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+# CSE4303 Introduction to Computer Security (Lecture 17)
+
+> Due to lack of my attention, this lecture note is generated by AI to create continuations of the previous lecture note. I kept this warning because the note was created by AI.
+
+#### Software security
+
+### Administrative notes
+
+#### Project details
+
+- Project plan
+  - Thursday, `4/9` at the end of class
+  - `5%`
+- Written document and presentation recording
+  - Thursday, `4/30` at `11:30 AM`
+  - `15%`
+- View peer presentations and provide feedback
+  - Wednesday, `5/6` at `11:59 PM`
+  - `5%`
+
+#### Upcoming schedule
+
+- This week (`3/20`)
+  - software security lecture
+  - studio
+  - some time for studio on Tuesday
+- Next week (`4/6`)
+  - fuzzing
+  - some time to discuss project ideas
+- `4/13`
+  - Web security
+- `4/20`
+  - Privacy and ethics overview
+  - time to work on projects
+  - course wrap-up
+
+### Overview
+
+#### Outline
+
+- Context
+- Prominent software vulnerabilities and exploits
+- Buffer overflows
+  - Background: C code, compilation, memory layout, execution
+  - Baseline exploit
+  - Challenges
+  - Defenses, countermeasures, counter-countermeasures
+
+Sources:
+- SEED lab book
+- Gilbert/Tamassia book
+- Slides from Bryant/O'Hallaron (CMU), Dan Boneh (Stanford), Michael Hicks (UMD)
+
+### Context
+
+#### Context: computing stack (informal)
+
+| Layer | Example |
+| --- | --- |
+| Application | web server, standalone app |
+| Compiler / assembler | `gcc`, `clang` |
+| OS: syscalls | `execve()`, `setuid()`, `write()`, `open()`, `fork()` |
+| OS: processes, mem layout | Linux virtual memory layout |
+| Architecture (ISA, execution) | x86, x86_64, ARM |
+| Hardware | Intel Sky Lake processor |
+
+- User control is strongest near the application / compiler level.
+- System control becomes more important as we move down toward OS, architecture, and hardware.
+
+### Prominent software vulnerabilities and exploits
+
+#### Software security: categories
+
+- Race conditions
+- Privilege escalation
+- Path traversal
+- Environment variable modification
+- Language-specific vulnerabilities
+  - Format string attack
+  - Buffer overflows
+
+#### Buffer Overflows (BoFs)
+
+- A buffer overflow is a bug that affects low-level code, typically in C and C++, with significant security implications.
+- Normally, a program with this bug will simply crash.
+- But an attacker can alter the situations that cause the program to do much worse.
+  - Steal private information
+    - e.g. Heartbleed
+  - Corrupt valuable information
+  - Run code of the attacker's choice
+
+#### Application behavior
+
+- Slide contains a figure only.
+- Intended point: normal application behavior can become attacker-controlled if input handling is unsafe.
+
+#### BoFs: why do we care?
+
+- Reference from slide: [IEEE Spectrum top programming languages 2025](https://spectrum.ieee.org/top-programming-languages-2025)
+
+#### Critical systems in C/C++
+
+- Most OS kernels and utilities
+  - `fingerd`
+  - X windows server
+  - shell
+- Many high-performance servers
+  - Microsoft IIS
+  - Apache `httpd`
+  - `nginx`
+  - Microsoft SQL Server
+  - MySQL
+  - `redis`
+  - `memcached`
+- Many embedded systems
+  - Mars rover
+  - industrial control systems
+  - automobiles
+
+A successful attack on these systems can be particularly dangerous.
+
+#### Morris Worm
+
+- Slide contains a figure / historical reference only.
+- It is included as an example of how memory-corruption vulnerabilities mattered in practice.
+
+#### Why do we still care?
+
+- The slide references the NVD search page: [NVD vulnerability search](https://nvd.nist.gov/vuln/search)
+- Why the drop?
+  - Memory-safe languages
+    - Rust
+    - Go
+  - Stronger defenses
+  - Fuzzing
+    - find bugs before release
+  - Change in development practices
+    - code review
+    - static analysis tools
+    - related engineering improvements
+
+#### MITRE Top 25 2025
+
+- Reference from slide: [MITRE CWE Top 25](http://cwe.mitre.org/top25/)
+
+### Buffer overflows
+
+#### Outline
+
+- System Basics
+  - Application memory layout
+  - How does function call work under the hood
+    - `32-bit x86` only
+    - `64-bit x86_64` similar, but with important differences
+- Buffer overflow
+  - Overwriting the return address pointer
+  - Point it to shell code injected
+
+#### Buffer Overflows (BoFs)
+
+- 2-minute version first, then all background / full version
+
+#### Process memory layout: virtual address space
+
+- Slide reference: [virtual address space reference](https://hungys.xyz/unix-prog-process-environment/)
+
+#### Process memory layout: function calls
+
+- Slide reference: [Tenouk function call figure 1](http://www.tenouk.com/Bufferoverflowc/Bufferoverflow2.html)
+- Slide reference: [Tenouk function call figure 2](http://www.tenouk.com/Bufferoverflowc/Bufferoverflow4.html)
+
+#### Process memory layout: compromised frame
+
+- Slide reference: [Tenouk compromised frame figure](http://www.tenouk.com/Bufferoverflowc/Bufferoverflow4.html)
+
+#### Computer System
+
+High-level examples used in the slide:
+
+```c
+car *c = malloc(sizeof(car));
+c->miles = 100;
+c->gals = 17;
+float mpg = get_mpg(c);
+free(c);
+```
+
+```java
+Car c = new Car();
+c.setMiles(100);
+c.setGals(17);
+float mpg = c.getMPG();
+```
+
+Assembly-language example used in the slide:
+
+```asm
+get_mpg:
+    pushq   %rbp
+    movq    %rsp, %rbp
+    ...
+    popq    %rbp
+    ret
+```
+
+- The same computation can be viewed at multiple levels:
+  - C / Java source
+  - assembly language
+  - machine code
+  - operating system context
+
+#### Little Theme 1: Representation
+
+- All digital systems represent everything as `0`s and `1`s.
+  - The `0` and `1` are really two different voltage ranges in wires.
+  - Or magnetic positions on a disk, hole depths on a DVD, or even DNA.
+- "Everything" includes:
+  - numbers
+    - integers and floating point
+  - characters
+    - building blocks of strings
+  - instructions
+    - directives to the CPU that make up a program
+  - pointers
+    - addresses of data objects stored in memory
+- These encodings are stored throughout the computer system.
+  - registers
+  - caches
+  - memories
+  - disks
+- They all need addresses.
+  - find an item
+  - find a place for a new item
+  - reclaim memory when data is no longer needed
+
+#### Little Theme 2: Translation
+
+- There is a big gap between how we think about programs / data and the `0`s and `1`s of computers.
+- We need languages to describe what we mean.
+- These languages must be translated one level at a time.
+- Example point from the slide:
+  - we know Java as a programming language
+  - but we must work down to the `0`s and `1`s of computers
+  - we try not to lose anything in translation
+  - we encounter Java bytecode, C, assembly, and machine code
+
+#### Little Theme 3: Control Flow
+
+- How do computers orchestrate everything they are doing?
+- Within one program:
+  - How are `if/else`, loops, and switches implemented?
+  - How do we track nested procedure calls?
+  - How do we know what to do upon `return`?
+- At the operating-system level:
+  - library loading
+  - sharing system resources
+    - memory
+    - I/O
+    - disks
+
+#### HW/SW Interface: Code / Compile / Run Times
+
+- Code time
+  - user program in C
+  - `.c` file
+- Compile time
+  - C compiler
+  - assembler
+- Run time
+  - executable `.exe` file
+  - hardware executes it
+- Note from slide:
+  - the compiler and assembler are themselves just programs developed using this same process
+
+#### Assembly Programmer's View
+
+- Programmer-visible CPU / memory state
+  - Program counter
+    - address of next instruction
+    - called `RIP` in x86-64
+  - Named registers
+    - heavily used program data
+    - together called the register file
+  - Condition codes
+    - store status information about most recent arithmetic operation
+    - used for conditional branching
+- Memory
+  - byte-addressable array
+  - contains code and user data
+  - includes the stack for supporting procedures
+
+#### Turning C into Object Code
+
+- Code in files `p1.c` and `p2.c`
+- Compile with:
+
+```bash
+gcc -Og p1.c p2.c -o p
+```
+
+- Notes from the slide
+  - `-Og` uses basic optimizations
+  - resulting machine code goes into file `p`
+- Translation chain
+  - C program -> assembly program -> object program -> executable program
+- Associated tools
+  - compiler
+  - assembler
+  - linker
+  - static libraries (`.a`)
+
+#### Machine Instruction Example
+
+- C code
+
+```c
+*dest = t;
+```
+
+- Meaning
+  - store value `t` where designated by `dest`
+- Assembly
+
+```asm
+movq %rsi, (%rdx)
+```
+
+- Interpretation
+  - move 8-byte value to memory
+  - operands
+    - `t` is in register `%rsi`
+    - `dest` is in register `%rdx`
+    - `*dest` means memory `M[%rdx]`
+- Object code
+
+```text
+0x400539: 48 89 32
+```
+
+- It is a 3-byte instruction stored at address `0x400539`.
+
+#### IA32 Registers - 32 bits wide
+
+- General-purpose register families shown in the slide
+  - `%eax`, `%ax`, `%ah`, `%al`
+  - `%ecx`, `%cx`, `%ch`, `%cl`
+  - `%edx`, `%dx`, `%dh`, `%dl`
+  - `%ebx`, `%bx`, `%bh`, `%bl`
+  - `%esi`, `%si`
+  - `%edi`, `%di`
+  - `%esp`, `%sp`
+  - `%ebp`, `%bp`
+- Roles highlighted in the slide
+  - accumulate
+  - counter
+  - data
+  - base
+  - source index
+  - destination index
+  - stack pointer
+  - base pointer
+
+#### Data Sizes
+
+- Slide is primarily a figure summarizing common integer widths and sizes.
+
+#### Assembly Data Types
+
+- "Integer" data of `1`, `2`, `4`, or `8` bytes
+  - data values
+  - addresses / untyped pointers
+- No aggregate types such as arrays or structures at the assembly level
+  - just contiguous bytes in memory
+- Two common syntaxes
+  - `AT&T`
+    - used in the course, slides, textbook, GNU tools
+  - `Intel`
+    - used in Intel documentation and Intel tools
+- Need to know which syntax you are reading because operand order may be reversed.
+
+#### Three Basic Kinds of Instructions
+
+- Transfer data between memory and register
+  - load
+    - `%reg = Mem[address]`
+  - store
+    - `Mem[address] = %reg`
+- Perform arithmetic on register or memory data
+  - examples: addition, shifting, bitwise operations
+- Control flow
+  - unconditional jumps to / from procedures
+  - conditional branches
+
+#### Abstract Memory Layout
+
+```text
+High addresses
+Stack              <- local variables, procedure context
+Dynamic Data       <- heap, new / malloc
+Static Data        <- globals / static variables
+Literals           <- large constants such as strings
+Instructions
+Low addresses
+```
+
+#### The ELF File Format
+
+- ELF = Executable and Linkable Format
+- One of the most widely used binary object formats
+- ELF is architecture-independent
+- ELF file types
+  - Relocatable
+    - must be fixed by the linker before execution
+  - Executable
+    - ready for execution
+  - Shared
+    - shared libraries with linking information
+  - Core
+    - core dumps created when a program terminates with a fault
+- Tools mentioned on slide
+  - `readelf`
+  - `file`
+  - `objdump -D`
+
+#### Process Memory Layout (32-bit x86 machine)
+
+- This slide is primarily a diagram.
+- Key idea: a `32-bit x86` process has a standard virtual memory layout with code, static data, heap, and stack arranged in distinct regions.
+
+We continue with the concrete runtime layout and the actual overflow mechanics in Lecture 18.

content/CSE4303/CSE4303_L18.md

@@ -0,0 +1,594 @@
+# CSE4303 Introduction to Computer Security (Lecture 18)
+
+> Due to lack of my attention, this lecture note is generated by AI to create continuations of the previous lecture note. I kept this warning because the note was created by AI.
+
+#### Software security
+
+### Overview
+
+#### Outline
+
+- Context
+- Prominent software vulnerabilities and exploits
+- Buffer overflows
+  - Background: C code, compilation, memory layout, execution
+  - Baseline exploit
+  - Challenges
+  - Defenses, countermeasures, counter-countermeasures
+
+### Buffer overflows
+
+#### All programs are stored in memory
+
+- The process's view of memory is that it owns all of it.
+- For a `32-bit` process, the virtual address space runs from:
+  - `0x00000000`
+  - to `0xffffffff`
+- In reality, these are virtual addresses.
+  - The OS and CPU map them to physical addresses.
+
+#### The instructions themselves are in memory
+
+- Program text is also stored in memory.
+- The slide shows instructions such as:
+
+```asm
+0x4c2 sub $0x224,%esp
+0x4c1 push %ecx
+0x4bf mov %esp,%ebp
+0x4be push %ebp
+```
+
+- Important point:
+  - code and data are both memory-resident
+  - control flow therefore depends on values stored in memory
+
+#### Data's location depends on how it's created
+
+- Static initialized data example
+
+```c
+static const int y = 10;
+```
+
+- Static uninitialized data example
+
+```c
+static int x;
+```
+
+- Command-line arguments and environment are set when the process starts.
+- Stack data appears when functions run.
+
+```c
+int f() {
+    int x;
+    ...
+}
+```
+
+- Heap data appears at runtime.
+
+```c
+malloc(sizeof(long));
+```
+
+- Summary from the slide
+  - Known at compile time
+    - text
+    - initialized data
+    - uninitialized data
+  - Set when process starts
+    - command line and environment
+  - Runtime
+    - stack
+    - heap
+
+#### We are going to focus on runtime attacks
+
+- Stack and heap grow in opposite directions.
+- Compiler-generated instructions adjust the stack size at runtime.
+- The stack pointer tracks the active top of the stack.
+- Repeated `push` instructions place values onto the stack.
+- The slides use the sequence:
+  - `push 1`
+  - `push 2`
+  - `push 3`
+  - `return`
+- Heap allocation is apportioned by the OS and managed in-process by `malloc`.
+- The lecture says: focusing on the stack for now.
+
+```text
+0x00000000                              0xffffffff
+Heap  --------------------------------->     <--------------------------------- Stack
+```
+
+#### Stack layout when calling functions
+
+Questions asked on the slide:
+
+- What do we do when we call a function?
+  - What data need to be stored?
+  - Where do they go?
+- How do we return from a function?
+  - What data need to be restored?
+  - Where do they come from?
+
+Example used in the slide:
+
+```c
+void func(char *arg1, int arg2, int arg3)
+{
+    char loc1[4];
+    int loc2;
+    int loc3;
+}
+```
+
+Important layout points:
+
+- Arguments are pushed in reverse order of code.
+- Local variables are pushed in the same order as they appear in the code.
+- The slide then introduces two unknown slots between locals and arguments.
+
+#### Accessing variables
+
+Example:
+
+```c
+void func(char *arg1, int arg2, int arg3)
+{
+    char loc1[4];
+    int loc2;
+    int loc3;
+    ...
+    loc2++;
+    ...
+}
+```
+
+Question from the slide:
+- Where is `loc2`?
+
+Step-by-step answer developed in the slides:
+
+- Its absolute address is undecidable at compile time.
+- We do not know exactly where `loc2` is in absolute memory.
+- We do not know how many arguments there are in general.
+- But `loc2` is always a fixed offset before the frame metadata.
+- This motivates the frame pointer.
+
+Definitions from the slide:
+
+- Stack frame
+  - the current function call's region on the stack
+- Frame pointer
+  - `%ebp`
+- Example answer
+  - `loc2` is at `-8(%ebp)`
+
+#### Notation
+
+- `%ebp`
+  - a memory address stored in the frame-pointer register
+- `(%ebp)`
+  - the value at memory address `%ebp`
+  - like dereferencing a pointer
+
+The slide sequence then shows:
+
+```asm
+pushl %ebp
+movl %esp, %ebp
+```
+
+- Meaning:
+  - first save the old frame pointer on the stack
+  - then set the new frame pointer to the current stack pointer
+
+#### Returning from functions
+
+Example caller:
+
+```c
+int main()
+{
+    ...
+    func("Hey", 10, -3);
+    ...
+}
+```
+
+Questions from the slides:
+
+- How do we restore `%ebp`?
+- How do we resume execution at the correct place?
+
+Slide answers:
+
+- Push `%ebp` before locals.
+- Set `%ebp` to current `%esp`.
+- Set `%ebp` to `(%ebp)` at return.
+- Push next `%eip` before `call`.
+- Set `%eip` to `4(%ebp)` at return.
+
+#### Stack and functions: Summary
+
+- Calling function
+  - push arguments onto the stack in reverse order
+  - push the return address
+    - the address of the instruction that should run after control returns
+  - jump to the function's address
+- Called function
+  - push old frame pointer `%ebp` onto the stack
+  - set frame pointer `%ebp` to current `%esp`
+  - push local variables onto the stack
+  - access locals as offsets from `%ebp`
+- Returning function
+  - reset previous stack frame
+    - `%ebp = (%ebp)`
+  - jump back to return address
+    - `%eip = 4(%ebp)`
+
+#### Quick overview (again)
+
+- Buffer
+  - contiguous set of a given data type
+  - common in C
+    - all strings are buffers of `char`
+- Overflow
+  - put more into the buffer than it can hold
+- Question
+  - where does the extra data go?
+- Slide answer
+  - now that we know memory layouts, we can reason about where the overwrite lands
+
+#### A buffer overflow example
+
+Example 1 from the slide:
+
+```c
+void func(char *arg1)
+{
+    char buffer[4];
+    strcpy(buffer, arg1);
+    ...
+}
+
+int main()
+{
+    char *mystr = "AuthMe!";
+    func(mystr);
+    ...
+}
+```
+
+Step-by-step effect shown in the slides:
+
+- Initial stack region includes:
+  - `buffer`
+  - saved `%ebp`
+  - saved `%eip`
+  - `&arg1`
+- First 4 bytes copied:
+  - `A u t h`
+- Remaining bytes continue writing:
+  - `M e ! \0`
+- Because `strcpy` keeps copying until it sees `\0`, bytes go past the end of the buffer.
+- In the example, upon return:
+  - `%ebp` becomes `0x0021654d`
+- Result:
+  - segmentation fault
+  - shown as `SEGFAULT (0x00216551)` in the slide sequence
+
+#### A buffer overflow example: changing control data vs. changing program data
+
+Example 2 from the slide:
+
+```c
+void func(char *arg1)
+{
+    int authenticated = 0;
+    char buffer[4];
+    strcpy(buffer, arg1);
+    if (authenticated) { ... }
+}
+
+int main()
+{
+    char *mystr = "AuthMe!";
+    func(mystr);
+    ...
+}
+```
+
+Step-by-step effect shown in the slides:
+
+- Initial stack contains:
+  - `buffer`
+  - `authenticated`
+  - saved `%ebp`
+  - saved `%eip`
+  - `&arg1`
+- Overflow writes:
+  - `A u t h` into `buffer`
+  - `M e ! \0` into `authenticated`
+- Result:
+  - code still runs
+  - user now appears "authenticated"
+
+Important lesson:
+- A buffer overflow does not need to crash.
+- It may silently change program data or logic.
+
+#### `gets` vs `fgets`
+
+Unsafe function shown in the slide:
+
+```c
+void vulnerable()
+{
+    char buf[80];
+    gets(buf);
+}
+```
+
+Safer version shown in the slide:
+
+```c
+void safe()
+{
+    char buf[80];
+    fgets(buf, 64, stdin);
+}
+```
+
+Even safer pattern from the next slide:
+
+```c
+void safer()
+{
+    char buf[80];
+    fgets(buf, sizeof(buf), stdin);
+}
+```
+
+Reference from slide:
+- [List of vulnerable C functions](https://security.web.cern.ch/security/recommendations/en/codetools/c.shtml)
+
+#### User-supplied strings
+
+- In the toy examples, the strings are constant.
+- In reality they come from users in many ways:
+  - text input
+  - packets
+  - environment variables
+  - file input
+- Validating assumptions about user input is extremely important.
+
+#### What's the worst that could happen?
+
+Using:
+
+```c
+char buffer[4];
+strcpy(buffer, arg1);
+```
+
+- `strcpy` will let you write as much as you want until a `\0`.
+- If attacker-controlled input is long enough, the memory past the buffer becomes "all ours" from the attacker's perspective.
+- That raises the key question from the slide:
+  - what could you write to memory to wreak havoc?
+
+#### Code injection
+
+- Title-only transition slide.
+- It introduces the move from accidental overwrite to deliberate attacker payloads.
+
+#### High-level idea
+
+Example used in the slide:
+
+```c
+void func(char *arg1)
+{
+    char buffer[4];
+    sprintf(buffer, arg1);
+    ...
+}
+```
+
+Two-step plan shown in the slides:
+
+- 1. Load my own code into memory.
+- 2. Somehow get `%eip` to point to it.
+
+The slide sequence draws this as:
+- vulnerable buffer on stack
+- attacker-controlled bytes placed in memory
+- `%eip` redirected toward those bytes
+
+#### This is nontrivial
+
+- Pulling off this attack requires getting a few things really right, and some things only sorta right.
+- The lecture says to think about what is tricky about the attack.
+- Main security idea:
+  - the key to defending it is to make the hard parts really hard
+
+#### Challenge 1: Loading code into memory
+
+- The attacker payload must be machine-code instructions.
+  - already compiled
+  - ready to run
+- We have to be careful in how we construct it.
+  - It cannot contain all-zero bytes.
+    - otherwise `sprintf`, `gets`, `scanf`, and similar routines stop copying
+  - It cannot make use of the loader.
+    - because we are injecting the bytes directly
+  - It cannot use the stack.
+    - because we are in the process of smashing it
+- The lecture then gives the name:
+  - shellcode
+
+#### What kind of code would we want to run?
+
+- Goal: full-purpose shell
+  - code to launch a shell is called shellcode
+  - it is nontrivial to write shellcode that works as injected code
+    - no zeroes
+    - cannot use the stack
+    - no loader dependence
+  - there are many shellcodes already written
+  - there are even competitions for writing the smallest shellcode
+- Goal: privilege escalation
+  - ideally, attacker goes from guest or non-user to root
+
+#### Shellcode
+
+High-level C version shown in the slides:
+
+```c
+#include <stdio.h>
+int main() {
+    char *name[2];
+    name[0] = "/bin/sh";
+    name[1] = NULL;
+    execve(name[0], name, NULL);
+}
+```
+
+Assembly version shown in the slides:
+
+```asm
+xorl %eax, %eax
+pushl %eax
+pushl $0x68732f2f
+pushl $0x6e69622f
+movl %esp, %ebx
+pushl %eax
+...
+```
+
+Machine-code bytes shown in the slides:
+
+```text
+"\x31\xc0"
+"\x50"
+"\x68""//sh"
+"\x68""/bin"
+"\x89\xe3"
+"\x50"
+...
+```
+
+Important point from the slide:
+- those machine-code bytes can become part of the attacker's input
+
+#### Challenge 2: Getting our injected code to run
+
+- We cannot insert a fresh "jump into my code" instruction.
+- We must use whatever code is already running.
+
+#### Hijacking the saved `%eip`
+
+- Strategy:
+  - overwrite the saved return address
+  - make it point into the injected bytes
+- Core idea:
+  - when the function returns, the CPU loads the overwritten return address into `%eip`
+
+Question raised by the slides:
+- But how do we know the address?
+
+Failure mode shown in the slide sequence:
+- if the guessed address is wrong, the CPU tries to execute data bytes
+- this is most likely not valid code
+- result:
+  - invalid instruction
+  - CPU "panic" / crash
+
+#### Challenge 3: Finding the return address
+
+- If we do not have the code, we may not know how far the buffer is from the saved `%ebp`.
+- One approach:
+  - try many different values
+- Worst case:
+  - `2^32` possible addresses on `32-bit`
+  - `2^64` possible addresses on `64-bit`
+- But without address randomization:
+  - the stack always starts from the same fixed address
+  - the stack grows, but usually not very deeply unless heavily recursive
+
+#### Improving our chances: nop sleds
+
+- `nop` is a single-byte instruction.
+- Definition:
+  - it does nothing except move execution to the next instruction
+- NOP sled idea:
+  - put a long sequence of `nop` bytes before the real malicious code
+  - now jumping anywhere in that region still works
+  - execution slides down into the payload
+
+Why this helps:
+- it increases the chance that an approximate address guess still succeeds
+- the slides explicitly state:
+  - now we improve our chances of guessing by a factor of `#nops`
+
+```text
+[padding][saved return address guess][nop nop nop ...][malicious code]
+```
+
+#### Putting it all together
+
+- Payload components shown in the slides:
+  - padding
+  - guessed return address
+  - NOP sled
+  - malicious code
+- Constraint noted by the lecture:
+  - input has to start wherever the vulnerable `gets` / similar function begins writing
+
+#### Buffer overflow defense #1: use secure bounds-checking functions
+
+- User-level protection
+- Replace unbounded routines with bounded ones.
+- Prefer secure languages where possible:
+  - Java
+  - Rust
+  - etc.
+
+#### Buffer overflow defense #2: Address Space Layout Randomization (ASLR)
+
+- Randomize starting address of program regions.
+- Goal:
+  - prevent attacker from guessing / finding the correct address to put in the return-address slot
+- OS-level protection
+
+#### Buffer overflow counter-technique: NOP sled
+
+- Counter-technique against uncertain addresses
+- By jumping somewhere into a wide sled, exact address knowledge becomes less necessary
+
+#### Buffer overflow defense #3: Canary
+
+- Put a guard value between vulnerable local data and control-flow data.
+- If overflow changes the canary, the program can detect corruption before returning.
+- OS-level / compiler-assisted protection in the lecture framing
+
+#### Buffer overflow defense #4: No-execute bits (NX)
+
+- Mark the stack as not executable.
+- Requires hardware support.
+- OS / hardware-level protection
+
+#### Buffer overflow counter-technique: ret-to-libc and ROP
+
+- Code in the C library is already stored at consistent addresses.
+- Attacker can find code in the C library that has the desired effect.
+  - possibly heavily fragmented
+- Then return to the necessary address or addresses in the proper order.
+- This is the motivation behind:
+  - `ret-to-libc`
+  - Return-Oriented Programming (ROP)
+
+We will continue from defenses / exploitation follow-ups in the next lecture.

content/CSE4303/_meta.js

@@ -20,5 +20,7 @@ export default {
     CSE4303_L13: "Introduction to Computer Security (Lecture 13)",
     CSE4303_L14: "Introduction to Computer Security (Lecture 14)",
     CSE4303_L15: "Introduction to Computer Security (Lecture 15)",
-    CSE4303_L16: "Introduction to Computer Security (Lecture 16)"
+    CSE4303_L16: "Introduction to Computer Security (Lecture 16)",
+    CSE4303_L17: "Introduction to Computer Security (Lecture 17)",
+    CSE4303_L18: "Introduction to Computer Security (Lecture 18)"
 }

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_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/Math4202_L29.md

@@ -0,0 +1,57 @@
+# Math4202 Topology II (Lecture 29)
+
+## Algebraic Topology
+
+### Fundamental Groups of Some Surfaces
+
+Recall from previous lecture, we talked about figure 8 shape.
+
+#### Lemma The fundamental group of figure-8 is not abelian
+
+The fundamental group of figure-8 is not abelian.
+
+<details>
+<summary>Proof</summary>
+
+Consider $U,V$ be two "fish shape" where $U\cup V$ is the figure-8 shape, and $U\cap V$ is $x$ shape.
+
+The $x$ shape is path connected, 
+
+$\pi_1(U,x_0)$ is isomorphic to $\pi_1(S^1,x_0)$, and $\pi_1(V,x_0)$ is isomorphic to $\pi_1(S^1,x_0)$.
+
+To show that is not abelian, we need to show that $\alpha*\beta\neq \beta*\alpha$.
+
+We will use covering map to do this.
+
+[Universal covering of figure-8](https://notenexta.trance-0.com/Math4202/universal-covering-of-figure-8.png)
+
+However, for proving our result, it is sufficient to use xy axis with loops on each integer lattice.
+
+And $\tilde{\alpha*\beta}(1)=(1,0)$ and $\tilde{\beta*\alpha}(1)=(0,1)$. By path lifting correspondence, the two loops are not homotopic.
+
+</details>
+
+#### Theorem for fundamental groups of double torus (Torus with genus 2)
+
+The fundamental group of Torus with genus 2 is not abelian.
+
+<details>
+<summary>Proof</summary>
+
+If we cut the torus in the middle, we can have $U,V$ is two "punctured torus", which is homotopic to the figure-8 shape.
+
+But the is trick is not enough to show that the fundamental group is not abelian.
+
+---
+
+First we use quotient map $q_1$ to map double torus to two torus connected at one point.
+
+Then we use quotient map $q_2$ to map two torus connected at one point to figure-8 shape.
+
+So $q=q_2\circ q_1$ is a quotient map from double torus to figure-8 shape.
+
+Then consider the inclusion map $i$ and let the double torus be $X$, we claim that $i_*:\pi_1(\infty,x_0)\to \pi_1(X,x_0)$ is injective.
+
+If $\pi_1(X,x_0)$ is abelian, then the figure 8 shape is abelian, that is contradiction.
+
+</details>

content/Math4202/Math4202_L30.md

@@ -0,0 +1,91 @@
+# Math4202 Topology II (Lecture 30)
+
+## Algebraic Topology
+
+We skipped a few chapters about Jordan curve theorem, which will be your final project soon. LOL, I will embedded the link once I'm done.
+
+### Seifert-Van Kampen Theorem
+
+#### The Seifert-Van Kampen Theorem
+
+Let $X=U\cup V$ be a union of two open subspaces. Suppose that $U\cap V$, $U,V$ are path connected. Fix $x_0\in U\cap V$.
+
+Let $H$ be a group (arbitrary). And now we assume $\phi_1,\phi_2$ be a group homomorphism, and $\phi_1:\pi_1(U,x_0)\to H$, and $\phi_2:\pi_1(V,x_0)\to H$.
+
+![Seifert-Van Kampen Theorem](https://notenextra.trance-0.com/Math4202/Math4202_L30/Seifert-Van-Kampen-Theorem.png)
+
+Let $i_1,i_2,j_1,j_2,i_{12}$ be group homomorphism induced by the inclusion maps.
+
+Assume this diagram commutes.
+
+$$
+\phi_1\circ i_1=\phi_2\circ i_2
+$$
+
+There is a group homomorphism $\Phi:\pi_1(X,x_0)\to H$ making the diagram commute. $\Phi\circ j_1=\phi_1$ and $\Phi\circ j_2=\phi_2$.
+
+We may change the base point using conjugations.
+
+<details>
+<summary>Side notes about free product of two groups</summary>
+
+Consider arbitrary group $G_1,G_2$, then $G_1\times G_2$ is a group.
+
+Note that the inclusion map $i_1:G_1\to G_1\times G_2$ is a group homomorphism and the inclusion map $i_2:G_2\to G_1\times G_2$ is a group homomorphism. The image of them commutes since $(e,g_2)(g_1,e)=(g_1,g_2)=(g_1,e)(e,g_2)$.
+
+#### The universal property
+
+Then we want to have a group $G$ such that for all group homomorphism $\phi:G_1\to H$ and $G_2\to H$, such that there always exists a map $\Phi: G\to H$ such that:
+
+- $\Phi\circ i_1=\phi_1$
+- $\Phi\circ i_2=\phi_2$
+
+#### How to construct the free group?
+
+We consider
+
+$$
+G_1*G_2=S=\{g_1h_1g_2h_2:g_1,g_2\in G_1,h_1,h_2\in G_2\}/\sim
+$$
+
+And we set $g_ie_{G_2}g_{i+1}\sim g_ig_{i+1}$ for $g_i\in G_1$ and $g_{i+1}\in G_2$.
+
+And $h_je_{G_1}h_{j+1}\sim h_jh_{j+1}$ for $h_j\in G_2$ and $h_{j+1}\in G_1$.
+
+And we define the group operation
+
+$$
+(g_1 h_1\cdots g_k h_k)*(h_1' g_1'\cdots h_l' g_l')=g_1 h_1\cdots g_k h_k g_1' h_2'\cdots h_l' g_l'
+$$
+
+And the inverse is defined
+
+$$
+(g_1 h_1\cdots g_k h_k)^{-1}=h_k^{-1} g_k^{-1}\cdots h_1^{-1} g_1^{-1}
+$$
+
+And $G=S$ is a well-defined group.
+
+The homeomorphism $G\to H$ is defined as
+
+$$
+\Phi((g_1 h_1\cdots g_k h_k))=\phi_1(g_1)\circ \phi_2(h_1)\circ \cdots \circ \phi_1(g_k)\circ \phi_2(h_k)
+$$
+
+Note $\circ$ is the group operation in $H$.
+
+> Group with such universal property is unique, so we don't need to worry for that too much.
+
+</details>
+
+Back to the Seifert-Van Kampen Theorem:
+
+Let $H=\pi_1(U,x_0)* \pi_1(V,x_0)$.
+
+Let $N$ be the **least normal subgroup** in the free product $H$, containing $i_1(g)i_2(g)^{-1}$, $\forall g\in \pi_1(U\cap V,x_0)$.
+
+Note $i_1(g)\in \pi_1(U,x_0)$ and $i_2(g)\in \pi_1(V,x_0)$. You may think of them as $G_1,G_2$ in the free group descriptions.
+
+#### Seifert-Van Kampen Theorem (classical version)
+
+There is an isomorphism between $\pi_1(U,x_0)* \pi_1(V,x_0)/N$ and $\pi_1(U\cup V,x_0)$.

content/Math4202/_meta.js

@@ -31,4 +31,8 @@ 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)",
+    Math4202_L29: "Topology II (Lecture 29)",
 }

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

@@ -1,5 +1,6 @@
 export default {
     index: "Course Description",
+    Exam_reviews: "Exam reviews",
     "---":{
         type: 'separator'
     },
@@ -29,4 +30,7 @@ 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)",
 }