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content/Math4302/Math4302_L24.md

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+# Math4302 Modern Algebra (Lecture 24)
+
+## Rings
+
+### 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.
+
+<details>
+<summary>Examples of rings</summary>
+
+$(\mathbb{Z},+,*)$, $(\mathbb{R},+,*)$ are rings.
+
+---
+
+$(2\mathbb{Z},+,\cdot)$ is a ring.
+
+---
+
+$(M_n(\mathbb{R}),+,\cdot)$ is a ring.
+
+---
+
+$(\mathbb{Z}_n,+,\cdot)$ is a ring, where $a\cdot b=a*b\mod n$.
+
+e.g. in $\mathbb{Z}_{12}, 4\cdot 8=8$.
+
+</details>
+
+> [!TIP]
+>
+> If $(R+,\cdot)$ is a ring, then $(R,\cdot)$ may not be necessarily a group.
+
+#### Properties of rings
+
+Let $0$ denote the identity of addition of $R$. $-a$ denote the additive inverse of $a$.
+
+- $0\cdot a=a\cdot 0=0$
+- $(-a)b=a(-b)=-(ab)$, $\forall a,b\in R$
+- $(-a)(-b)=ab$, $\forall a,b\in R$
+
+<details>
+<summary>Proof</summary>
+
+1) $0\cdot a=(0+0)\cdot a=0\cdot a+0\cdot a$, by cancellation, $0\cdot a=0$.  
+Similarly, $a\cdot 0=0\cdot a=0$.
+
+2) $(a+(-a))\cdot b=0\cdot b=0$ by (1), So $a\cdot b +(-a)\cdot b=0$, $(-a)\cdot b=-(ab)$. Similarly, $a\cdot (-b)=-(ab)$.
+
+3) $(-a)(-b)=(a(-b))$ by (2), apply (2) again, $-(-(ab))=ab$.
+</details>
+
+#### Definition of commutative ring
+
+A ring $(R,+,\cdot)$ is commutative if $a\cdot b=b\cdot a$, $\forall a,b\in R$.
+
+<details>
+<summary>Example of non commutative ring</summary>
+
+$(M_n(\mathbb{R}),+,\cdot)$ is not commutative.
+
+</details>
+
+#### Definition of unity element
+
+A ring $R$ has unity element if there is an element $1\in R$ such that $a\cdot 1=1\cdot a=a$, $\forall a\in R$.
+
+> [!NOTE]
+>
+> Unity element is unique.
+>
+> Suppose $1,1'$ are unity elements, then $1\cdot 1'=1'\cdot 1=1$, $1=1'$.
+
+<details>
+<summary>Example of field have no unity element</summary>
+
+$(2\mathbb{Z},+,\cdot)$ does not have unity element.
+</details>
+
+#### 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 $a$.
+
+> [!TIP]
+>
+> If $a$ is a unit, then its inverse is unique. If $b,b'$ are inverses of $a$, then $b'=1b'=bab'=b1=b$.
+
+We use $a^{-1}$ or $\frac{1}{a}$ to represent the inverse of $a$.
+
+Let $R$ be a ring with unity, then $0$ is not a unit. (identity of addition has no multiplicative inverse)
+
+If $0b=b0=1$, then $\forall a\in R$, $a=1a=0a=0$.
+
+#### 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.
+
+<details>
+<summary>Example of field</summary>
+
+$(\mathbb{R},+,\cdot)$ is a field.
+
+---
+
+$(\mathbb{Z}_p,+,\cdot)$ is a field, where $p$ is a prime number.
+</details>
+
+#### Lemma $\mathbb{Z}_p$ is a field
+
+$\mathbb{Z}_p$ is a field if and only if $p$ is prime.
+
+<details>
+<summary>Proof</summary>
+
+If $\mathbb{Z}_n$ is a field, then $n$ is prime.
+
+<!-- This is equivalent to the statement that: If $\mathbb{Z}_p$ is a field and $1\leq m\leq n-1$, then $\operatorname{gcd}(m,n)=1$.
+
+We $\operatorname{gcd}(m,n)=d>1$,  -->
+
+We proceed by contradiction. Suppose $n$ is not a prime, then $d|n$ for some $2\leq d\leq n-1$, then $[d]$ does not have inverse.
+
+If $[d][x]=[1]$, then $dx\equiv 1\mod n$, so $dx-1=ny$ for some $y\in \mathbb{Z}$, but $d|dx$, and $d|ny$, so $d|1$ which is impossible.
+
+Therefore, $n$ is prime.
+
+---
+
+If $p$ is prime, then $\mathbb{Z}_p$ is a field.
+
+Since $p$ is a prime, then $\operatorname{gcd}(m,n)=1$ for $1\leq m\leq n-1$. So $1=mx+ny$ for some $x,y\in \mathbb{Z}_p$. Then $[x]$ (the remainder of $x$ when divided by $p$) is the multiplicative inverse of $[m]$. $[m][x]=[mx]=[1-ny]=[1]$.
+
+</details>

content/Math4302/_meta.js

@@ -26,4 +26,5 @@ export default {
     Math4302_L21: "Modern Algebra (Lecture 21)",
     Math4302_L22: "Modern Algebra (Lecture 22)",
     Math4302_L23: "Modern Algebra (Lecture 23)",
+    Math4302_L24: "Modern Algebra (Lecture 24)",
 }