiction.
\item \( a^2 + b^2 = 1 \), and \( c^2 + d^2 = p^2 \)
This is impossible. \( a^2 + b^2 = 1 \implies a + ib \in \{ \pm 1, \pm i \} \) so \( \gcd(p, u + i) = 1 \) which is a contradiction.
\end{itemize}
\end{itemize}
\end{proof}
@@ -892,11 +892,11 @@ We get unique factorization, i.e., prime iff irreducible. \[
\begin{itemize}
\item \( \pi \in \Z \) and \( p = \pi \) is prime in \( \Z \)
This is the case where \( p = \pi \equiv 3 \pmod{4} \) and \( p \) is prime in \( \Z \).
This is the case where \( p = \pi \equiv 3 \pmod{4} \) and \( p \) is prime in \( \Z \).
\item \( \pi \bar{\pi} \in \Z \) and \( p = \pi \bar{\pi} \) is prime in \( \Z \), where \( \bar{\pi} \) is the conjugate of \( \pi \).
This is the case where \( p = \pi \bar{\pi} \equiv 1 \pmod{4} \) and \( p \) is prime in \( \Z \).
This is the case where \( p = \pi \bar{\pi} \equiv 1 \pmod{4} \) and \( p \) is prime in \( \Z \).
\end{itemize}
\end{theorem}
@@ -906,26 +906,26 @@ We get unique factorization, i.e., prime iff irreducible. \[
\begin{enumerate}
\item Suppose \( \pi \in \Z \) is a prime. WTS \( \pi \) is a prime in \( \Z \).
Suppose \( \pi | ab \), \( a, b \in \Z \).
Suppose \( \pi | ab \), \( a, b \in \Z \).
Since \( \pi \) a prime in \( \Z[i] \), \( \pi | a \) or \( \pi | b \) in \( \Z[i] \).
Since \( \pi \) a prime in \( \Z[i] \), \( \pi | a \) or \( \pi | b \) in \( \Z[i] \).
This means that there exists an \( w \in \Z[i] \) such that \( \pi w = a \) or \( \pi w = b \).
This means that there exists an \( w \in \Z[i] \) such that \( \pi w = a \) or \( \pi w = b \).
Since \( \pi, a, b \in \R \), we must have \( w \in \R \cap \Z[i] = \Z \).
Since \( \pi, a, b \in \R \), we must have \( w \in \R \cap \Z[i] = \Z \).
This means that \( \pi | a \) or \( \pi | b \) in \( \Z \).
This means that \( \pi | a \) or \( \pi | b \) in \( \Z \).
\item Suppose \( \pi \notin \Z \). WTS \( \pi \bar{\pi} \) is prime in \( \Z \).
Known that \( \pi \) and \( \bar{\pi} \) are irreducible.
Let \( p = \pi \bar{\pi} \). WTS \( p \) is irreducible in \( \Z \).
Known that \( \pi \) and \( \bar{\pi} \) are irreducible.
If \( p \) were reducible, then \[
1 < | \gcd(p, \pi) | < p \qquad \text{ or } \qquad 1 < | \gcd(p, \bar{\pi}) | < p
\]
Either \( \gcd(p, \pi) \) or \( \gcd(p, \bar{\pi}) \) is a proper factor of \( \pi \) or \( \bar{\pi} \), contradicting the fact that \( \pi \) and \( \bar{\pi} \) are irreducible.
Let \( p = \pi \bar{\pi} \). WTS \( p \) is irreducible in \( \Z \).
If \( p \) were reducible, then \[
1 < | \gcd(p, \pi) | < p \qquad \text{ or } \qquad 1 < | \gcd(p, \bar{\pi}) | < p
\]
Either \( \gcd(p, \pi) \) or \( \gcd(p, \bar{\pi}) \) is a proper factor of \( \pi \) or \( \bar{\pi} \), contradicting the fact that \( \pi \) and \( \bar{\pi} \) are irreducible.
\end{enumerate}
\end{proof}
@@ -939,4 +939,126 @@ We get unique factorization, i.e., prime iff irreducible. \[
\begin{note}[Exercise]
Take \( \omega = \frac{1 + \sqrt{-3}}{2} \), \( \omega^3 = 1 \), show that \( \Z[\omega] \) is a Euclidean domain.
\end{note}
\end{note}
\subsection{Characteristics of \( \F_p \)}
\begin{definition}[Additive Character]\index{Additive Character}
A function \( \psi: \F_p \to \C^x \) is an additive character if
\begin{itemize}
\item \( \psi(x + y) = \phi(x) \phi(y) \)
\item \( \psi(0) \neq 1 \)
\item \( \psi(xn) = \phi(x)^n \) for \( n \in \Z \)
\end{itemize}
\end{definition}
\begin{definition}
Take \( p = 3 \), \( \F_3 = \{ 0, 1, 2 \} \), \( \omega = e^{\frac{2 \pi i}{3}} \).
\begin{itemize}
\item \( 1 \to \omega \) or \( 1 \to \omega^2 \) or \( 1 \to 1 \)
\item \( 2 \to \omega^2 \) or \( 2 \to 1 \) or \( 2 \to (\omega^2)^2 = \omega \)
\end{itemize}
\end{definition}
\begin{remark}
\( px = 0 \) in \( \F_p \) means \[
\psi(px) = \psi(x)^p \qquad \text{ and } \qquad \psi(0) = 1
\] so \( \psi(x) \) hs order dividing \( p \).
\( \psi(1) = e^{\frac{2 \pi i k}{p}} \implies \psi(x) = e^{\frac{2 \pi i k n}{p}} \) for some \( n \in \Z \).
\end{remark}
\begin{definition}[Multiplicative Character]\index{Multiplicative Character}
A map \( \chi: \F_p^x \to \C^x \) is a multiplicative character if \[
\chi(x, y) = \chi(x) \chi(y) \left( \implies \chi(x^n) = \chi(x)^n \right)
\] and \[
\chi(1) = 1
\]
\end{definition}
\begin{remark}
We will extend \( \chi \) to \( \F_p \) by setting \( \chi(0) = 0 \).
\end{remark}
\begin{theorem}[Gauss, sum]
Fix \( \mathcal{X}, \psi \) as above, and \( p \). \[
g(\chi, \psi) = \sum_{x \in \F_p} \chi(x) \psi(x)
\]
\end{theorem}
\begin{example}
Take \( p = 3 \), \( \chi(x) = \left( \frac{x}{3} \right) \), \( \psi(x) = \omega^x \).
\begin{align*}
g(\chi, \psi)
& = \left( \frac{1}{3} \right) \omega + \left( \frac{2}{3} \right) \omega^2
\\
& = \omega - \omega^2
\\
& = 1 \omega + 1
\\
& = \sqrt{-3}
\end{align*}
What if \( \psi(x) = \omega^{2x} \)?
\begin{align*}
g(\chi, \psi)
& = \left( \frac{1}{3} \right) \omega^2 + \left( \frac{2}{3} \right) \omega
\\
& = \omega^2 - \omega
\end{align*}
If \( \psi(x) = 1 \), \( g(\chi, \phi) = \left( \frac{1}{3} \right) + \left( \frac{2}{3} \right) = 0 \).
\end{example}
\begin{lemma}
If \( \chi \) is non-trivial, that is, \( \exists x, \chi(x) \neq 1 \), nad \( \psi \) is trivial, that is, \( \forall x, \psi(x) = 1 \), then \[
g(\chi, \psi) = 0 \iff \sum_{x \in \F_p^x} \chi(x) = 0
\]
\end{lemma}
\begin{example}
Take \( p = 5 \), \( \chi(x) = \left( \frac{x}{5} \right) \), \( \psi(1) = e^{\frac{2 \pi i}{5}} = \zeta_5 \).
\begin{align*}
g(\chi, \psi)
& = \left( \frac{1}{5} \right) \zeta_5 + \left( \frac{2}{5} \right) \zeta_5^2 + \left( \frac{3}{5} \right) \zeta_5^3 + \left( \frac{4}{5} \right) \zeta_5^4
\\
& = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4
\end{align*}
\end{example}
\begin{proof}
Take \[
S = \sum_{x \in \F_p^x} \chi(x)
\] \( y \in \F_p^x \) is such that \( \chi(y) \neq 1 \). \[
\chi(y)S = \sum_{x \in \F_p^x} \chi(x) \chi(y) = \sum_{x \in \F_p^x} \chi(xy) = S
\] because \( y \F_p^x = \F_p^x \). This implies \( [\chi(y) - 1] S = 0 \) which means \( S = 0 \).
\end{proof}
\begin{remark}
This of \( \chi \) as a generated Legendre symbol if it keeps track of whether \( x \in \F_p^x \) is a k-t power for som e\( k \).
\end{remark}
\begin{example}
Suppose \( g(\chi, \psi) = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4 \). What is \( g^2 \)?
\begin{align*}
\left( \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4 \right)^2
& = \zeta_5^2 + \zeta_5^4 + \zeta_5^6 + \zeta_5^8 - 2 \zeta_5 \left(
-\zeta_5^2 - \zeta_5^3 + \zeta_5^4
\right) - 2 \zeta_5^2 \left( -\zeta_5^3 + \zeta_5^4 \right) + 2 \zeta_5^3 \left( -\zeta_5^4 \right)
\\
& = -1 \zeta_5 - 1\zeta_5^2 - 1\zeta_5^3 - 1\zeta_5^4 + 4
\\
& = -1( \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4) + 4
\end{align*}
and since \[
1 + \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = 0
\] we have \[
\zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = -1
\] so \[
g^2 = 5
\]
\end{example}