From 604b48ca704e69d855c3d7d37df19a9568501c96 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Mon, 23 Mar 2026 13:50:46 -0500
Subject: [PATCH] updates

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 content/Math4302/_meta.js        |   1 +
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+# Math4302 Modern Algebra (Lecture 26)
+
+## Rings
+
+### Integral Domains
+
+Recall from last lecture, we consider $\mathbb{Z}_p$ and $\mathbb{Z}_p^*$ denote the group of units in $\mathbb{Z}_p$ with multiplication.
+
+$$
+\mathbb{Z}_p^* = \{1,2,\cdots,p-1\}, \quad |\mathbb{Z}_p^*| = p-1
+$$
+
+Let $[a]\in \mathbb{Z}_p^*$, then $[a]^{p-1}=[1]$, this implies that $a^{p-1}\mod p=1$. 
+
+Now if $m\in \mathbb{Z}$ and $a=$ remainder of $m$ by $p$, $[a]\in \mathbb{Z}_p$, implies $m\equiv a\mod p$.
+
+Then $m^{p-1}\equiv a^{p-1}\mod p$.
+
+So
+
+#### Fermat’s little theorem
+
+If $p$ is not a divisor of $m$, then $m^{p-1}\equiv 1\mod p$.
+
+#### Corollary of Fermat’s little theorem
+
+If $m\in \mathbb{Z}$, then $m^p\equiv m\mod p$.
+
+<details>
+<summary>Proof</summary>
+
+If $p|m$, then $m^{p-1}\equiv 0\equiv m\mod p$.
+
+If $p\not|m$, then by Fermat’s little theorem, $m^{p-1}\equiv 1\equiv m\mod p$, so $m^p\equiv m\mod p$.
+
+</details>
+
+<details>
+<summary>Example</summary>
+
+Find the remainder of $40^{100}$ by $19$.
+
+$40^{100}\equiv 2^{100}\mod 19$
+
+$2^{100}\equiv 2^{10}\mod 19$ (Fermat’s little theorem $2^18\equiv 1\mod 19, 2^{90}\equiv 1\mod 19$)
+
+$2^10\equiv (-6)^2\mod 19\equiv 36\mod 19\equiv 17\mod 19$
+
+---
+
+For every integer $n$, $15|(n^{33}-n)$.
+
+$15=3\cdot 5$, therefore enough to show that $3|(n^{33}-n)$ and $5|(n^{33}-n)$.
+
+Apply the corollary of Fermat’s little theorem to $p=3$: $n^3\equiv n\mod 3$, $(n^3)^11\equiv n^{11}\equiv (n^3)^3 n^2=n^3 n^2\equiv n^3\mod 3\equiv n\mod 3$.
+
+Therefore $3|(n^{33}-n)$.
+
+Apply the corollary of Fermat’s little theorem to $p=5$: $n^5\equiv n\mod 5$, $n^30 n^3\equiv (n^5)^6 n^3\equiv n^6 n^3\equiv n^5\mod 5\equiv n\mod 5$.
+
+Therefore $5|(n^{33}-n)$.
+
+</details>
+
+#### Euler’s totient function
+
+Consider $\mathbb{Z}_6$, by definition for the group of units, $\mathbb{Z}_6^*=\{1,5\}$.
+
+$$
+\phi(n)=|\mathbb{Z}_n^*|=|\{1\leq x\leq n:gcd(x,n)=1\}|
+$$
+
+<details>
+<summary>Example</summary>
+
+$\phi(8)=|\{1,3,5,7\}|=4$
+
+</details>
+
+If $[a]\in \mathbb{Z}_n^*$, then $[a]^{\phi(n)}=[1]$. So $a^{\phi(n)}\equiv 1\mod n$.
+
+#### Theorem
+
+If $m\in \mathbb{Z}$, and $gcd(m,n)=1$, then $m^{\phi(n)}\equiv 1\mod n$.
+
+<details>
+<summary>Proof</summary>
+
+If $a$ is the remainder of $m$ by $n$, then $m\equiv a\mod n$, and $\operatorname{gcd}(a,n)=1$, so $m^{\phi(n)}\equiv a^{\phi(n)}\equiv 1\mod n$.
+
+</details>
+
+#### Applications on solving modular equations
+
+Solving equations of the form $ax\equiv b\mod n$.
+
+Not always have solution, $2x\equiv 1\mod 4$ has no solution since $1$ is odd.
+
+Solution for $2x\equiv 1\mod 3$
+
+- $x\equiv 0\implies 2x\equiv 0\mod 3$
+- $x\equiv 1\implies 2x\equiv 2\mod 3$
+- $x\equiv 2\implies 2x\equiv 1\mod 3$
+
+So solution for $2x\equiv 1\mod 3$ is $\{3k+2|k\in \mathbb{Z}\}$.
+
+#### Theorem for solving modular equations
+
+$ax\equiv b\mod n$ has a solution if and only if $\operatorname{gcd}(a,n)|b$ and in that case the equation has $d$ solutions in $\mathbb{Z}_n$.
+
+Proof on next lecture.
\ No newline at end of file
diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js
index 5406c99..830e780 100644
--- a/content/Math4302/_meta.js
+++ b/content/Math4302/_meta.js
@@ -28,4 +28,5 @@ export default {
     Math4302_L23: "Modern Algebra (Lecture 23)",
     Math4302_L24: "Modern Algebra (Lecture 24)",
     Math4302_L25: "Modern Algebra (Lecture 25)",
+    Math4302_L26: "Modern Algebra (Lecture 26)",
 }