Merge branch 'main' of https://git.trance-0.com/Trance-0/NoteNextra-origin

content/Math401/Math401_H1.md

@@ -5,8 +5,8 @@ I made this little book for my Honor Thesis, showing the relevant parts of my wo
 Contents updated as displayed and based on my personal interest and progress with Prof.Feres.
 
 
-<iframe src="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/main.pdf" width="100%" height="600px" style="border: none;" title="Embedded PDF Viewer">
+<iframe src="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/latex/main.pdf" width="100%" height="600px" style="border: none;" title="Embedded PDF Viewer">
   <!-- Fallback content for browsers that do not support iframes or PDFs within them -->
-  <iframe src="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/main.pdf" width="100%" height="500px">
-  <p>Your browser does not support iframes. You can <a href="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/main.pdf">download the PDF</a> file instead.</p>
+  <iframe src="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/latex/main.pdf" width="100%" height="500px">
+  <p>Your browser does not support iframes. You can <a href="https://git.trance-0.com/Trance-0/HonorThesis/raw/branch/main/latex/main.pdf">download the PDF</a> file instead.</p>
 </iframe>

content/Math4202/Math4202_L25.md

@@ -76,7 +76,7 @@ $$
 
 </details>
 
-#### Lemma
+#### Lemma of homotopy equivalence
 
 Let $f,g:X\to Y$ be continuous maps. let $f(x_0)=y_0$ and $g(x_0)=y_1$. If $f$ and $g$ are homotopic, then there is a path $\alpha:I\to Y$ such that $\alpha(0)=y_0$ and $\alpha(1)=y_1$.
 

content/Math4202/Math4202_L26.md

@@ -0,0 +1,90 @@
+# Math4202 Topology II (Lecture 26)
+
+## Algebraic Topology
+
+### Deformation Retracts and Homotopy Type
+
+#### Lemma of homotopy equivalence
+
+Let $f,g:X\to Y$ be continuous maps. let
+
+$$
+f_*=\pi_1(X,f(x_0))\quad\text{and}\quad g_*=\pi_1(Y,g(x_0))
+$$
+
+And $H:X\times I\to Y$ is a homotopy from $f$ to $g$ with a path $H(x_0,t)=\alpha(t)$ for all $t\in I$.
+
+Then $\hat{\alpha}\circ f_*=[\bar{\alpha}*(f\circ \gamma)*\alpha]=[g\circ \gamma]=g_*$. where $\gamma$ is a loop in $X$ based at $x_0$.
+
+<details>
+<summary>Proof</summary>
+$I\times I\xrightarrow{\gamma_{id}} X\times I\xrightarrow{H} Y$
+
+- $I\times \{0\}\mapsto f\circ\gamma$
+- $I\times \{1\}\mapsto g\circ\gamma$
+- $\{0\}\times I\mapsto \alpha$
+- $\{1\}\times I\mapsto \alpha$
+
+As $I\times I$ is convex, $I\times \{0\}\simeq (\{0\}\times I)*(I\times \{1\})*(\{1\}\times I)$.
+
+</details>
+
+#### Corollary for homotopic continuous maps
+
+Let $h,k$ be homotopic continuous maps. And let $h(x_0)=y_0,k(x_0)=y_1$. If $h_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is injective, then $k_*:\pi_1(X,x_0)\to \pi_1(Y,y_1)$ is injective.
+
+<details>
+<summary>Proof</summary>
+
+$\hat{\alpha}$ is an isomorphism of $\pi_1(Y,y_0)$ to $\pi_1(Y,y_1)$.
+
+</details>
+
+#### Corollary for nulhomotopic maps
+
+Let $h:X\to Y$ be nulhomotopic. Then $h_*:\pi_1(X,x_0)\to \pi_1(Y,h(x_0))$ is a trivial group homomorphism (mapping to the constant map on $h(x_0)$).
+
+#### Theorem for fundamental group isomorphism by homotopy equivalence
+
+Let $f:X\to Y$ be a continuous map. Let $f(x_0)=y_0$. If $f$ is a [homotopy equivalence](https://notenextra.trance-0.com/Math4202/Math4202_L25/#definition-of-homotopy-equivalence) ($\exists g:Y\to X$ such that $fg\simeq id_X$, $gf\simeq id_Y$), then
+
+$$
+f_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)
+$$
+is an isomorphism.
+
+<details>
+<summary>Proof</summary>
+
+Let $g:Y\to X$ be the homotopy inverse of $f$. 
+
+Then,
+
+$f_*\circ g_*=\alpha \circ id_{\pi_1(Y,y_0)}=\alpha$
+
+And $g_*\circ f_*=\bar{\alpha}\circ id_{\pi_1(X,x_0)}=\bar{\alpha}$
+
+So $f_*\circ (g_*\circ \hat{\alpha}^-1)=id_{\pi_1(X,x_0)}$
+
+And $g_*\circ (f_*\circ \hat{\alpha}^-1)=id_{\pi_1(Y,y_0)}$
+
+So $f_*$ is an isomorphism (have left and right inverse).
+</details>
+
+### Fundamental group of higher dimensional sphere
+
+$\pi_1(S^n,x_0)=\{e\}$ for $n\geq 2$.
+
+We can decompose the sphere to the union of two hemisphere and compute $\pi_1(S^n_+,x_0)=\pi_1(S^n_-,x_0)=\{e\}$
+
+But for $n\geq 2$, $S^n_+\cap S^n_-=S^{n-1}$, where $S^1_+\cap S^1_-$ is two disjoint points.
+
+#### Theorem for "gluing" fundamental group
+
+Suppose $X=U\cup V$, where $U$ and $V$ are open subsets of $X$. Suppose that $U\cap V$ is path connected, and $x\in U\cap V$. Let $i,j$ be the inclusion maps of $U$ and $V$ into $X$, the images of the induced homomorphisms
+
+$$
+i_*:\pi_1(U,x_0)\to \pi_1(X,x_0)\quad j_*:\pi_1(V,x_0)\to \pi_1(X,x_0)
+$$
+
+The image of the two map generate $\pi_1(X,x_0)$.

content/Math4202/Math4202_L27.md

@@ -0,0 +1,69 @@
+# Math4202 Topology II (Lecture 27)
+
+## Algebraic Topology
+
+### Fundamental Groups for Higher Dimensional Sphere
+
+#### Theorem for "gluing" fundamental group
+
+Suppose $X=U\cup V$, where $U$ and $V$ are open subsets of $X$. Suppose that $U\cap V$ is path connected, and $x\in U\cap V$. Let $i,j$ be the inclusion maps of $U$ and $V$ into $X$, the images of the induced homomorphisms
+
+$$
+i_*:\pi_1(U,x_0)\to \pi_1(X,x_0)\quad j_*:\pi_1(V,x_0)\to \pi_1(X,x_0)
+$$
+
+The image of the two map generate $\pi_1(X,x_0)$.
+
+$G$ is a group, and let $S\subseteq G$, where $G$ is generated by $S$, if $\forall g\in G$, $\exists s_1,s_2,\ldots,s_n\in S$ such that $g=s_1s_2\ldots s_n\in G$. (We can write $G$ as a word of elements in $S$.)
+
+<details>
+<summary>Proof</summary>
+
+Let $f$ be a loop in $X$, $f\simeq g_1*g_2*\ldots*g_n$, where $g_i$ is a loop in $U$ or $V$.
+
+For example, consider the function, $f=f_1*f_2*f_3*f_4$, where $f_1\in S_+$, $f_2\in S_-$, $f_3\in S_+$, $f_4\in S_-$.
+
+Take the functions $\bar{\alpha_1}*\alpha_1\simeq e_{x_1}$ where $x_1$ is the intersecting point on $f_1$ and $f_2$.
+
+Therefore,
+
+$$
+\begin{aligned}
+f&=f_1*f_2*f_3*f_4\\
+&(f_1*\bar{\alpha})*(\alpha_1*f_2*\bar{\alpha_2})*(\alpha_2*f_3*\bar{\alpha_3})*(\alpha_4*f_4)
+\end{aligned}
+$$
+
+This decompose $f$ into a word of elements in either $S_+$ or $S_-$.
+
+---
+
+Note that $f$ is a continuous function $I\to X$, for $t\in I$, $\exists I_t$ being a small neighborhood of $t$ such that $f(I_t)\subseteq U$ or $f(I_t)\subseteq V$.
+
+Since $U_{t\in I}I_t=I$, then $\{I_t\}_{t\in I}$ is an open cover of $I$.
+
+By compactness of $I$, there is a finite subcover $\{I_{t_1},\ldots,I_{t_n}\}$.
+
+Therefore, we can create a partition of $I$ into $[s_i,s_{i+1}]\subseteq I_{t_k}$ for some $k$.
+
+Then with the definition of $I_{t_k}$, $f([s_i,s_{i+1}])\subseteq U$ or $V$.
+
+Then we can connect $x_0$ to $f(s_i)$ with a path $\alpha_i\subseteq U\cap V$.
+
+$$
+\begin{aligned}
+f&=f|_{[s_0,s_1]}*f|_{[s_1,s_2]}*\ldots**f|_{[s_{n-1},s_n]}\\
+&\simeq f|_{[s_0,s_1]}*(\bar{\alpha_1}*\alpha_1)*f|_{[s_1,s_2]}*(\bar{\alpha_2}*\alpha_2)*\ldots*f|_{[s_{n-1},s_n]}*(\bar{\alpha_n}*\alpha_n
+)\\
+&=(f|_{[s_0,s_1]}*\bar{\alpha_1})*(\alpha_1*f|_{[s_1,s_2]}*\bar{\alpha_2})*\ldots*(\alpha_{n-1}*f|_{[s_{n-1},s_n]}*\bar{\alpha_n})\\
+&=g_1*g_2*\ldots*g_n
+\end{aligned}
+$$
+
+</details>
+
+#### Corollary in higher dimensional sphere
+
+Since $S^n_+$ and $S^n_-$ are homeomorphic to open balls $B^n$, then $\pi_1(S^n_+,x_0)=\pi_1(S^n_-,x_0)=\pi_1(B^n,x_0)=\{e\}$ for $n\geq 2$.
+
+> Preview: Van Kampen Theorem

content/Math4202/Math4202_L28.md

@@ -0,0 +1,72 @@
+# Math4202 Topology II (Lecture 28)
+
+## Algebraic Topology
+
+### Fundamental Groups of Some Surfaces
+
+Recall from last week, we will see the fundamental group of $T^2=S^1\times S^1$, and $\mathbb{R}P^2$, Torus with genus $2$.
+
+Some of them are abelian, and some are not.
+
+#### Theorem for fundamental groups of product spaces
+
+Let $X,Y$ be two manifolds. Then the fundamental group of $X\times Y$ is the direct product of their fundamental groups, 
+
+i.e.
+
+$$
+\pi_1(X\times Y,(x_0,y_0))=\pi_1(X,x_0)\times \pi_1(Y,y_0)
+$$
+
+<details>
+<summary>Proof</summary>
+
+We need to find group homomorphism: $\phi:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(X,x_0)\times \pi_1(Y,y_0)$.
+
+Let $P_x,P_y$ be the projection from $X\times Y$ to $X$ and $Y$ respectively.
+
+$$
+(P_x)_*:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(X,x_0)
+$$
+
+$$
+(P_y)_*:\pi_1(X\times Y,(x_0,y_0))\to \pi_1(Y,y_0)
+$$
+
+Given $\alpha\in \pi_1(X\times Y,(x_0,y_0))$, then $\phi(\alpha)=((P_x)_*\alpha,(P_y)_*\alpha)\in \pi_1(X,x_0)\times \pi_1(Y,y_0)$.
+
+Since $(P_x)_*$ and $(P_y)_*$ are group homomorphism, so $\phi$ is a group homomorphism.
+
+**Then we need to show that $\phi$ is bijective.** Then we have the isomorphism of fundamental groups.
+
+To show $\phi$ is injective, then it is sufficient to show that $\ker(\phi)=\{e\}$.
+
+Given $\alpha\in \ker(\phi)$, then $(P_x)_*\alpha=\{e_x\}$ and $(P_y)_*\alpha=\{e_y\}$, so we can find a path homotopy $P_X(\alpha)\simeq e_x$ and $P_Y(\alpha)\simeq e_y$.
+
+So we can build $(H_x,H_y):X\times Y\times I\to X\times I$ by $(x,y,t)\mapsto (H_x(x,t),H_y(y,t))$ is a homotopy from $\alpha$ and $e_x\times e_y$.
+
+So $[\alpha]=[(e_x\times e_y)]$. $\ker(\phi)=\{[(e_x\times e_y)]\}$.
+
+Next, we show that $\phi$ is surjective.
+
+Given $(\alpha,\beta)\in \pi_1(X,x_0)\times \pi_1(Y,y_0)$, then $(\alpha,\beta)$ is a loop in $X\times Y$ based at $(x_0,y_0)$. and $(P_x)_*([\alpha,\beta])=[\alpha]$ and $(P_y)_*([\alpha,\beta])=[\beta]$.
+</details>
+
+#### Corollary for fundamental groups of $T^2$
+
+The fundamental group of $T^2=S^1\times S^1$ is $\mathbb{Z}\times \mathbb{Z}$.
+
+#### Theorem for fundamental groups of $\mathbb{R}P^2$
+
+$\mathbb{R}P^2$ is a compact 2-dimensional manifold with the universal covering space $S^2$ and a $2-1$ covering map $q:S^2\to \mathbb{R}P^2$.
+
+#### Corollary for fundamental groups of $\mathbb{R}P^2$
+
+$\pi_1(\mathbb{R}P^2)=\#q^{-1}(\{x_0\})=\{a,b\}=\mathbb{Z}/2\mathbb{Z}$
+
+Using the path-lifting correspondence.
+
+#### Lemma for The fundamental group of figure-8
+
+The fundamental group of figure-8 is not abelian.
+

content/Math4202/_meta.js

@@ -31,4 +31,7 @@ export default {
     Math4202_L23: "Topology II (Lecture 23)",
     Math4202_L24: "Topology II (Lecture 24)",
     Math4202_L25: "Topology II (Lecture 25)",
+    Math4202_L26: "Topology II (Lecture 26)",
+    Math4202_L27: "Topology II (Lecture 27)",
+    Math4202_L28: "Topology II (Lecture 28)",
 }

content/Math4302/Exam_reviews/Math4302_E2.md

@@ -0,0 +1,439 @@
+# Math 4302 Exam 2 Review
+
+## Groups
+
+
+### Direct products
+
+$\mathbb{Z}_m\times \mathbb{Z}_n$ is cyclic if and only if $m$ and $n$ have greatest common divisor $1$.
+
+More generally, for $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times \cdots \times \mathbb{Z}_{n_k}$, if $n_1,n_2,\cdots,n_k$ are pairwise coprime, then the direct product is cyclic.
+
+
+If $n=p_1^{m_1}\ldots p_k^{m_k}$, where $p_i$ are distinct primes, then the group 
+
+$$
+G=\mathbb{Z}_n=\mathbb{Z}_{p_1^{m_1}}\times \mathbb{Z}_{p_2^{m_2}}\times \cdots \times \mathbb{Z}_{p_k^{m_k}}
+$$
+
+is cyclic.
+
+### Structure of finitely generated abelian groups
+
+#### Theorem for finitely generated abelian groups
+
+Every finitely generated abelian group $G$ is isomorphic to 
+
+$$
+Z_{p_1}^{n_1}\times Z_{p_2}^{n_2}\times \cdots \times Z_{p_k}^{n_k}\times\underbrace{\mathbb{Z}\times \ldots \times \mathbb{Z}}_{m\text{ times}}
+$$
+
+#### Corollary for divisor size of abelian subgroup
+
+If $g$ is abelian and $|G|=n$, then for every divisor $m$ of $n$, $G$ has a subgroup of order $m$.
+
+> [!WARNING]
+>
+> This is not true if $G$ is not abelian.
+>
+> Consider $A_4$ (alternating group for $S_4$) does not have a subgroup of order 6.
+
+
+### Cosets
+
+#### Definition of Cosets
+
+Let $G$ be a group and $H$ its subgroup.
+
+Define a relation on $G$ and $a\sim b$ if $a^{-1}b\in H$.
+
+This is an equivalence relation.
+
+- Reflexive: $a\sim a$: $a^{-1}a=e\in H$
+- Symmetric: $a\sim b\Rightarrow b\sim a$: $a^{-1}b\in H$, $(a^{-1}b)^{-1}=b^{-1}a\in H$
+- Transitive: $a\sim b$ and $b\sim c\Rightarrow a\sim c$ : $a^{-1}b\in H, b^{-1}c\in H$, therefore their product is also in $H$, $(a^{-1}b)(b^{-1}c)=a^{-1}c\in H$
+
+So we get a partition of $G$ to equivalence classes.
+
+Let $a\in G$, the equivalence class containing $a$
+
+$$
+aH=\{x\in G| a\sim x\}=\{x\in G| a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$
+
+This is called the coset of $a$ in $H$.
+
+#### Definition of Equivalence Class
+
+Let $a\in H$, and the equivalence class containing $a$ is defined as:
+
+$$
+aH=\{x|a\simeq x\}=\{x|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$
+
+#### Properties of Equivalence Class
+
+$aH=bH$ if and only if $a\sim b$.
+
+#### Lemma for size of cosets
+
+Any coset of $H$ has the same cardinality as $H$.
+
+Define $\phi:H\to aH$ by $\phi(h)=ah$.
+
+$\phi$ is an bijection, if $ah=ah'\implies h=h'$, it is onto by definition of $aH$.
+
+#### Corollary: Lagrange's Theorem
+
+If $G$ is a finite group, and $H\leq G$, then $|H|\big\vert |G|$. (size of $H$ divides size of $G$)
+
+### Normal Subgroups
+
+#### Definition of Normal Subgroup
+
+A subgroup $H\leq G$ is called a normal subgroup if $aH=Ha$ for all $a\in G$. We denote it by $H\trianglelefteq G$
+
+
+#### Lemma for equivalent definition of normal subgroup
+
+The following are equivalent:
+
+1. $H\trianglelefteq G$
+2. $aHa^{-1}=H$ for all $a\in G$
+3. $aHa^{-1}\subseteq H$ for all $a\in G$, that is $aha^{-1}\in H$ for all $a\in G$
+
+### Factor group
+
+Consider the operation on the set of left coset of $G$, denoted by $S$. Define
+
+$$
+(aH)(bH)=abH
+$$
+
+#### Condition for operation
+
+The operation above is well defined if and only if $H\trianglelefteq G$.
+
+#### Definition of factor (quotient) group
+
+If $H\trianglelefteq G$, then the set of cosets with operation:
+
+$$
+(aH)(bH)=abH
+$$
+
+is a group denoted by $G/H$. This group is called the quotient group (or factor group) of $G$ by $H$.
+
+#### Fundamental homomorphism theorem (first isomorphism theorem)
+
+If $\phi:G\to G'$ is a homomorphism, then the function $f:G/\ker(\phi)\to \phi(G)$, ($\phi(G)\subseteq G'$) given by $f(a\ker(\phi))=\phi(a)$, $\forall a\in G$, is an well-defined isomorphism.
+
+> - If $G$ is abelian, $N\leq G$, then $G/N$ is abelian.
+> - If $G$ is finitely generated and $N\trianglelefteq G$, then $G/N$ is finitely generated.
+
+#### Definition of simple group
+
+$G$ is simple if $G$ has no proper ($H\neq G,\{e\}$), normal subgroup.
+
+### Center of a group
+
+Recall from previous lecture, the center of a group $G$ is the subgroup of $G$ that contains all elements that commute with all elements in $G$.
+
+$$
+Z(G)=\{a\in G\mid \forall g\in G, ag=ga\}
+$$
+
+this subgroup is normal and measure the "abelian" for a group.
+
+#### Definition of the commutator of a group
+
+Let $G$ be a group and $a,b\in G$, the commutator $[a,b]$ is defined as $aba^{-1}b^{-1}$.
+
+$[a,b]=e$ if and only if $a$ and $b$ commute.
+
+Some additional properties:
+
+- $[a,b]^{-1}=[b,a]$
+
+#### Definition of commutator subgroup
+
+Let $G'$ be the subgroup of $G$ generated by all commutators of $G$.
+
+$$
+G'=\{[a_1,b_1][a_2,b_2]\ldots[a_n,b_n]\mid a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n\in G\}
+$$
+
+Then $G'$ is the subgroup of $G$.
+
+- Identity: $[e,e]=e$
+- Inverse: $([a_1,b_1],\ldots,[a_n,b_n])^{-1}=[b_n,a_n],\ldots,[b_1,a_1]$
+
+Some additional properties:
+
+- $G$ is abelian if and only if $G'=\{e\}$
+- $G'\trianglelefteq G$
+- $G/G'$ is abelian
+- If $N$ is a normal subgroup of $G$, and $G/N$ is abelian, then $G'\leq N$.
+
+### Group acting on a set
+
+#### Definition for group acting on a set
+
+Let $G$ be a group, $X$ be a set, $X$ is a $G$-set or $G$ acts on $X$ if there is a map
+
+$$
+G\times X\to X
+$$
+$$
+(g,x)\mapsto g\cdot x\, (\text{ or simply }g(x))
+$$
+
+such that
+
+1. $e\cdot x=x,\forall x\in X$
+2. $g_2\cdot(g_1\cdot x)=(g_2 g_1)\cdot x$
+
+#### Group action is a homomorphism
+
+Let $X$ be a $G$-set, $g\in G$, then the function
+
+$$
+\sigma_g:X\to X,x\mapsto g\cdot x
+$$
+
+is a bijection, and the function $\phi:G\to S_X, g\mapsto \sigma_g$ is a group homomorphism.
+
+
+#### Definition of orbits
+
+We define the equivalence relation on $X$
+
+$$
+x\sim y\iff y=g\cdot x\text{ for some }g
+$$
+
+So we get a partition of $X$ into equivalence classes: orbits
+
+$$
+Gx\coloneqq \{g\cdot x|g\in G\}=\{y\in X|x\sim y\}
+$$
+
+is the orbit of $X$.
+
+$x,y\in X$ either $Gx=Gy$ or $Gx\cap Gy=\emptyset$.
+
+$X=\bigcup_{x\in X}Gx$.
+
+#### Definition of isotropy subgroup
+
+Let $X$ be a $G$-set, the stabilizer (or isotropy subgroup) corresponding to $x\in X$ is
+
+$$
+G_x=\{g\in G|g\cdot x=x\}
+$$
+
+$G_x$ is a subgroup of $G$. $G_x\leq G$.
+
+- $e\cdot x=x$, so $e\in G_x$
+- If $g_1,g_2\in G_x$, then $(g_1g_2)\cdot x=g_1\cdot(g_2\cdot x)=g_1 \cdot x$, so $g_1g_2\in G_x$
+- If $g\in G_x$, then $g^{-1}\cdot g=x=g^{-1}\cdot x$, so $g^{-1}\in G_x$
+
+#### Orbit-stabilizer theorem
+
+If $X$ is a $G$-set and $x\in X$, then
+
+$$
+|Gx|=(G:G_x)=\text{ number of left cosets of }G_x=\frac{|G|}{|G_x|}
+$$
+
+#### Theorem for orbit with prime power groups
+
+Suppose $X$ is a $G$-set, and $|G|=p^n$ for some prime $p$. Let $X_G$ be the set of all elements in $X$ whose orbit has size $1$. (Recall the orbit divides $X$ into disjoint partitions.) Then $|X|\equiv |X_G|\mod p$.
+
+#### Corollary: Cauchy's theorem
+
+If $p$ is prime and $p|(|G|)$, then $G$ has a subgroup of order $p$.
+
+> This does not hold when $p$ is not prime.
+>
+> Consider $A_4$ with order $12$, and $A_4$ has no subgroup of order $6$.
+
+#### Corollary: Center of prime power group is non-trivial
+
+If $|G|=p^m$, then $Z(G)$ is non-trivial. ($Z(G)\neq \{e\}$)
+
+#### Proposition: Prime square group is abelian
+
+If $|G|=p^2$, where $p$ is a prime, then $G$ is abelian.
+
+
+### Classification of small order
+
+Let $G$ be a group
+
+- $|G|=1$
+  - $G=\{e\}$
+- $|G|=2$
+  - $G\simeq\mathbb{Z}_2$ (prime order)
+- $|G|=3$
+  - $G\simeq\mathbb{Z}_3$ (prime order)
+- $|G|=4$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_4$
+- $|G|=5$
+  - $G\simeq\mathbb{Z}_5$ (prime order)
+- $|G|=6$
+  - $G\simeq S_3$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_2\simeq \mathbb{Z}_6$
+<details>
+<summary>Proof</summary>
+
+$|G|$ has an element of order $2$, namely $b$, and an element of order $3$, namely $a$.
+
+So $e,a,a^2,b,ba,ba^2$ are distinct.
+
+Therefore, there are only two possibilities for value of $ab$. ($a,a^2$ are inverse of each other, $b$ is inverse of itself.)
+
+If $ab=ba$, then $G$ is abelian, then $G\simeq \mathbb{Z}_2\times \mathbb{Z}_3$.
+
+If $ab=ba^2$, then $G\simeq S_3$.
+</details>
+
+- $|G|=7$
+  - $G\simeq\mathbb{Z}_7$ (prime order)
+- $|G|=8$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_4\times \mathbb{Z}_2$
+  - $G\simeq\mathbb{Z}_8$
+  - $G\simeq D_4$
+  - $G\simeq$ quaternion group $\{e,i,j,k,-1,-i,-j,-k\}$ where $i^2=j^2=k^2=-1$, $(-1)^2=1$. $ij=l$, $jk=i$, $ki=j$, $ji=-k$, $kj=-i$, $ik=-j$.
+- $|G|=9$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_3$
+  - $G\simeq\mathbb{Z}_9$ (apply the corollary, $9=3^2$, these are all the possible cases)
+- $|G|=10$
+  - $G\simeq\mathbb{Z}_5\times \mathbb{Z}_2\simeq \mathbb{Z}_{10}$
+  - $G\simeq D_5$
+- $|G|=11$
+  - $G\simeq\mathbb{Z}_11$ (prime order)
+- $|G|=12$
+  - $G\simeq\mathbb{Z}_3\times \mathbb{Z}_4$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3$
+  - $A_4$
+  - $D_6\simeq S_3\times \mathbb{Z}_2$
+  - ??? One more
+- $|G|=13$
+  - $G\simeq\mathbb{Z}_{13}$ (prime order)
+- $|G|=14$
+  - $G\simeq\mathbb{Z}_2\times \mathbb{Z}_7$
+  - $G\simeq D_7$
+
+
+#### Lemma for group of order $2p$ where $p$ is prime
+
+If $p$ is prime, $p\neq 2$, and $|G|=2p$, then $G$ is either abelian $\simeq \mathbb{Z}_2\times \mathbb{Z}_p$ or $G\simeq D_p$
+
+## Ring
+
+
+### Definition of ring
+
+A ring is a set $R$ with binary operation $+$ and $\cdot$ such that:
+
+- $(R,+)$ is an abelian group.
+- Multiplication is associative: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$.
+- Distribution property: $a\cdot (b+c)=a\cdot b+a\cdot c$, $(b+c)\cdot a=b\cdot a+c\cdot a$. (Note that $\cdot$ may not be abelian, may not even be a group, therefore we need to distribute on both sides.)
+
+> [!NOTE]
+>
+> $a\cdot b=ab$ will be used for the rest of the sections.
+
+#### Properties of rings
+
+Let $0$ denote the identity of addition of $R$. $-a$ denote the additive inverse of $a$.
+
+- $0\cdot a=a\cdot 0=0$
+- $(-a)b=a(-b)=-(ab)$, $\forall a,b\in R$
+- $(-a)(-b)=ab$, $\forall a,b\in R$
+
+#### Definition of commutative ring
+
+A ring $(R,+,\cdot)$ is commutative if $a\cdot b=b\cdot a$, $\forall a,b\in R$.
+
+#### Definition of unity element
+
+A ring $R$ has unity element if there is an element $1\in R$ such that $a\cdot 1=1\cdot a=a$, $\forall a\in R$.
+
+#### Definition of unit
+
+Suppose $R$ is a ring with unity element. An element $a\in R$ is called a unit if there is $b\in R$ such that $a\cdot b=b\cdot a=1$.
+
+In this case $b$ is called the inverse of $a$.
+
+#### Definition of division ring
+
+If every $a\neq 0$ in $R$ has a multiplicative inverse (is a unit), then $R$ is called a division ring.
+
+#### Definition of field
+
+A commutative division ring is called a field.
+
+#### Units in $\mathbb{Z}_n$ is coprime to $n$
+
+More generally, $[m]\in \mathbb{Z}_n$ is a unit if and only if $\operatorname{gcd}(m,n)=1$.
+
+### Integral Domains
+
+#### Definition of zero divisors
+
+If $a,b\in R$ with $a,b\neq 0$ and $ab=0$, then $a,b$ are called zero divisors.
+
+#### Zero divisors in $\mathbb{Z}_n$
+
+$[m]\in \mathbb{Z}_n$ is a zero divisor if and only if $\operatorname{gcd}(m,n)>1$ ($m$ is not a unit).
+
+#### Corollaries of integral domain
+
+If $R$ is a integral domain, then we have cancellation property $ab=ac,a\neq 0\implies b=c$.
+
+#### Units with multiplication forms a group
+
+If $R$ is a ring with unity, then the units in $R$ forms a group under multiplication.
+
+### Fermat’s and Euler’s Theorems
+
+#### 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 $m^p\equiv m\mod p$.
+
+#### Euler’s totient function
+
+Consider $\mathbb{Z}_6$, by definition for the group of units, $\mathbb{Z}_6^*=\{1,5\}$.
+
+$$
+\phi(n)=|\mathbb{Z}_n^*|=|\{1\leq x\leq n:gcd(x,n)=1\}|
+$$
+
+#### Euler’s Theorem
+
+If $m\in \mathbb{Z}$, and $gcd(m,n)=1$, then $m^{\phi(n)}\equiv 1\mod n$.
+
+#### Theorem for existence of solution of modular equations
+
+$ax\equiv b\mod n$ has a solution if and only if $d=\operatorname{gcd}(a,n)|b$ And if there is a solution, then there are exactly $d$ solutions in $\mathbb{Z}_n$.
+
+### Ring homomorphisms
+
+#### Definition of ring homomorphism
+
+Let $R,S$ be two rings, $f:R\to S$ is a ring homomorphism if $\forall a,b\in R$,
+
+- $f(a+b)=f(a)+f(b)\implies f(0)=0, f(-a)=-f(a)$
+- $f(ab)=f(a)f(b)$
+
+#### Definition of ring isomorphism
+
+If $f$ is a ring homomorphism and a bijection, then $f$ is called a ring isomorphism.

content/Math4302/Math4302_L26.md

@@ -2,7 +2,7 @@
 
 ## Rings
 
-### Integral Domains
+### Fermat’s and Euler’s Theorems
 
 Recall from last lecture, we consider $\mathbb{Z}_p$ and $\mathbb{Z}_p^*$ denote the group of units in $\mathbb{Z}_p$ with multiplication.
 
@@ -79,7 +79,7 @@ $\phi(8)=|\{1,3,5,7\}|=4$
 
 If $[a]\in \mathbb{Z}_n^*$, then $[a]^{\phi(n)}=[1]$. So $a^{\phi(n)}\equiv 1\mod n$.
 
-#### Theorem
+#### Euler’s Theorem
 
 If $m\in \mathbb{Z}$, and $gcd(m,n)=1$, then $m^{\phi(n)}\equiv 1\mod n$.
 
@@ -104,7 +104,7 @@ Solution for $2x\equiv 1\mod 3$
 
 So solution for $2x\equiv 1\mod 3$ is $\{3k+2|k\in \mathbb{Z}\}$.
 
-#### Theorem for solving modular equations
+#### Theorem for existence of solution of 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 $\mathbb{Z}_n$.
 

content/Math4302/Math4302_L27.md

@@ -0,0 +1,126 @@
+# Math4302 Modern Algebra (Lecture 27)
+
+## Rings
+
+### Fermat’s and Euler’s Theorems
+
+Recall from last lecture, $ax\equiv b \mod n$, if $x\equiv y\mod n$, then $x$ is a solution if and only if $y$ is a solution.
+
+#### Theorem for existence of solution of modular equations
+
+$ax\equiv b\mod n$ has a solution if and only if $d=\operatorname{gcd}(a,n)|b$ And if there is a solution, then there are exactly $d$ solutions in $\mathbb{Z}_n$.
+
+<details>
+<summary>Proof</summary>
+
+For the forward direction, we proved if $ax\equiv b\mod n$ then $ax-b=ny$, $y\in\mathbb{Z}$.
+
+then $b=ax-ny$, $d|(ax-ny)$ implies that $d|b$.
+
+---
+
+For the backward direction, assume $d=\operatorname{gcd}(a,n)=1$. Then we need to show, there is exactly $1$ solution between $0$ and $n-1$.
+
+If $ax\equiv b\mod n$, then in $\mathbb{Z}_n$, $[a][x]=[b]$. (where $[a]$ denotes the remainder of $a$ by $n$ and $[b]$ denotes the remainder of $b$ by $n$)
+
+Since $\operatorname{gcd}(a,n)=1$, then $[a]$ is a unit in $\mathbb{Z}_n$, so we can multiply the above equation by the inverse of $[a]$. and get $[x]=[a]^{-1}[b]$.
+
+Now assume $d=\operatorname{gcd}(a,n)$ where $n$ is arbitrary. Then $a=a'd$, then $n=n'd$, with $\operatorname{gcd}(a',n')=1$.
+
+Also $d|b$ so $b=b'd$. So
+
+$$
+\begin{aligned}
+ax\equiv b \mod n&\iff n|(ax-b)\\
+&\iff n'd|(a'dx-b'd)\\
+&\iff n'|(a'x-b')\\
+&\iff a'x\equiv b'\mod n'
+\end{aligned}
+$$.
+
+Since $\operatorname{gcd}(a',n')=1$, there is a unique solution $x_0\in \mathbb{Z}_{n'}$. $0\leq x_0\leq n'+1$. Other solution in $\mathbb{Z}$ are of the form $x_0+kn'$ for $k\in \mathbb{Z}$.
+
+And there will be $d$ solutions in $\mathbb{Z}_n$,
+
+</details>
+
+<details>
+<summary>Examples</summary>
+
+Solve $12x\equiv 25\mod 7$.
+
+$12\equiv 5\mod 7$, $25\equiv 4\mod 7$. So the equation becomes $5x\equiv 4\mod 7$.
+
+$[5]^{-1}=3\in \mathbb{Z}_7$, so $[5][x]\equiv [4]$ implies $[x]\equiv [3][4]\equiv [5]\mod 7$.
+
+So solution in $\mathbb{Z}$ is $\{5+7k:k\in \mathbb{Z}\}$.
+
+---
+
+Solve $6x\equiv 32\mod 20$.
+
+$\operatorname{gcd}(6,20)=2$, so $6x\equiv 12\mod 20$ if and only if $3x\equiv 6\mod 10$.
+
+$[3]^{-1}=[7]\in \mathbb{Z}_{10}$, so $[3][x]\equiv [6]$ implies $[x]\equiv [7][6]\equiv [2]\mod 10$.
+
+So solution in $\mathbb{Z}_{20}$ is $[2]$ and $[12]$
+
+So solution in $\mathbb{Z}$ is $\{2+10k:k\in \mathbb{Z}\}$
+
+</details>
+
+### Ring homomorphisms
+
+#### Definition of ring homomorphism
+
+Let $R,S$ be two rings, $f:R\to S$ is a ring homomorphism if $\forall a,b\in R$,
+
+- $f(a+b)=f(a)+f(b)\implies f(0)=0, f(-a)=-f(a)$
+- $f(ab)=f(a)f(b)$
+
+#### Definition of ring isomorphism
+
+If $f$ is a ring homomorphism and a bijection, then $f$ is called a ring isomorphism.
+
+<details>
+<summary>Example</summary>
+Let $f:(\mathbb{Z},+,\times)\to(2\mathbb{Z},+,\times)$ by $f(a)=2a$.
+
+Is not a ring homomorphism since $f(ab)\neq f(a)f(b)$ in general.
+
+---
+
+Let $f:(\mathbb{Z},+,\times)\to(\mathbb{Z}_n,+,\times)$ by $f(a)=a\mod n$
+
+Is a ring homomorphism.
+
+</details>
+
+### Integral domains and their file fo fractions.
+
+Let $R$ be an integral domain: (i.e. $R$ is commutative with unity and no zero divisors).
+
+#### Definition of field of fractions
+
+If $R$ is an integral domain, we can construct a field containing $R$ called the field of fractions (or called field of quotients) of $R$.
+
+$$
+S=\{(a,b)|a,b\in R, b\neq 0\}
+$$
+
+a relation on $S$ is defined as follows:
+
+$(a,b)\sim (c,d)$ if and only if $ad=bc$.
+
+<details>
+<summary>This equivalence relation is well defined</summary>
+
+- Reflectivity: $(a,b)\sim (a,b)$ $ab=ab$
+- Symmetry: $(a,b)\sim (c,d)\Rightarrow (c,d)\sim (a,b)$
+- Transitivity: $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)\Rightarrow (a,b)\sim (e,f)$
+  - $ad=bc$, and $cf=ed$, we want to conclude that $af=be$. since $ad=bc$, then $adf=bcf$, since $cf=ed$, then $cfb=edb$, therefore $adf=edb$.
+  - Then $d(af-be)=0$ since $d\neq 0$ then $af=be$.
+
+</details>
+
+Then $S/\sim$ is a field.

content/Math4302/Math4302_L28.md

@@ -0,0 +1,153 @@
+# Math4302 Modern Algebra (Lecture 28)
+
+## Rings
+
+### Field of quotients
+
+Let $R$ be an integral domain ($R$ has unity and commutative with no zero divisors).
+
+Consider the pair $S=\{(a,b)|a,b\in R, b\neq 0\}$.
+
+And define the equivalence relation on $S$ as follows:
+
+$(a,b)\sim (c,d)$ if and only if $ad=bc$.
+
+We denote $[(a,b)]$ as set of all elements in $S$ equivalent to $(a,b)$.
+
+Let $F$ be the set of all equivalent classes. We define addition and multiplication on $F$ as follows:
+
+$$
+[(a,b)]+[(c,d)]=[(ad+bc,bd)]
+$$
+
+$$
+[(a,b)]\cdot[(c,d)]=[(ac,bd)]
+$$
+
+<details>
+<summary>The multiplication and addition is well defined </summary>
+
+Addition:
+
+If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ad+bc,bd)\sim (a'd+c'd,b'd)$.
+
+Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$,
+
+So $ab'dd'=a'bdd'$, and $cd'bb'=dc'bb'$.
+
+ $adb'd'+bcb'd'=a'd'bd+b'c'bd$, therefore $(ad+bc,bd)\sim (a'd+c'd,b'd)$.
+
+---
+
+Multiplication:
+
+If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ac,bd)\sim (a'c',b'd')$.
+
+Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$, so $(ac,bd)\sim (a'c',b'd')$
+
+</details>
+
+#### Claim (F,+,*) is a field
+
+- additive identity: $(0,1)\in F$
+- additive inverse: $(a,b)\in F$, then $(-a,b)\in F$ and $(-a,b)+(a,b)=(0,1)\in F$
+- additive associativity: bit long.
+
+- multiplicative identity: $(1,1)\in F$
+- multiplicative inverse: $[(a,b)]$ is non zero if and only if $a\neq 0$, then $a^{-1}=[(b,a)]\in F$.
+- multiplicative associativity: bit long
+
+- distributivity: skip, too long.
+
+Such field is called a quotient field of $R$.
+
+And $F$ contains $R$ by $\phi:R\to F$, $\phi(a)=[(a,1)]$.
+
+This is a ring homomorphism.
+
+- $\phi(a+b)=[(a+b,1)]=[(a,1)][(b,1)]\phi(a)+\phi(b)$
+- $\phi(ab)=[(ab,1)]=[(a,1)][(b,1)]\phi(a)\phi(b)$
+
+and $\phi$ is injective.
+
+If $\phi(a)=\phi(b)$, then $a=b$.
+
+<details>
+<summary>Example</summary>
+
+Let $D\subset \mathbb R$ and
+
+$$
+\mathbb Z \subset D\coloneqq \{a+b\sqrt{2}:a,b\in \mathbb Z\}
+$$
+
+Then $D$ is a subring of $\mathbb R$, and integral domain, with usual addition and multiplication.
+
+$$
+(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}
+$$
+
+$$
+-(a+b\sqrt{2})=(-a)+(-b)\sqrt{2})
+$$
+
+...
+
+$D$ is a integral domain since $\mathbb R$ has no zero divisors, therefore $D$ has no zero divisors.
+
+Consider the field of quotients of $D$. $[(a+b\sqrt{2},c+d\sqrt{2})]$. This is isomorphic to $\mathbb Q(\sqrt2)=\{r+s\sqrt{2}:r,s\in \mathbb Q\}$
+
+$$
+m+n\sqrt{2}=\frac{m}{n}+\frac{m'}{n'}\sqrt{2}\mapsto [(mn'+nm'\sqrt{2},nn')]
+$$
+
+And use rationalization on the forward direction.
+
+</details>
+
+#### Polynomial rings
+
+Let $R$ be a ring, a polynomial with coefficients in $R$ is a sum
+
+$$
+a_0+a_1x+\cdots+a_nx^n
+$$
+
+where $a_i\in R$. $x$ is indeterminate, $a_0,a_1,\cdots,a_n$ are called coefficients. $a_0$ is the constant term.
+
+If $f$ is a non-zero polynomial, then the degree of $f$ is defined as the largest $n$ such that $a_n\neq 0$.
+
+<details>
+<summary>Example</summary>
+
+Let $f=1+2x+0x^2-1x^3+0x^4$, then $deg f=3$
+
+</details>
+
+If $R$ has a unity $1$, then we write $x^m$ instead of $1x^m$.
+
+Let $R[x]$ denote the set of all polynomials with coefficients in $R$.
+
+We define multiplication and addition on $R[x]$.
+
+$f:a_0+a_1x+\cdots+a_nx^n$
+
+$g:b_0+b_1x+\cdots+b_mx^m$
+
+Define,
+
+$$
+f+g=a_0+b_0+a_1x+b_1x+\cdots+a_nx^n+b_mx^m
+$$
+
+$$
+fg=(a_0b_0)+(a_1b_0)x+\cdots+(a_nb_m)x^m
+$$
+
+In general, the coefficient of $x^m=\sum_{i=0}^{m}a_ix^{m-i}$.
+
+> [!CAUTION]
+>
+> The field $R$ may not be commutative, follow the order of computation matters.
+
+We will show that this is a ring and explore additional properties.

content/Math4302/Math4302_L29.md

@@ -0,0 +1,145 @@
+# Math4302 Modern Algebra (Lecture 29)
+
+## Rings
+
+### Polynomial Rings
+
+$$
+R[x]=\{a_0+a_1x+\cdots+a_nx^n:a_0,a_1,\cdots,a_n\in R,n>1\}
+$$
+
+Then $(R[x],+,\cdot )$ is a ring.
+
+If $R$ has a unity $1$, then $R[x]$ has a unity $1$.
+
+If $R$ is commutative, then $(R[x],+,\cdot )$ is commutative.
+
+#### Definition of evaluation map
+
+Let $F$ be a field, and $F[x]$. Fix $\alpha\in F$. $\phi_\alpha:F[x]\to F$ defined by $f(x)\mapsto f(\alpha)$ (the evaluation map).
+
+Then $\phi_\alpha$ is a ring homomorphism. $\forall f,g\in F[x]$,
+
+- $(f+g)(\alpha)=f(\alpha)+g(\alpha)$
+- $(fg)(\alpha)=f(\alpha)g(\alpha)$ (use commutativity of $\cdot$ of $F$, $f(\alpha)g(\alpha)=\sum_{k=0}^{n+m}c_k x^k$, where $c_k=\sum_{i=0}^k a_ib_{k-i}$)
+
+#### Definition of roots
+
+Let $\alpha\in F$ is zero (or root) of $f\in F[x]$, if $f(\alpha)=0$.
+
+<details>
+<summary>Example</summary>
+
+$f(x)=x^3-x, F=\mathbb{Z}_3$
+
+$f(0)=f(1)=0$, $f(2)=8-2=2-2=0$
+
+but note that $f(x)$ is not zero polynomial $f(x)=0$, but all the evaluations are zero.
+
+</details>
+
+#### Factorization of polynomials
+
+Division algorithm. Let $F$ be a field, $f(x),g(x)\in F[x]$ with $g(x)$ non-zero. Then there are unique polynomials $q(x),r(x)\in F[x]$ such that
+
+$f(x)=q(x)g(x)+r(x)$
+
+where $f(x)=a_0+a_1x+\cdots+a_nx^n$ and $g(x)=b_0+b_1x+\cdots+b_mx^m$, $r(x)=c_0+c_1x+\cdots+c_tx^t$, and $a^n,b^m,c^t\neq 0$.
+
+$r(x)$ is the zero polynomial or $\deg r(x)<\deg g(x)$.
+
+<details>
+<summary>Proof</summary>
+
+Uniqueness: exercise
+
+---
+
+Existence:
+
+Let $S=\{f(x)-h(x)g(x):h(x)\in F[x]\}$.
+
+If $0\in S$, then we are done. Suppose $0\notin S$.
+
+Let $r(x)$ be the polynomial with smallest degree in $S$.
+
+$f(x)-h(x)g(x)=r(x)$ implies that $f(x)=h(x)g(x)+r(x)$.
+
+If $\deg r(x)<\deg g(x)$, then we are done; we set $q(x)=h(x)$.
+
+If $\deg r(x)\geq\deg g(x)$, we get a contradiction, let $t=\deg r(x)$.
+
+$m=\deg g(x)$. (so $m\leq t$) Look at $f(x)-(h(x)+\frac{c_t}{b_m}x^{t-m})g(x)$.
+
+then $f(x)-(h(x)+\frac{c_t}{b_m}x^{t-m})g(x)=f(x)-h(x)g(x)-\frac{c_t}{b_m}x^{t-m}g(x)$.
+
+And $f(x)-h(x)g(x)=r(x)=c_0+c_1x+\cdots+c_tx^t$, $c_t\neq 0$.
+
+$\frac{c_t}{b_m}x^{t-m}g(x)=\frac{c_0c_t}{b_m}x^{t-m}+\cdots+c_t x^t$
+
+That the largest terms cancel, so this gives a polynomial of degree $<t$, which violates that $r(x)$ has smallest degree.
+
+</details>
+
+<details>
+<summary>Example</summary>
+
+$F=\mathbb{Z}_5=\{0,1,2,3,4\}$
+
+Divide $3x^4+2x^3+x+2$ by $x^2+4$ in $\mathbb{Z}_5[x]$.
+
+$$
+3x^4+2x^3+x+2=(3x^3+2x-2)(x^2+4)+3x
+$$
+
+So $q(x)=3x^3+2x-2$, $r(x)=3x$.
+
+</details>
+
+#### Some corollaries
+
+$a\in F$ is a zero of $f(x)$ if and only if $(x-a)|f(x)$.
+
+That is, the remainder of $f(x)$ when divided by $(x-a)$ is zero.
+
+<details>
+<summary>Proof</summary>
+
+If $(x-a)|f(x)$, then $f(a)=0$.
+
+If $f(x)=(x-a)q(x)$, then $f(a)=(a-a)q(a)=0$.
+
+---
+
+If $a$ is a zero of $f(x)$, then $f(x)$ is divisible by $(x-a)$.
+
+We divide $f(x)$ by $(x-a)$.
+
+$f(x)=q(x)(x-a)+r(x)$, where $r(x)$ is a constant polynomial (by degree of division).
+
+Evaluate at $f(a)=0=0+r$, therefore $r=0$.
+
+</details>
+
+#### Another corollary
+
+If $f(x)\in F[x]$ and $\deg f(x)=0$, then $f(x)$ has at most $n$ zeros.
+
+<details>
+<summary>Proof</summary>
+
+We proceed by induction on $n$, if $n=1$, this is clear. $ax+b$ have only root $x=-\frac{b}{a}$.
+
+Suppose $n\geq 2$.
+
+If $f(x)$ has no zero, done.
+
+If $f(x)$ has at least $1$ zero, then $f(x)=(x-a)q(x)$ (by our first corollary), where degree of $q(x)$ is $n-1$.
+
+So zeros of $f(x)=\{a\}\cup$ zeros of $q(x)$, and such set has at most $n$ elements.
+
+Done.
+
+</details>
+
+Preview: How to know if a polynomial is irreducible? (On Friday)

content/Math4302/_meta.js

@@ -29,4 +29,7 @@ export default {
     Math4302_L24: "Modern Algebra (Lecture 24)",
     Math4302_L25: "Modern Algebra (Lecture 25)",
     Math4302_L26: "Modern Algebra (Lecture 26)",
+    Math4302_L27: "Modern Algebra (Lecture 27)",
+    Math4302_L28: "Modern Algebra (Lecture 28)",
+    Math4302_L29: "Modern Algebra (Lecture 29)",
 }