## Algebra - Rings and fields

### 1. Rings, fields and integral domains

• Definition 1.1: A non-empty set $R$ together with a binary operation $+$ is called an abelian group denoted by $(R,+)$ if the following holds:
(1) Closure law: $\forall$ $a,b \in R$ $\Rightarrow$ $a+b \in R$
(2) Associative law: $a+(b+c)=(a+b)+c$ $\forall$ $a,b,c \in R$
(3) Existence of additive identity: $\exists$ $e \in R$ such that $a+0=0+a=a$ $\forall$ $a \in R$
(4) Existence of additive inverse: For each $a \in R$ $\exists$ $-a \in R$ such that $a+(-a)=(-a)+a=0$
(5) Commutative law: $a+b=b+a$ $\forall$ $a,b \in R$

• Definition 1.2: A non-empty set $R$ together with a binary operation $\cdot$ is called an abelian group denoted by $(R,\cdot)$ if the following holds:
(1) Closure law: $\forall$ $a,b \in R$ $\Rightarrow$ $a\cdot b \in R$
(2) Associative law: $a\cdot(b\cdot c)=(a\cdot b)\cdot c$ $\forall$ $a,b,c \in R$
(3) Existence of multiplicative identity: $\exists$ $e \in R$ such that $a\cdot 1=1\cdot a=a$ $\forall$ $a \in R$
(4) Existence of multiplicative inverse: For each non zero element $a \in R$ $\exists$ $\frac{1}{a} \in R$ such that $a\cdot\frac{1}{a}=\frac{1}{a}\cdot a=1$
(5) Commutative law: $a\cdot b=b\cdot a$ $\forall$ $a,b \in R$

• Definition 1.3: A non-empty set $R$ together with binary operations $*_1$ and $*_2$ is called a ring if the following holds:
(1) $(R,*_1)$ is an abelian group.
(2) Closure law: $\forall$ $a,b \in R$, we have $a*_2b \in R$
(3) Associative law: $\forall$ $a,b,c \in R$, we have $a*_2(b*_2c)=(a*_2b)*_2c$
(4) Distributive law: $a*_2(b*_1c)=(a*_2b)*_1(a*_2c)$ and $(a*_1b)*_2c=(a*_2c)*_1(b*_2c)$, $\forall$ $a,b,c \in R$

• Notation: A ring $R$ with binary operations $*_1$ and $*_2$ is denoted by $(R,*_1,*_2)$

• Definition 1.4: A non-empty set $R$ together with binary operations $+$ and $\cdot$ is called a ring if the following holds:
(1) $(R,+)$ is an abelian group.
(2) Closure law: $\forall$ $a,b \in R$, we have $a\cdot b \in R$
(3) Associative law: $\forall$ $a,b,c \in R$, we have $a\cdot (b\cdot c)=(a\cdot b)\cdot c$
(4) Distributive law: $a\cdot(b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$, $\forall$ $a,b,c \in R$

• Notation: A ring $R$ with binary operations $+$ and $\cdot$ is denoted by $(R,+,\cdot)$
We simply write $(R,+,\cdot)$ as $R$

• Definition 1.5: A ring $R$ is said to be a commutative ring if $\forall$ $a,b \in R$, we have $a\cdot b=b\cdot a$

• Definition 1.6: A ring $R$ is said to be a ring with identity if $\exists$ an element $e \in R$, such that $a\cdot e=e \cdot a=a$ $\forall$ $a \in R$

• Definition 1.7: A ring with identity $R$ is said to be a skew field or division ring if every non zero element in $R$ has a multiplicative inverse in $R$ i.e., for each non zero $a \in R$ $\exists$ $\frac{1}{a} \in R$ such that $a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=e$

• Definition 1.8: A non-empty set $F$ with binary operations $+$ and $\cdot$ is said to be a field if the following holds:
(1) $(F,+)$ is an abelian group
(2) $(F,\cdot)$ is an abelian group
(3) Distributive law: $a\cdot(b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$ , $\forall$ $a,b,c \in F$

• Definition 1.9: A non zero element $a$ in a ring $R$ is called a zero divisor, if $\exists$ a non zero element $b$ in $R$, such that $a\cdot b=0$

• Definition 1.10: A commutative ring with identity having no zero divisors is called integral domain.

• Problem 1.1: Show that the set of integers is an integral domain with identity under usual addition and multiplication.
Solution: Consider $\mathbb{Z}=\{0,\pm 1,\pm 2, \pm 3,.......\}$ Let $x,y,z \in \mathbb{Z}$
(1.) Closure law: $\forall$ $x,y \in$ $\mathbb{Z}$, we have $x+y \in$ $\mathbb{Z}$
(2.) Associative law: $\forall$ $x,y,z \in$ $\mathbb{Z}$, we have $(x+y)+z=x+(y+z)$
(3.) Existence of additive identity: $\exists$ an element $0 \in$ $\mathbb{Z}$, such that $x+0=0+x=x$, $\forall$ $x \in$ $\mathbb{Z}$
(4.) Existence of additive inverse: For each $x \in$ $\mathbb{Z}$, $\exists$ an element $-x$ $\in$ $\mathbb{Z}$, such that $x+(-x)=(-x)+x=0$
(5.) Commutative law: $\forall$ $x,y \in$ $\mathbb{Z}$, we have $x+y=y+x$
(6.) Closure law: $\forall$ $x,y \in \mathbb{Z}$, we have $x\cdot y \in$ $\mathbb{Z}$
(7.) Associative law: $\forall$ $x,y,z \in$ $\mathbb{Z}$, we have $x\cdot (y\cdot z)=(x\cdot y)\cdot z$
(8.) Distributive law: $x\cdot (y+z)=x\cdot y+x\cdot z$ and $(x+y)\cdot z=x\cdot z+y\cdot z$, $\forall$ $x,y,z \in$ $\mathbb{Z}$
$\therefore$ $(\mathbb{Z},+,.)$ is a ring.
(9.) Commuatative law: $\forall$ $x,y \in$ $\mathbb{Z}$, we have $x\cdot y=y\cdot x$
$\therefore$ $(\mathbb{Z},+,.)$ is a commutative ring.}
(10.) Existence of multiplicative identity: $\exists$ an element $1 \in$ $\mathbb{Z}$, such that $x\cdot 1=1\cdot x=x$, $\forall$ $a \in$ $\mathbb{Z}$
$\therefore$ $(\mathbb{Z},+,.)$ is a commutative ring with identity.
(11.) Consider $x\cdot y=0$ $\Rightarrow$ $x=0$ or $y=0$ Hence, zero divisors are absent.
$\therefore$ $(\mathbb{Z},+,.)$ is an integral domain.

• Remark: $\mathbb{Z}$ is an integral domain but not a field. For $6$ $\in$ $\mathbb{Z}$ $\nexists$ $\frac{1}{6}$ $\in$ $\mathbb{Z}$

• Problem 1.2: Show that the set of real numbers is a field under usual addition and multiplication.
Solution: Let $x,y,z \in \mathbb{R}$
(1.) Closure law: $\forall$ $x,y \in$ $\mathbb{R}$, we have $x+y \in$ $\mathbb{R}$
(2.) Associative law: $\forall$ $x,y,z \in$ $\mathbb{R}$,we have $(x+y)+z=x+(y+z)$
(3.) Existence of additive identity: $\exists$ an element $0 \in$ $\mathbb{R}$, such that $x+0=0+x=x$, $\forall$ $x \in$ $\mathbb{R}$
(4.) Existence of additive inverse: For each $x \in$ $\mathbb{R}$, $\exists$ an element $-x$ $\in$ $\mathbb{R}$, such that $x+(-x)=(-x)+x=0$
(5.) Commutative law: $\forall$ $x,y \in$ $\mathbb{R}$, we have $x+y=y+x$
(6.) Closure law: $\forall$ $x,y \in \mathbb{R}$, we have $x\cdot y \in$ $\mathbb{R}$
(7.) Associative law: $\forall$ $x,y,z \in$ $\mathbb{R}$, we have $x\cdot(y\cdot z)=(x\cdot y)\cdot z$
(8.) Distributive law: $x\cdot (y+z)=x\cdot y+x\cdot z$ and $(x+y)\cdot z=x\cdot z+y\cdot z$, $\forall$ $x,y,z \in$ $\mathbb{R}$
$\therefore$ $(\mathbb{R},+,.)$ is a ring.
(9.) Commuatative law: $\forall$ $x,y \in$ $\mathbb{R}$, we have $x\cdot y=y\cdot x$
$\therefore$ $(\mathbb{R},+,.)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\exists$ an element $1 \in$ $\mathbb{R}$, such that $x\cdot 1=1\cdot x=x$, $\forall$ $x \in$ $\mathbb{R}$
$\therefore$ $(\mathbb{R},+,.)$ is a commutative ring with identity.
(11.) Existence of multiplicative inverse: For each non zero $x$ $\in$ $\mathbb{R}$ $\exists$ $\frac{1}{x}$ $\in$ $\mathbb{R}$ such that $x\cdot\frac{1}{x}=\frac{1}{x}\cdot x=1$
$\therefore$ $(\mathbb{R},+,.)$ is a field.

• Problem 1.3: Show that the set of real numbers is an integeral domain under usual addition and multiplication.
Solution: Let $x,y,z \in \mathbb{R}$ (First show that $\mathbb{R}$ is a commutative ring with idenity. Refer problem 2)
(11.) Consider $x\cdot y=0$ $\Rightarrow$ $x=0$ or $y=0$. Hence zero divisors are absent.
$\therefore$ $(\mathbb{R},+,.)$ is a integral domain.

• Problem 1.4: Show that the set of complex numbers is a field under usual addition and multiplication.
Solution: Consider $\mathbb{C}=\{a+ib$ /$a,b \in \mathbb{R}$ and $i=\sqrt{-1}\}$ Let $x,y,z \in \mathbb{C}$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{R}$ and $i=\sqrt{-1}$
(1.) Closure law: Consider, $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2) \in$ $\mathbb{C}$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $x+y \in \mathbb{C}$
(2.) Associative law: Consider, $(x+y)+z$
$=[(a_1+ib_1)+(a_2+ib_2)]+(a_3+ib_3)$
$=[(a_1+a_2)+i(b_1+b_2)]+(a_3+ib_3)$
$=(a_1+a_2+a_3)+i(b_1+b_2+bz_3)$
$=(a_1+ib_1)+[(a_2+a_3)+i(b_2+b_3)]$
$=(a_1+ib_1)+[(a_2+ib_2)]+(a_3+ib_3)]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $x,y,z \in$ $\mathbb{C}$
(3.) Existence of additive identity: $\exists$ an element $0=0+i0 \in$ $\mathbb{C}$, such that $x+0=0+x=x$ $\forall$ $x \in$ $\mathbb{C}$
(4.) Existence of additive inverse: For each $x=(a_1+ib_1) \in$ $\mathbb{C}$, $\exists$ an element $-x=-(a_1+ib_1)=-a_1-ib_1$ $\in$ $\mathbb{C}$, such that $x+(-x)=(-x)+x=0$ $\forall$ $x \in$ $\mathbb{C}$
(5.) Commutative law: Consider, $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2)$
$=(a_2+a_1)+i(b_2+b_1)$
$=(a_2+ib_2)+(a_1+ib_1)$
$=y+x$
$\therefore$ $x+y=y+z$ $\forall$ $x,y \in$ $\mathbb{C}$
(6.) Closure law: Consider, $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2+i^2b_1b_2$
$=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1) \in$ $\mathbb{C}$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $x\cdot y \in \mathbb{C}$
(7.) Associative law: Consider, $x(yz)$
$=(a_1+ib_1)[(a_2a_3-b_2b_3)+i(a_2b_3+a_3b_2)]$
$=a_1(a_2a_3)-a_1(b_2b_3)+a_1(ia_2b_3)+a_1(ia_3b_2)+ib_1(a_2a_3)-ib_1(b_2b_3)-b_1(a_2b_3)-b_1(a_3b_2)$
$=(a_1a_2)a_3-(a_1b_2)b_3+(a_1ia_2)b_3+(a_1ib_2)a_3+(ib_1a_2)a_3-(ib_1b_2)b_3-(b_1a_2)b_3-(b_1b_2)a_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(-a_1b_2+a_1a_2i-ib_1b_2-b_1a_2)b_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(ia_1b_2+a_1a_2-b_1b_2+ib_1a_2)ib_3$
$=(a_1a_2+ia_1b_2+ia_2b_1-b_1b_2)(a_3+ib_3)$
$=[(a_1+ib_1)(a_2+ib_2)](a_3+ib_3)$
$=(xy)z$
$\therefore$ $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ $\forall$ $x,y,z \in \mathbb{C}$
(8.) Distributive law: Consider, $x(y+z)$
$=(a_1+ib_1)[(a_2+ib_2)+(a_3+ib_3)]$
$=(a_1+ib_1)(a_2+a_3+ib_2+ib_3)$
$=a_1(a_2+a_3+ib_2+ib_3)+ib_1(a_2+a_3+ib_2+ib_3)$
$=a_1a_2+a_1a_3+a_1ib_2+a_1ib_3+ib_1a_2+ib_1a_3-b_1b_2-b_1b_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)+(a_1a_3+ia_1b_3+ib_1a_3-b_1b_3)$
$=(a_1+ib_1)(a_2+ib_2)+(a_1+ib_1)(a_3+ib_3)$
$=xy+xz$
Similarly, we have $(x+y)z=xz+yz$
$\therefore$ $x\cdot(y+z)=x\cdot y+x\cdot z$ and $(x+y)\cdot z=x\cdot z+y\cdot z$ $\forall$ $x,y,z \in$ $\mathbb{C}$
$\therefore$ $(\mathbb{C},+,.)$ is a ring.
(9.) Commuatative law: Consider, $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2-b_1b_2$
$=a_2a_1+ib_2a_1+ia_2b_1-b_2b_1$
$=(a_2+ib_2)(a_1+ib_1)$
$=yx$
$\therefore$ $x\cdot y=y\cdot x$ $\forall$ $x,y \in \mathbb{C}$
$\therefore$ $(\mathbb{C},+,.)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\exists$ an element $1=1+i0 \in$ $\mathbb{C}$, such that $x\cdot 1=1\cdot x=x$ $\forall$ $x \in$ $\mathbb{C}$
$\therefore$ $(\mathbb{C},+,.)$ is a commutative ring with identity.
(11.) Existence of multiplicative inverse: Consider $\frac{1}{x}$
$=\frac{1}{a_1+ib_1}=\frac{1}{a_1+ib_1}\frac{a_1-ib_1}{a_1-ib_1}=\frac{a_1-ib_1}{(a_1+ib_1)(a_1-ib_1)}=\frac{a_1-ib_1}{a_1^2+b_1^2}=\frac{a_1}{a_1^2+b_1^2}+i\frac{-b_1}{a_1^2+b_1^2} \in \mathbb{C}$
$\therefore$ For each non zero $x \in \mathbb{C}$ $\exists$ $\frac{1}{x} \in \mathbb{C}$ such that $x\cdot\frac{1}{x}=\frac{1}{x}\cdot x=1$ $\forall$ $x \in \mathbb{C}$
$\therefore$ $(\mathbb{C},+,.)$ is a field.

• Problem 1.5: Show that the set of complex numbers is an integral domain under usual addition and multiplication.
Solution: Consider $\mathbb{C}=\{a+ib$ /$a,b \in \mathbb{R}$ and $i=\sqrt{-1}\}$ Let $x,y,z \in \mathbb{C}$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{R}$ and $i=\sqrt{-1}$ (First show that $\mathbb{C}$ is an commutative ring with idnetity. Refer problem 4)
(11.) Consider $x\cdot y=0 \Rightarrow (a_1+ib_1)(a_2+ib_2)=0$
$\Rightarrow$ $(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)=0$
$\Rightarrow$ $a_1a_2-b_1b_2=0$ and $a_1b_2+a_2b_1=0$
$\Rightarrow$ $a_1a_2b_2-b_1b_2^2=0$ (Multiply by $b_2$) and $a_1a_2b_2+a_2^2b_1=0$ (Multiply by $a_2$)
$\Rightarrow$ $a_1a_2b_2+a_2^2b_1-a_1a_2b_2+b_1b_2^2=0$
$\Rightarrow$ $a_2^2b_1+b_1b_2^2=0$ $\Rightarrow$ $b_1(a_2^2+b_2^2)=0$
$\Rightarrow$ $b_1=0$ or $a_2^2+b_2^2=0$
If $a_2^2+b_2^2=0$ then $a_2^2=-b_2^2$ $\Rightarrow$ $\frac{a_2}{b_2}=\sqrt{-1}$ which is not possible.
So $b_1=0$ $\Rightarrow$ $a_1a_2=0$ and $a_1b_2=0$
$\Rightarrow$ $a_1=0$ or $a_2=0$ and $a_1=0$ or $b_2=0$
Hence $x=a_1+ib_1=0$ and $y=a_2+ib_2=0$. Zero divisors are absent.
$\therefore$ $(\mathbb{C},+,\cdot)$ is an integral domain. $\spadesuit$

• Problem 1.6: Show that the set of Gaussian integers is an integral domain under complex addition and complex multiplication.
Solution: Consider $\mathbb{Z}[i]=\{a+ib$ /$a,b \in \mathbb{Z}$ and $i=\sqrt{-1}\}$ Let $x,y,z \in \mathbb{Z}[i]$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$, $z=a_3+ib_3$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Z}$ and $i=\sqrt{-1}$
(1.) Closure law: Consider, $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2) \in$ $\mathbb{Z}[i]$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $x+y \in \mathbb{Z}[i]$
(2.) Associative law: Consider, $(x+y)+z$
$=[(a_1+ib_1)+(a_2+ib_2)]+(a_3+ib_3)$
$=[(a_1+a_2)+i(b_1+b_2)]+(a_3+ib_3)$
$=(a_1+a_2+a_3)+i(b_1+b_2+b_3)$
$=(a_1+ib_1)+[(a_2+a_3)+i(b_2+b_3)]$
$=(a_1+ib_1)+[(a_2+ib_2)]+(a_3+ib_3)]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $x,y,z \in$ $\mathbb{Z}[i]$
(3.) Existence of additive identity: $\exists$ an element $0=0+i0 \in$ $\mathbb{Z}[i]$, such that $x+0=0+x=x$ $\forall$ $x \in$ $\mathbb{Z}[i]$
(4.) Existence of additive inverse: For each $x=(a_1+ib_1) \in$ $\mathbb{Z}[i]$, $\exists$ an element $-x=-(a_1+ib_1)=-a_1-ib_1$ $\in$ $\mathbb{Z}[i]$, such that $x+(-x)=(-x)+x=0$ $\forall$ $x \in$ $\mathbb{Z}[i]$
(5.) Commutative law: Consider, $x+y$
$=(a_1+ib_1)+(a_2+ib_2)$
$=(a_1+a_2)+i(b_1+b_2)$
$=(a_2+a_1)+i(b_2+b_1)$
$=(a_2+ib_2)+(a_1+ib_1)$
$=y+x$
$\therefore$ $x+y=y+z$ $\forall$ $x,y \in$ $\mathbb{Z}[i]$
(6.) Closure law: Consider, $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2+i^2b_1b_2$
$=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1) \in$ $\mathbb{Z}[i]$
$\therefore$ $\forall$ $x,y \in$ $\mathbb{C}$ $\Rightarrow$ $x\cdot y \in \mathbb{Z}[i]$
(7.) Associative law: Consider, $x(yz)$
$=(a_1+ib_1)[(a_2a_3-b_2b_3)+i(a_2b_3+a_3b_2)]$
$=a_1(a_2a_3)-a_1(b_2b_3)+a_1(ia_2b_3)+a_1(ia_3b_2)+ib_1(a_2a_3)-ib_1(b_2b_3)-b_1(a_2b_3)-b_1(a_3b_2)$
$=(a_1a_2)a_3-(a_1b_2)b_3+(a_1ia_2)b_3+(a_1ib_2)a_3+(ib_1a_2)a_3-(ib_1b_2)b_3-(b_1a_2)b_3-(b_1b_2)a_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(-a_1b_2+a_1a_2i-ib_1b_2-b_1a_2)b_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)a_3+(ia_1b_2+a_1a_2-b_1b_2+ib_1a_2)ib_3$
$=(a_1a_2+ia_1b_2+ia_2b_1-b_1b_2)(a_3+ib_3)$
$=[(a_1+ib_1)(a_2+ib_2)](a_3+ib_3)$
$=(xy)z$
$\therefore$ $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ $\forall$ $x,y,z \in \mathbb{Z}[i]$
(8.) Distributive law: Consider, $x(y+z)$
$=(a_1+ib_1)[(a_2+ib_2)+(a_3+ib_3)]$
$=(a_1+ib_1)(a_2+a_3+ib_2+ib_3)$
$=a_1(a_2+a_3+ib_2+ib_3)+ib_1(a_2+a_3+ib_2+ib_3)$
$=a_1a_2+a_1a_3+a_1ib_2+a_1ib_3+ib_1a_2+ib_1a_3-b_1b_2-b_1b_3$
$=(a_1a_2+ia_1b_2+ib_1a_2-b_1b_2)+(a_1a_3+ia_1b_3+ib_1a_3-b_1b_3)$
$=(a_1+ib_1)(a_2+ib_2)+(a_1+ib_1)(a_3+ib_3)$
$=xy+xz$
Similarly, we have $(x+y)z=xz+yz$
$\therefore$ $x\cdot(y+z)=x\cdot y+x\cdot z$ and $(x+y)\cdot z=x\cdot z+y\cdot z$ $\forall$ $x,y,z \in$ $\mathbb{Z}[i]$
$\therefore$ $(\mathbb{Z}[i],+,\cdot)$ is a ring.
(9.) Commuatative law: Consider, $xy$
$=(a_1+ib_1)(a_2+ib_2)$
$=a_1a_2+ib_1a_2+ia_1b_2-b_1b_2$
$=a_2a_1+ib_2a_1+ia_2b_1-b_2b_1$
$=(a_2+ib_2)(a_1+ib_1)$
$=yx$
$\therefore$ $x\cdot y=y\cdot x$ $\forall$ $x,y \in \mathbb{Z}[i]$
$\therefore$ $(\mathbb{Z}[i],+,\cdot)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\exists$ an element $1=1+i0 \in$ $\mathbb{Z}[i]$, such that $x\cdot 1=1\cdot x=x$ $\forall$ $x \in$ $\mathbb{Z}[i]$
$\therefore$ $(\mathbb{Z}[i],+,\cdot)$ is a commutative ring with identity.
(11.) Consider $x\cdot y=0 \Rightarrow (a_1+ib_1)(a_2+ib_2)=0$
$\Rightarrow$ $(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)=0$
$\Rightarrow$ $a_1a_2-b_1b_2=0$ and $a_1b_2+a_2b_1=0$
$\Rightarrow$ $a_1a_2b_2-b_1b_2^2=0$ (Multiply by $b_2$) and $a_1a_2b_2+a_2^2b_1=0$ (Multiply by $a_2$)
$\Rightarrow$ $a_1a_2b_2+a_2^2b_1-a_1a_2b_2+b_1b_2^2=0$
$\Rightarrow$ $a_2^2b_1+b_1b_2^2=0$ $\Rightarrow$ $b_1(a_2^2+b_2^2)=0$
$\Rightarrow$ $b_1=0$ or $a_2^2+b_2^2=0$
If $a_2^2+b_2^2=0$ then $a_2^2=-b_2^2$ $\Rightarrow$ $\frac{a_2}{b_2}=\sqrt{-1}$ which is not possible.
So $b_1=0$ $\Rightarrow$ $a_1a_2=0$ and $a_1b_2=0$
$\Rightarrow$ $a_1=0$ or $a_2=0$ and $a_1=0$ or $b_2=0$
Hence $x=a_1+ib_1=0$ and $y=a_2+ib_2=0$. Zero divisors are absent.
$\therefore$ $(\mathbb{Z}[i],+,.)$ is an integral domain. $\spadesuit$

• Remark: $\mathbb{Z}[i]$ is an integral domain but not a field. For $5+0i \in \mathbb{Z}[i]$ but $\nexists$ $\frac{1}{5+0i} \in \mathbb{Z}[i]$.

• Problem 1.7: Show that the set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are integers is an integral domain.
Solution: Set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are integers, is denoted by $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}$ /$a,b \in \mathbb{Z}$ \}
Let $x,y,z \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}, y=a_2+b_2\sqrt{2}$ and $z=a_3+b_3\sqrt{2}$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Z}$
(1.) Closure law: Consider, $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $x+y \in \mathbb{Z}[\sqrt{2}]$
(2.) Associative law: Consider, $(x+y)+z$
$=[(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})$
$=[(a_1+a_2)+(b_1+b_2)\sqrt{2}]+(a_3+b_3\sqrt{2})$
$=(a_1+a_2+a_3)+(b_1+b_2+bz_3)\sqrt{2}$
$=(a_1+b_1)\sqrt{2}+[(a_2+a_3)+(b_2+b_3)\sqrt{2}]$
$=(a_1+b_1\sqrt{2})+[(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $\in \mathbb{Z}[\sqrt{2}]$
(3.) Existence of additive identity: $\exists$ an element $0=0+0\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$, such that $x+0=0+x=x$ $\forall$ $x \in \mathbb{Z}[\sqrt{2}]$
(4.) Existence of additive inverse: For each $x=a_1+b_1\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$, $\exists$ an element $-x=-(a_1+b_1\sqrt{2})=-a_1-b_1\sqrt{2}$ $\in$ $\mathbb{Z}[\sqrt{2}]$, such that $x+(-x)=(-x)+x=0$ $\forall$ $x \in \mathbb{Z}[\sqrt{2}]$
(5.) Commutative law: Consider, $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2}$
$=(a_2+a_1)+(b_2+b_1)\sqrt{2}$
$=(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})$
$=y+x$
$\therefore$ $x+y=y+x$ $\forall$ $x,y \in \mathbb{Z}[\sqrt{2}]$
(6.) Closure law: Consider, $xy$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=(a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $x\cdot y \in \mathbb{Z}[\sqrt{2}]$
(7.) Associative law: Consider, $x(yz)$
$=(a_1+b_1\sqrt{2})[(a_2+b_2\sqrt{2})(a_3+b_3\sqrt{2})]$
$=(a_1+b_1\sqrt{2})[(a_2a_3+2b_2b_3)+(a_2b_3+a_3b_2)\sqrt{2}]$
$=a_1a_2a_3+a_1a_2b_3\sqrt{2}+a_1a_3b_2\sqrt{2}+2a_1b_2b_3+b_1a_2a_3\sqrt{2}+2a_2b_1b_3+2a_3b_1b_2+2\sqrt{2}b_1b_2b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2\sqrt{2}+2a_1b_2+a_2b_1\sqrt{2}+2\sqrt{2}b_1b_2)b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)\sqrt{2}b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)(a_3+b_3\sqrt{2})$
$=[(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})](a_3+b_3\sqrt{2})$
$=(xy)z$
$\therefore$ $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ $\forall$ $x,y,z \in \mathbb{Z}[\sqrt{2}]$
(8.) Distributive law: $x(y+z)$
$=(a_1+b_1\sqrt{2})[(a_2+b_2\sqrt{2})+(a_3+b_3\sqrt{2})]$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2}+a_3+b_3\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2+a_1a_3+b_1a_3\sqrt{2}+a_1b_3\sqrt{2}+2b_1b_3$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})(a_3+b_3\sqrt{3})$
$=xy+xz$
Similarly, we have $(x+y)z=xz+yz$
$\therefore$ $x\cdot(y+z)=x\cdot y+x\cdot z$ and $(x+y)z=x\cdot z+y\cdot z$ $\forall$ $x,y,z$ $\mathbb{Z}[\sqrt{2}]$
$\therefore$ $(\mathbb{Z}[\sqrt{2}],+,\cdot)$ is a ring.
(9.) Commuatative law: Consider, $xy$
$=(a_1+b_2\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=a_2a_1+b_2a_1\sqrt{2}+a_2b_1\sqrt{2}+2b_2b_1$
$=(a_2+b_2\sqrt{2})(a_1+b_1\sqrt{2})$
$=yx$
$\therefore$ $x\cdot y=y\cdot x$ $\forall$ $x,y \in \mathbb{Z}[\sqrt{2}]$
$\therefore$ $(\mathbb{Z}[\sqrt{2}],+,.)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\exists$ an element $1=1+0\sqrt{2} \in$ $\mathbb{Z}[\sqrt{2}]$, such that $x\cdot 1=1\cdot x=x$ $\forall$ $x \in \mathbb{Z}[\sqrt{2}]$
$\therefore$ $(\mathbb{Z}[\sqrt{2}],+,\cdot)$ is a commutative ring with identity.
(11.) Consider $xy=0$
$\Rightarrow$ $(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})=0$
$\Rightarrow (a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2}=0$
$\Rightarrow (a_1a_2+2b_1b_2)=0$ and $(a_1b_2+a_2b_1)=0$
$\Rightarrow a_1a_2b_2+2b_1b_2^2=0$ (Multiply by $b_2$) and $a_1a_2b_2+a_2^2b_1=0$ (Multiply by $a_2$)
$\Rightarrow a_1a_2b_2+a_2^2b_1-a_1a_2b_2-2b_1b_2^2=0$ $\Rightarrow$ $b_1(a_2^2-2b_2^2)=0$
$\Rightarrow$ $b_1=0$ or $a_2^2-2b_2^2=0$
If $a_2^2-2b_2^2=0$ $\Rightarrow$ $a_2^2=2b_2^2$ $\Rightarrow$ $\frac{a_2}{b_2}=\sqrt{2}$ which is not possible.
So $b_1=0$ we have $a_1a_2=0$ and $a_1b_2=0$
$\Rightarrow$ $a_1=0$ or $a_2=0$ and $a_1=0$ or $b_2=0$
Thus, $x=a_1+b_1\sqrt{2}=0$ and $y=a_2+b_2\sqrt{2}=0$ Hence, Zero divisors are absent.
$\therefore$ $(\mathbb{Z}[\sqrt{2}],+,\cdot)$ is an integral domain. $\spadesuit$

• Remark: $(\mathbb{Z}[\sqrt{2}],+,\cdot)$ is not a field. For $5+0\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ $\nexists$ $\frac{1}{5+0\sqrt{2}} \in \mathbb{Z}[\sqrt{2}]$

• Problem 1.8: Show that the set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are rationals is an field and integral domain.
Solution: Set of numbers of the form $a+b\sqrt{2}$, where $a$ and $b$ are rationals, is denoted by $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}$ /$a,b \in \mathbb{Q}$ \} Let $x,y,z \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}, y=a_2+b_2\sqrt{2}$ and $z=a_3+b_3\sqrt{2}$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in \mathbb{Q}$
(1.) Closure law: Consider, $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x+y \in \mathbb{Q}[\sqrt{2}]$
(2.) Associative law: Consider, $(x+y)+z$
$=[(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})$
$=[(a_1+a_2)+(b_1+b_2)\sqrt{2}]+(a_3+b_3\sqrt{2})$
$=(a_1+a_2+a_3)+(b_1+b_2+bz_3)\sqrt{2}$
$=(a_1+b_1)\sqrt{2}+[(a_2+a_3)+(b_2+b_3)\sqrt{2}]$
$=(a_1+b_1\sqrt{2})+[(a_2+b_2\sqrt{2})]+(a_3+b_3\sqrt{2})]$
$=x+(y+z)$
$\therefore$ $(x+y)+z=x+(y+z)$ $\forall$ $\in \mathbb{Q}[\sqrt{2}]$
(3.) Existence of additive identity: $\exists$ an element $0=0+0\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$, such that $x+0=0+x=x$ $\forall$ $x \in \mathbb{Q}[\sqrt{2}]$
(4.) Existence of additive inverse: For each $x=a_1+b_1\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$, $\exists$ an element $-x=-(a_1+b_1\sqrt{2})=-a_1-b_1\sqrt{2}$ $\in$ $\mathbb{Q}[\sqrt{2}]$, such that $x+(-x)=(-x)+x=0$ $\forall$ $x \in \mathbb{Q}[\sqrt{2}]$
(5.) Commutative law: Consider, $x+y$
$=(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})$
$=(a_1+a_2)+(b_1+b_2)\sqrt{2}$
$=(a_2+a_1)+(b_2+b_1)\sqrt{2}$
$=(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})$
$=y+x$
$\therefore$ $x+y=y+x$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$
(6.) Closure law: Consider, $xy$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=(a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$
$\therefore$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x\cdot y \in \mathbb{Q}[\sqrt{2}]$
(7.) Associative law: Consider, $x(yz)$
$=(a_1+b_1\sqrt{2})[(a_2+b_2\sqrt{2})(a_3+b_3\sqrt{2})]$
$=(a_1+b_1\sqrt{2})[(a_2a_3+2b_2b_3)+(a_2b_3+a_3b_2)\sqrt{2}]$
$=a_1a_2a_3+a_1a_2b_3\sqrt{2}+a_1a_3b_2\sqrt{2}+2a_1b_2b_3+b_1a_2a_3\sqrt{2}+2a_2b_1b_3+2a_3b_1b_2+2\sqrt{2}b_1b_2b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2\sqrt{2}+2a_1b_2+a_2b_1\sqrt{2}+2\sqrt{2}b_1b_2)b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)a_3+(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)\sqrt{2}b_3$
$=(a_1a_2+a_1b_2\sqrt{2}+a_2b_1\sqrt{2}+2b_1b_2)(a_3+b_3\sqrt{2})$
$=[(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})](a_3+b_3\sqrt{2})$
$=(xy)z$
$\therefore$ $x\cdot(y\cdot z)=(x\cdot y)\cdot z$ $\forall$ $x,y,z \in \mathbb{Q}[\sqrt{2}]$
(8.) Distributive law: $x(y+z)$
$=(a_1+b_1\sqrt{2})[(a_2+b_2\sqrt{2})+(a_3+b_3\sqrt{2})]$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2}+a_3+b_3\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2+a_1a_3+b_1a_3\sqrt{2}+a_1b_3\sqrt{2}+2b_1b_3$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})+(a_1+b_1\sqrt{2})(a_3+b_3\sqrt{3})$
$=xy+xz$
Similarly, we have $(x+y)z=xz+yz$
$\therefore$ $x\cdot(y+z)=x\cdot y+x\cdot z$ and $(x+y)z=x\cdot z+y\cdot z$ $\forall$ $x,y,z$ $\mathbb{Q}[\sqrt{2}]$
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is a ring.
(9.) Commuatative law: Consider, $xy$
$=(a_1+b_2\sqrt{2})(a_2+b_2\sqrt{2})$
$=a_1a_2+b_1a_2\sqrt{2}+a_1b_2\sqrt{2}+2b_1b_2$
$=a_2a_1+b_2a_1\sqrt{2}+a_2b_1\sqrt{2}+2b_2b_1$
$=(a_2+b_2\sqrt{2})(a_1+b_1\sqrt{2})$
$=yx$
$\therefore$ $x\cdot y=y\cdot x$ $\forall$ $x,y \in \mathbb{Q}[\sqrt{2}]$
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\exists$ an element $1=1+0\sqrt{2} \in$ $\mathbb{Q}[\sqrt{2}]$, such that $x\cdot 1=1\cdot x=x$ $\forall$ $x \in \mathbb{Q}[\sqrt{2}]$
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is a commutative ring with identity.
(11.) Existence of multiplicative inverse: Consider $\frac{1}{x}$
$=\frac{1}{a_1+b_1\sqrt{2}}=\frac{1}{a_1+b_1\sqrt{2}}\frac{a_1-b_1\sqrt{2}}{a_1-b_1\sqrt{2}}=\frac{a_1-b_1\sqrt{2}}{a_1^2-2b_1^2}=\frac{a_1}{a_1^2-2b_1^2}+\frac{-b_1\sqrt{2}}{a_1^2-2b_1^2}$ $\in \mathbb{Q}[\sqrt{2}]$
$\therefore$ For each non zero $x \in \mathbb{Q}[\sqrt{2}]$ $\exists$ $\frac{1}{x} \in \mathbb{Q}[\sqrt{2}]$ such that $x\cdot\frac{1}{x}=\frac{1}{x}\cdot x=1$
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is a field.
(12.) Consider $xy=0$
$\Rightarrow$ $(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})=0$
$\Rightarrow (a_1a_2+2b_1b_2)+(a_1b_2+a_2b_1)\sqrt{2}=0$
$\Rightarrow (a_1a_2+2b_1b_2)=0$ and $(a_1b_2+a_2b_1)=0$
$\Rightarrow a_1a_2b_2+2b_1b_2^2=0$ (Multiply by $b_2$) and $a_1a_2b_2+a_2^2b_1=0$ (Multiply by $a_2$)
$\Rightarrow a_1a_2b_2+a_2^2b_1-a_1a_2b_2-2b_1b_2^2=0$ $\Rightarrow$ $b_1(a_2^2-2b_2^2)=0$
$\Rightarrow$ $b_1=0$ or $a_2^2-2b_2^2=0$
If $a_2^2-2b_2^2=0$ $\Rightarrow$ $a_2^2=2b_2^2$ $\Rightarrow$ $\frac{a_2}{b_2}=\sqrt{2}$ which is not possible.
So $b_1=0$ we have $a_1a_2=0$ and $a_1b_2=0$
$\Rightarrow$ $a_1=0$ or $a_2=0$ and $a_1=0$ or $b_2=0$
Thus, $x=a_1+b_1\sqrt{2}=0$ and $y=a_2+b_2\sqrt{2}=0$ Hence, Zero divisors are absent.
$\therefore$ $(\mathbb{Q}[\sqrt{2}],+,\cdot)$ is an integral domain. $\spadesuit$

• Problem 1.9: Show that the set of even intergers is a commutative ring without identity under usual addition and multiplication.
Solution: Consider the set of even integers, $2\mathbb{Z}=\{0,\pm 2,\pm 4, \pm 6,.......\}$ Let $a,b,c \in 2\mathbb{Z}$
(1.) Closure law: $\forall$ $a,b \in$ $2\mathbb{Z}$, we have $a+b \in$ $2\mathbb{Z}$
(2.) Associative law: $\forall$ $a,b,c \in$ $2\mathbb{Z}$, we have $(a+b)+c=a+(b+c)$
(3.) Existence of additive identity: $\exists$ an element $0 \in$ $2\mathbb{Z}$, such that $a+0=0+a=a$, $\forall$ $a \in$ $2\mathbb{Z}$
(4.) Existence of additive inverse: For each $a \in$ $2\mathbb{Z}$, $\exists$ an element $-a$ $\in$ $2\mathbb{Z}$, such that $a+(-a)=(-a)+a=0$
(5.) Commutative law: $\forall$ $a,b \in$ $2\mathbb{Z}$, we have $a+b=b+a$
(6.) Closure law: $\forall$ $a,b \in 2\mathbb{Z}$, we have $a\cdot b \in$ $2\mathbb{Z}$
(7.) Associative law: $\forall$ $a,b,c \in$ $2\mathbb{Z}$, we have $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
(8.) Distributive law: $a\cdot(b+c)=a\cdot b+a\cdot c$ and $(a+b)\cdot c=a\cdot c+b\cdot c$ , $\forall$ $a,b,c \in$ $2\mathbb{Z}$
$\therefore$ $(2\mathbb{Z},+,\cdot)$ is a ring.
(9.) Commuatative law: $\forall$ $a,b \in$ $2\mathbb{Z}$, we have $a.b=b.a$
$\therefore$ $(2\mathbb{Z},+,\cdot)$ is a commutative ring.
(10.) Existence of multiplicative identity: $\nexists$ an element $1 \in$ $2\mathbb{Z}$
$\therefore$ $(2\mathbb{Z},+,\cdot)$ is a commutative ring without identity. $\spadesuit$

• Problem 1.10: Show that the set of all $2 \times 2$ matrices over the set of real numbers form a non commutative ring with identity under matrix addition and matrix multiplication.
Solution: Consider the set of $2\times 2$ matrices $\mathbb{M}=\Bigg\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \big/ a,b,c,d \in \mathbb{R} \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A= \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$, $B=\begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix}$ and $C=\begin{bmatrix} z_1 & z_2 \\ z_3 & z_4 \end{bmatrix}$ where $x_i$'s, $y_i$'s, $z_i$'s $\in \mathbb{R}$ for $1\leq i \leq 4$
(1.)] Closure law: Consider $A+B$
$=\begin{bmatrix} x_1&x_2\\x_3&x_4 \end{bmatrix} + \begin{bmatrix} y_1&y_2\\y_3&y_4 \end{bmatrix}$
$=\begin{bmatrix} x_1+y_1&x_2+y_2\\x_3+y_3&x_4+y_4 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M} \Rightarrow A+B \in \mathbb{M}$
(2.) Associative law: Consider, $A+[B+C]$
$=\begin{bmatrix} x_1 & x_2\\ x_3 & x_4 \end{bmatrix}+\Bigg[ \begin{bmatrix} y_1 & y_2\\ y_3 & y_4 \end{bmatrix}+\begin{bmatrix} z_1 & z_2\\ z_3 & z_4 \end{bmatrix}\Bigg]$
$=\begin{bmatrix} x_1 & x_2\\ x_3 & x_4 \end{bmatrix}+\begin{bmatrix} y_1+z_1 & y_2+z_2\\ y_3+z_3 & y_4+z_4 \end{bmatrix} =\begin{bmatrix} x_1+y_1+z_1 & x_2+y_2+z_2\\ x_3+y_3+z_3 & x_4+y_4+z_4 \end{bmatrix}$
$=\begin{bmatrix} x_1+y_1 & x_2+y_2\\ x_3+y_3 & x_4+y_4 \end{bmatrix}+\begin{bmatrix} z_1 & z_2\\ z_3 & z_4 \end{bmatrix}$
$=\Bigg[\begin{bmatrix} x_1 & x_2\\ x_3 & x_4 \end{bmatrix}+\begin{bmatrix} y_1 & y_2\\ y_3 & y_4 \end{bmatrix}\Bigg]+\begin{bmatrix} z_1 & z_2\\ z_3 & z_4 \end{bmatrix}$
$=[A+B]+C$
$\therefore$ $A+[B+C]=[A+B]+C$ $\forall$ $A,B,C \in \mathbb{M}$
(3.) Existence of additive identity: $\exists$ $0=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} \in \mathbb{M}$ such that $A+0=0+A=A$ $\forall$ $A \in \mathbb{M}$
(4.) Existence of additive inverse: For each $A \in \mathbb{M}$ $\exists$ $-A=-\begin{bmatrix} x_1&x_2\\x_3&x_4 \end{bmatrix}=\begin{bmatrix} -x_1&-x_2\\-x_3&-x_4 \end{bmatrix} \in \mathbb{M}$ such that $A+(-A)=(-A)+A=0$
(5.) Commutative law: Consider $A+B$
$=\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4\end{bmatrix}+\begin{bmatrix} y_1 & y_2 \\ y_3 & y_4\end{bmatrix}$
$=\begin{bmatrix}x_1+y_1 & x_2+y_2 \\ x_3+y_3 & x_4+y_4 \end{bmatrix}$
$=\begin{bmatrix} y_1+x_1 & y_2+x_2 \\ y_3+x_3 & y_4+x_4 \end{bmatrix}$
$=\begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix}+\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$
$=B+A$
$\therefore$ $A+B=B+A$ $\forall$ $A,B \in \mathbb{M}$
(6.) Closure law: Consider $AB$
$=\begin{bmatrix}x_1&x_2\\x_3&x_4\end{bmatrix}\begin{bmatrix}y_1&y_2\\y_3&y_4\end{bmatrix}$
$=\begin{bmatrix}x_1y_1+x_2y_3&x_1y_2+x_2y_4\\x_3y_1+x_4y_3&x_3y_2+x_3y_4\end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M}$ $\Rightarrow$ $A\cdot B \in \mathbb{M}$
(7.) Associative law: $A\cdot(B\cdot C)=(A\cdot B)\cdot C$ $\forall$ $A,B,C \in \mathbb{M}$
(8.) Distributive law: Consider $A(B+C)$
$=\begin{bmatrix} a_1&b_1\\c_1&d_1 \end{bmatrix} \begin{bmatrix} a_2+a_3&b_2+b_3\\c_2+c_3&d_2+d_3 \end{bmatrix}$
$=\begin{bmatrix} a_1a_2+a_1a_3+b_1c_2+b_1c_3&a_1b_2+a_1b_3+b_1d_2+b_1d_3\\ c_1a_2+c_1a_3+d_1c_2+d_2c_3&c_1b_2+c_1b_3+d_1d_2+d_1d_3 \end{bmatrix}$
$=\begin{bmatrix} a_1a_2+b_1c_2&a_1b_2+b_1d_2\\c_1a_2+d_1c_2&c_1b_2+d_1d_2 \end{bmatrix}+$ $\begin{bmatrix} a_1a_3+b_1c_3&a_1b_3+b_1d_3\\c_1a_3+d_1c_3&c_1b_3+d_1d_3 \end{bmatrix}$
$=AB+AC$
Hence, $A\cdot(B+C)=A\cdot B+A\cdot C$
Similarly we have $(A+B)\cdot C=A\cdot C+B\cdot C$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring.
(9.) Existence of identity: $\exists$ $I=\begin{bmatrix} 1&0\\0&1 \end{bmatrix} \in \mathbb{M}$ such that $A\cdot I=I\cdot A=A$ $\forall$ $A \in \mathbb{M}$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring with identity.
(10.) Commutative law: Consider $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right],\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right] \in$ $\mathbb{M}$, we have $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]=\left[ \begin{array}{cc} 4 & -1 \\ 2 & -2 \end{array} \right]$ and $\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]=\left[ \begin{array}{cc} 2 & 3 \\ 2 & 0 \end{array} \right]$
Hence $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]\left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right] \neq \left[ \begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array} \right]\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$
$\therefore$ $(2\mathbb{M},+,\cdot)$ is a non commutative ring with identity. $\spadesuit$

• Problem 1.11: Show that the set of matrices of the form $\left[ \begin{array}{cc} 0&x\\ 0&y \end{array} \right]$ where $x$ and $y$ are rational numbers, form a non-commuative ring without identity under matrix addition and multiplication.
Solution: Consider the set of $2\times 2$ matrices $\mathbb{M}=\Bigg\{ \begin{bmatrix} 0 & x \\ 0 & y \end{bmatrix} \big/ x,y \in \mathbb{R} \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A= \begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}$, $B=\begin{bmatrix} 0&x_2 \\ 0&y_2 \end{bmatrix}$ and $C=\begin{bmatrix} 0&x_3 \\ 0&y_3 \end{bmatrix}$ where $x_i$'s, $y_i$'s $\in \mathbb{R}$ for $1\leq i \leq 3$
(1.) Closure law: Consider $A+B$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix} + \begin{bmatrix} 0&x_2 \\ 0&y_2 \end{bmatrix}$
$=\begin{bmatrix} 0&x_1+x_2 \\ 0&y_1+y_2 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M} \Rightarrow A+B \in \mathbb{M}$
(2.) Associative law: Consider, $A+[B+C]$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}+\Bigg[ \begin{bmatrix} 0&x_2 \\ 0&y_2 \end{bmatrix}+\begin{bmatrix} 0&x_3 \\ 0&y_3 \end{bmatrix}\Bigg]$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1\end{bmatrix}+\begin{bmatrix} 0&x_2+x_3 \\ 0&y_2+y_3\end{bmatrix} =\begin{bmatrix} 0&x_1+x_2+x_3 \\ 0&y_1+y_2+y_3 \end{bmatrix}$
$=\begin{bmatrix} 0&x_1+x_2 \\ 0&y_1+y_2 \end{bmatrix}+\begin{bmatrix} 0&x_3 \\ 0&y_3\end{bmatrix}$
$=\Bigg[\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}+\begin{bmatrix} 0&x_2 \\ 0&y_2\end{bmatrix}\Bigg]+\begin{bmatrix} 0&x_3 \\ 0&y_3\end{bmatrix}$
$=[A+B]+C$
$\therefore$ $A+[B+C]=[A+B]+C$ $\forall$ $A,B,C \in \mathbb{M}$
(3.) Existence of additive identity: $\exists$ $0=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} \in \mathbb{M}$ such that $A+0=0+A=A$ $\forall$ $A \in \mathbb{M}$
(4.) Existence of additive inverse: For each $A \in \mathbb{M}$ $\exists$ $-A=-\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}=\begin{bmatrix} 0&-x_1 \\ 0&-y_1 \end{bmatrix} \in \mathbb{M}$ such that $A+(-A)=(-A)+A=0$
(5.) Commutative law: Consider $A+B$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}+\begin{bmatrix} 0&x_2 \\ 0&y_2 \end{bmatrix}$
$=\begin{bmatrix} 0&x_1+x_2 \\ 0&y_1+y_2 \end{bmatrix}$
$=\begin{bmatrix} 0&x_2+x_1 \\ 0&y_2+y_1 \end{bmatrix}$
$=\begin{bmatrix} 0&x_2 \\ 0&y_2 \end{bmatrix}+\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}$
$=B+A$
$\therefore$ $A+B=B+A$ $\forall$ $A,B \in \mathbb{M}$
(6.) Closure law: Consider $AB$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix}\begin{bmatrix}0&x_2 \\ 0&y_2\end{bmatrix}$
$=\begin{bmatrix} 0&x_1y_2 \\ 0&y_1y_2 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M}$ $\Rightarrow$ $A\cdot B \in \mathbb{M}$
(7.) Associative law: $A\cdot(B\cdot C)=(A\cdot B)\cdot C$ $\forall$ $A,B,C \in \mathbb{M}$
(8.) Distributive law: Consider $A(B+C)$
$=\begin{bmatrix} 0&x_1 \\ 0&y_1 \end{bmatrix} \begin{bmatrix} 0&x_2+x_3 \\ 0&y_2+y_3 \end{bmatrix}$
$=\begin{bmatrix} 0&x_1y_2+x_1y_3 \\ 0&y_1y_2+y_1y_3 \end{bmatrix}$
$=\begin{bmatrix} 0&x_1y_2 \\ 0&y_1y_2 \end{bmatrix}+$ $\begin{bmatrix} 0&x_1y_3 \\ 0&y_1y_3 \end{bmatrix}$
$=AB+AC$
Hence, $A\cdot(B+C)=A\cdot B+A\cdot C$
Similarly we have $(A+B)\cdot C=A\cdot C+B\cdot C$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring.
(9.) Existence of identity: $\nexists$ $I=\begin{bmatrix} 1&0\\0&1 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring with without identity.
(10.) Commutative law: Consider $\left[ \begin{array}{cc} 0&1 \\ 0&2 \end{array} \right],\left[ \begin{array}{cc} 0&-1 \\ 0&2 \end{array} \right] \in$ $\mathbb{M}$, we have $\left[ \begin{array}{cc} 0&1 \\ 0&2 \end{array} \right]\left[ \begin{array}{cc} 0&-1 \\ 0&2 \end{array} \right]=\left[ \begin{array}{cc} 0&2 \\ 0&4 \end{array} \right]$ and $\left[ \begin{array}{cc} 0&-1 \\ 0&2 \end{array} \right]\left[ \begin{array}{cc} 0&1 \\ 0&2 \end{array} \right]=\left[ \begin{array}{cc} 0&-2 \\ 0&4 \end{array} \right]$
Hence $\left[ \begin{array}{cc} 0&1 \\ 0&2 \end{array} \right]\left[ \begin{array}{cc} 0&-1 \\ 0&2 \end{array} \right] \neq \left[ \begin{array}{cc} 0&-1 \\ 0&2 \end{array} \right]\left[ \begin{array}{cc} 0&1 \\ 0&2 \end{array} \right]$
$\therefore$ $(2\mathbb{M},+,\cdot)$ is a non commutative ring without identity. $\spadesuit$

• Problem 1.12 Show that the set of matrices of the form $\begin{bmatrix} x & y \\ -y & x \end{bmatrix}$ where $x,y \in \mathbb{R}$ is a field with respect to matrix addtion and matrix multiplication.
Solution: Consider the set of $2\times 2$ matrices $\mathbb{M}=\Bigg\{ \begin{bmatrix} x & y \\ -y & x \end{bmatrix} \big/ x,y \in \mathbb{R} \Bigg\}$
Let $A,B,C \in \mathbb{M}$ $\Rightarrow$ $A= \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}$, $B=\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$ and $C=\begin{bmatrix} x_3 & y_3 \\ -y_3 & x_3 \end{bmatrix}$ where $x_i$'s, $y_i$'s $\in \mathbb{R}$ for $1\leq i \leq 3$
(1.) Closure law: Consider $A+B$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} + \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$
$= \begin{bmatrix} x_1+x_2 & y_1+y_2 \\ -y_1-y_2 & x_1+x_2 \end{bmatrix}$
$= \begin{bmatrix} x_1+x_2 & y_1+y_2 \\ -(y_1+y_2) & x_1+x_2 \end{bmatrix} \in \mathbb{M}$
$\therefore$ $\forall$ $A,B \in \mathbb{M} \Rightarrow A+B \in \mathbb{M}$
(2.) Associative law: Consider, $A+[B+C]$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}+\Bigg[ \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}+\begin{bmatrix} x_3 & y_3 \\ -y_3 & x_3 \end{bmatrix}\Bigg]$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}+\begin{bmatrix} x_2+x_3 & y_2+y_3 \\ -y_2-y_3 & x_2+x_3 \end{bmatrix} =\begin{bmatrix} x_1+x_2+x_3 & y_1+y_2+y_3 \\ -y_1-y_2-y_3 & x_1+x_2+x_3 \end{bmatrix}$
$=\begin{bmatrix} x_1+x_2 & y_1+y_2 \\ -y_1-y_2 & x_1+x_2 \end{bmatrix}+\begin{bmatrix} x_3 & y_3 \\ -y_3 & x_3 \end{bmatrix}$
$=\Bigg[\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}+\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}\Bigg]+\begin{bmatrix} x_3 & y_3 \\ -y_3 & x_3 \end{bmatrix}$
$=[A+B]+C$
$\therefore$ $A+[B+C]=[A+B]+C$ $\forall$ $A,B,C \in \mathbb{M}$
(3.) Existence of additive identity: $\exists$ $0=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} \in \mathbb{M}$ such that $A+0=0+A=A$ $\forall$ $A \in \mathbb{M}$
(4.) Existence of additive inverse: For each $A \in \mathbb{M}$ $\exists$ $-A=-\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}=\begin{bmatrix} -x_1 & -y_1 \\ y_1 & -x_1 \end{bmatrix}=\begin{bmatrix} -x_1 & -y_1 \\ -(-y_1) & -x_1 \end{bmatrix} \in \mathbb{M}$ such that $A+(-A)=(-A)+A=0$
(5.) Commutative law: Consider $A+B$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}+\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$
$=\begin{bmatrix} x_1+x_2 & y_1+y_2 \\ -y_1-y_2 & x_1+x_2 \end{bmatrix}$
$=\begin{bmatrix} x_2+x_1 & y_2+y_1 \\ -y_2-y_1 & x_2+x_1 \end{bmatrix}$
$=\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}+\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}$
$=B+A$
$\therefore$ $A+B=B+A$ $\forall$ $A,B \in \mathbb{M}$
(6.) Closure law: Consider $AB$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$
$=\begin{bmatrix} x_1x_2-y_1y_2 & x_1y_2+x_2y_1 \\ -x_2y_1-x_1y_2 & -y_1y_2+x_2x_1 \end{bmatrix}$
$=\begin{bmatrix} x_1x_2-y_1y_2 & x_1y_2+x_2y_1 \\ -(x_2y_1+x_1y_2) & x_1x_2-y_1y_2 \end{bmatrix} \in \mathbb{M}$ $......(1)$
$\therefore$ $\forall$ $A,B \in \mathbb{M}$ $\Rightarrow$ $A\cdot B \in \mathbb{M}$
(7.) Associative law: $A\cdot(B\cdot C)=(A\cdot B)\cdot C$ $\forall$ $A,B,C \in \mathbb{M}$
(8.) Distributive law: Consider $A(B+C)$
$=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} \begin{bmatrix} x_2+x_3 & y_2+y_3 \\ -y_2-y_3 & x_2+x_3 \end{bmatrix}$
$=\begin{bmatrix} x_1x_2+x_1x_3-y_1y_2-y_1y_3&x_1y_2+x_1y_3+y_1x_2+y_1x_3\\ -y_1x_2-y_1x_3-x_1y_2-x_1y_3&-y_1y_2-y_1y_3+x_1x_2+x_1x_3 \end{bmatrix}$
$=\begin{bmatrix} x_1x_2-y_1y_2&x_1y_2+y_1x_2\\-y_1x_2-x_1y_2&-y_1y_2+x_1x_2 \end{bmatrix}+$ $\begin{bmatrix} x_1x_3-y_1y_3&x_1y_3+y_1x_3\\-y_1x_3-x_1y_3&-y_1y_3+x_1x_3 \end{bmatrix}$
$=AB+AC$
Hence, $A\cdot(B+C)=A\cdot B+A\cdot C$
Similarly we have $(A+B)\cdot C=A\cdot C+B\cdot C$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring.
(9.) Existence of identity: $\exists$ $I=\begin{bmatrix} 1&0\\0&1 \end{bmatrix} \in \mathbb{M}$ such that $A\cdot I=I\cdot A=A$ $\forall$ $A \in \mathbb{M}$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a ring with identity.
(10.) Commutative law: Consider $BA$
$=\begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}$
$=\begin{bmatrix} x_2x_1-y_2y_1 & x_2y_1+x_1y_2 \\ -x_1y_2-x_2y_1 & -y_2y_1+x_1x_2 \end{bmatrix}$ $......(2)$
From (1) and (2) we have $AB=BA$
$\therefore$ $A\cdot B=B\cdot A$ $\forall$ $A,B \in \mathbb{M}$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a commutative ring with identity.
(11.) Existence of inverse: Consider $A=\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix}$ then $A^{-1}=\frac{1}{|A|}$adj$A=\frac{1}{x_1^2+y_1^2}\begin{bmatrix} x_1 & -y_1 \\ y_1 & x_1 \end{bmatrix}$ exists as $x_1^2+y_1^2 \neq 0$ (For if $x_1^2+y_1^2=0$ $\Rightarrow$ $\frac{x_1}{y_1}=i$ which is not possible.)
Hence, for each non zero $A$ $\in \mathbb{M}$ $\exists$ $\frac{1}{A}=A^{-1}$ $\in \mathbb{M}$ such that $AA^{-1}=A^{-1}A=I$
$\therefore$ $(\mathbb{M},+,\cdot)$ is a field. $\spadesuit$

• Problem 1.13: Show that the set $(\mathbb{Z}_6,\oplus_6,\otimes_6)$ is a commutative ring with identity. Is it an integral domain?
Solution: $\mathbb{Z}_6=\{0,1,2,3,4,5\}$

(1.) Closure law: Since all possible sums belong to $\mathbb{Z}_6$, $\mathbb{Z}_6$ is closed w.r.t $\oplus_6$.
$\therefore$ $\forall$ $a,b \in \mathbb{Z}_6$ $\Rightarrow$ $a\oplus_6 b \in \mathbb{Z}_6$
(2.) Associative law: $a\oplus_6(b\oplus_6c)=(a\oplus_6b)\oplus_6c$ $\forall$ $a,b,c \in \mathbb{Z}_6$
(3.) Existence of additive identity: The row headed by $0$ is same as the top row, So $0 \in \mathbb{Z}_6$ is the additive identity.
(4.) Existence of additive inverse: From the table, we have
additive inverse of 0 is 0, additive inverse of 1 is 5, additive inverse of 2 is 4, additive inverse of 3 is 3 , additive inverse of 4 is 2, additive inverse of 5 is 1
$\therefore$ additive inverse of every element in $\mathbb{Z}_6$ exists in $\mathbb{Z}_6$
(5.) Commuatative law: $a\oplus_6b=b\oplus_6a$ $\forall$ $a,b \in \mathbb{Z}_6$

(6.) Closure law: Since all possible product belong to $\mathbb{Z}_6$, $\mathbb{Z}_6$ is closed w.r.t $\otimes_6$.
$\therefore$ $\forall$ $a,b \in \mathbb{Z}_6$ $\Rightarrow$ $a\otimes_6 b \in \mathbb{Z}_6$
(7.) Associative law: $a\otimes_6(b\otimes_6c)=(a\otimes_6b)\otimes_6c$ $\forall$ $a,b,c \in \mathbb{Z}_6$
(8.) Distributive law: $a\otimes_6(b\oplus_6c)=(a\otimes_6b)\oplus_6(a\otimes_6c)$ and $(a\oplus_6b)\otimes_6c=(a\otimes_6c)\oplus_6(b\otimes_6c)$ $\forall$ $a,b,c \in \mathbb{Z}_6$
$\therefore$ $(\mathbb{Z}_6$,$\oplus_6,\otimes_6)$ is a ring.
(9.) Commutative law: $a\otimes_6b=b\otimes_6a$ $\forall$ $a,b \in \mathbb{Z}_6$
$\therefore$ $(\mathbb{Z}_6$,$\oplus_6,\otimes_6)$ is a commutative ring.
(10.) Existence of multiplicative identity: The row headed by $1$ is same as the top row, So $1 \in \mathbb{Z}_6$ is the multiplicative identity.
$\therefore$ $(\mathbb{Z}_6$,$\oplus_6,\otimes_6)$ is a commutative ring with identity.
(11.) From the table we have $4 \otimes_6 3 =0$ Hence zero divisors are present in $\mathbb{Z}_6$
$\therefore$ $(\mathbb{Z}_6$,$\oplus_6,\otimes_6)$ is not an integral domain. $\spadesuit$

• Problem 1.14: Show that the set $\mathbb{Z}_5,\oplus_5,\otimes_5$ is a field.
Solution: $\mathbb{Z}_5=\{0,1,2,3,4\}$

(1.) Closure law: Since all possible sums belong to $\mathbb{Z}_5$, $\mathbb{Z}_5$ is closed w.r.t $\oplus_5$.
$\therefore$ $\forall$ $a,b \in \mathbb{Z}_5$ $\Rightarrow$ $a\oplus_5 b \in \mathbb{Z}_5$
(2.) Associative law: $a\oplus_5(b\oplus_5c)=(a\oplus_5b)\oplus_5c$ $\forall$ $a,b,c \in \mathbb{Z}_5$
(3.) Existence of additive identity: The row headed by $0$ is same as the top row, So $0 \in \mathbb{Z}_5$ is the additive identity.
(4.) Existence of additive inverse: From the table, we have additive inverse of 0 is 0, additive inverse of 1 is 4, additive inverse of 2 is 3, additive inverse of 3 is 2, additive inverse of 4 is 1
$\therefore$ additive inverse of every element in $\mathbb{Z}_5$ exists in $\mathbb{Z}_5$
(5.) Commuatative law: $a\oplus_5b=b\oplus_5a$ $\forall$ $a,b \in \mathbb{Z}_5$

(6.) Closure law: Since all possible products belong to $\mathbb{Z}_5$, $\mathbb{Z}_5$ is closed w.r.t $\otimes_5$.
$\therefore$ $\forall$ $a,b \in \mathbb{Z}_5$ $\Rightarrow$ $a\otimes_5 b \in \mathbb{Z}_5$
(7.) Associative law: $a\otimes_5(b\otimes_5c)=(a\otimes_5b)\otimes_5c$ $\forall$ $a,b,c \in \mathbb{Z}_5$
(8.) Distributive law: $a\otimes_5(b\oplus_5c)=(a\otimes_5b)\oplus_5(a\otimes_5c)$ and $(a\oplus_5b)\otimes_5c=(a\otimes_5c)\oplus_5(b\otimes_5c)$ $\forall$ $a,b,c \in \mathbb{Z}_5$
$\therefore$ $(\mathbb{Z}_5$,$\oplus_5,\otimes_5)$ is a ring.
(9.) Commutative law: $a\otimes_5b=b\otimes_5a$ $\forall$ $a,b \in \mathbb{Z}_5$
$\therefore$ $(\mathbb{Z}_5$,$\oplus_5,\otimes_5)$ is a commutative ring.
(10.) Existence of multiplicative identity: The row headed by $1$ is same as the top row So $1 \in \mathbb{Z}_5$ is the multiplicative identity.
$\therefore$ $(\mathbb{Z}_5$,$\oplus_5,\otimes_5)$ is a commutative ring with identity.
(11.) Existence of multiplicative inverse: From the table, we have multiplicative inverse of 1 is 1, multiplicative inverse of 2 is 3, multiplicative inverse of 3 is 2, multiplicative inverse of 4 is 4
$\therefore$ multiplicative inverse of every non zero element in $\mathbb{Z}_5$ exists in $\mathbb{Z}_5$
$\therefore$ $(\mathbb{Z}_5$,$\oplus_5,\otimes_5)$ is a field. $\spadesuit$

• Problem 1.15: Show that the set $\{0,2,4,6,8\}$ is a field under $\oplus_{10}$ and $\otimes_{10}$
Solution: Let $A=\{0,2,4,6,8\}$

(1.) Closure law: Since all possible sums belong to $A$, $A$ is closed w.r.t $\oplus_{10}$.
$\therefore$ $\forall$ $a,b \in A$ $\Rightarrow$ $a\oplus_{10}b\in A$
(2.) Associative law: $a\oplus_{10}(b\oplus_{10}c)=(a\oplus_{10}b)\oplus_{10}c$ $\forall$ $a,b,c \in A$
(3.) Existence of additive identity: The row headed by $0$ is same as the top row, So $0 \in A$ is the additive identity.
(4.) Existence of additive inverse: From the table, we have additive inverse of 0 is 0, additive inverse of 2 is 8, additive inverse of 4 is 6, additive inverse of 6 is 4, additive inverse of 8 is 2
$\therefore$ additive inverse of every element in $A$ exists in $A$
(5.) Commuatative law: $a\oplus_{10}b=b\oplus_{10}a$ $\forall$ $a,b \in A$

(6.) Closure law: Since all possible product belong to $A$, $A$ is closed w.r.t $\otimes_{10}$.
$\therefore$ $\forall$ $a,b \in A$ $\Rightarrow$ $a\otimes_{10}b\in A$
(7.) Associative law: $a\otimes_{10}(b\otimes_{10}c)=(a\otimes_{10}b)\otimes_{10}c$ $\forall$ $a,b,c \in A$
(8.) Distributive law: $a\otimes_{10}(b\oplus_{10}c)=(a\otimes_{10}b)\oplus_{10}(a\otimes_{10}c)$ and $(a\oplus_{10}b)\otimes_{10}c=(a\otimes_{10}c)\oplus_{10}(b\otimes_{10}c)$ $\forall$ $a,b,c \in A$
$\therefore$ $(A$,$\oplus_5,\otimes_{10})$ is a ring.
(9.) Commutative law: $a\otimes_{10}b=b\otimes_{10}a$ $\forall$ $a,b \in A$
$\therefore$ $(A$,$\oplus_{10},\otimes_{10})$ is a commutative ring.
(10.) Existence of multiplicative identity: The row headed by $6$ is same as the top row, So $6 \in A$ is the multiplicative identity.
$\therefore$ $(A$,$\oplus_{10},\otimes_{10})$ is a commutative ring with identity.
(11.) Existence of multiplicative inverse: From the table, we have multiplicative inverse of 2 is 8, multiplicative inverse of 4 is 4, the multiplicative inverse of 6 is 6, multiplicative inverse of 8 is 2
$\therefore$ multiplicative inverse of every non zero element in $A$ exists in $A$
$\therefore$ $(A$,$\oplus_{10},\otimes_{10})$ is a field. $\spadesuit$

• Exercise 1.1: Let $\mathbb{R}$ be a ring of real numbers under usual additon and multiplication. If $\oplus$ and $\otimes$ are defined on $\mathbb{R}$ such that, $a\oplus b=a+b+1$ and $a \otimes b=a+b+ab$ $\forall$ $a,b \in \mathbb{R}$. Then prove that $(\mathbb{R},\oplus,\otimes)$ is a commutative ring.

• Exercise 1.2: Show that the set $C$ of all real valued continuous functions defined in $[0,1]$ is a commutative ring with unity w.r.t the addition and multiplication defined by $(f+g)(x)=f(x)+g(x)$ and $(fg)(x)=f(x)g(x)$.

• Exercise 1.3: Prove that the set of matrices of the form $\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix}$ where $x$ and $y$ are real numbers is a ring under matrix addition and matrix multiplication. Is it a commutative ring with identity? Justify

• Exercise 1.4: Show that the set $(\mathbb{Z}_7,\oplus_7,\otimes_7)$ is a commutative ring with unity.

• Exercise 1.5: Show that the set $\mathbb{Z}[\sqrt{7}]$ is a commutative ring with respect to usual additon and multiplication.

• Exercise 1.6: Show that the set of rationals is a field under usual addition and multiplication.

• Exercise 1.7: Show that the set of rationals is a commutative ring with identity under usual addition and multiplication.

• Exercise 1.8: Show that the set of rationals is an domain under usual addition and multiplication.

• Proposition 1.1: If $a,b,c$ are any elements of ring $R$, then
(1.) $a.0=0.a=0$
(2.) $a.(-b)=-(a.b)=(-a).b$
(3.) $(-a).(-b)=a.b$
(4.) $a.(b-c)=a.b-a.c$
(5.) $(a-b).c=a.c-b.c$
Proof: (1.) Consider, $0+0=0$
$a.(0+0)=a.0$ [by pre multiplying $a$]
$a.0+a.0=a.0$ [by distributive law in ring $R$]
$a.0+a.0=a.0+0$
$a.0=0$ [by left cancellation law in ring $R$]
Now again consider, $0+0=0$
$(0+0).a=0.a$ [by post multiplying $a$]
$0.a+0.a=0.a$ [by distributive law in ring $R$]
$0.a+0.a=0.a+0$
$0.a=0$ [by left cancellation law in ring $R$]
Hence, we have $a.0=0.a=0$
(2.) Consider, $a.(-b)+a.b$
$=a.[(-b)+b]$
$=a.0=0$ $\Rightarrow$ $a.(-b)+a.b=0$ $\Rightarrow$ $a.(-b)=-(a.b)$
Now consider, $(-a).b+a.b$
$=[(-a)+a].b$
$=0.b=0$ $\Rightarrow$ $(-a).b+a.b=0$ $\Rightarrow$ $(-a).b=-(a.b)$
Hence, we have $a.(-b)=-(a.b)=(-a).b$
(3.) Consider, $(-a).(-b)$
$=-[a.(-b)]$ [$\because$ $(-a).b=-(a.b)$]
$=-[-(a.b)]$ [$\because$ $a.(-b)=-(a.b)$]
$=a.b$
Hence, we have $(-a).(-b)=a.b$
(4.) Consider, $a.(b-c)$
$=a.[b+(-c)]$
$=a.b+a.(-c)$ [by distributive law in ring $R$]
$=a.b+[-(a.c)]$ [$\because$ $a.(-b)=-(a.b)$]
$=a.b-a.c$
Hence, we have $a.(b-c)=a.b-a.c$
(5.) Consider, $(a-b).c$
$=[a+(-b)].c$
$=a.c+(-b).c$ [by distributive law in ring $R$]
$=a.c+[-(b.c)]$ [$\because$ $a.(-b)=-(a.b)$]
$=a.c-b.c$ $\blacksquare$

• Proposition 1.2: Let $R$ be a ring with identity and $a \in R$, then
(1.) $(-1).a=a(-1)=-a$
(2.) $(-1).(-1)=1$
Proof: (1.) Consider, $a.(-1)+a.1$
$=a.[(-1)+1]$ [by distributive law in ring $R$]
$=a.0$
$=0$ [$\because$ $a.0=0$]
So, we have $a.(-1)+a.1=0$ $\Rightarrow$ $a.(-1)=-a$
Now Consider, $(-1).a+1.a$
$=[(-1)+1].a$ [by distributive law in ring $R$]
$=0.a$
$=0$ \hfill[$\because$ $0.a=0$]
So, we have $(-1).a+1.a=0$ $\Rightarrow$ $(-1).a=-a$
Hence, $(-1).a=a(-1)=-a$
(2.) Put $a=-1$ in the above proof. $\blacksquare$

• Theorem 1.1: A ring $R$ is without zero divisors if and only if the cancellation law holds in $R$.
Proof: Let $R$ be a ring without zero divisors and let $a,b,c \in R$ such that $ab=ac$ with $a \neq 0$
Then $ab-ac=0$
$\Rightarrow$ $a(b-c)=0$
$\Rightarrow$ $a=0$ or $b-c=0$ [$\because$ $R$ is without zero divisors]
$\Rightarrow$ $b-c=0$ [$\because a \neq 0$]
$\Rightarrow$ $b=c$
Thus $a.b=a.c$ $\Rightarrow$ $b=c$ i.e., left cancellation holds in $R$.
Now let $a,b,c \in R$ such that $ba=ca$ with $a\neq 0$
Then $ba-ca=0$
$\Rightarrow$ $(b-c)a=0$
$\Rightarrow$ $b-c=0$ or $a=0$
$\Rightarrow$ $b-c=0$
$\Rightarrow$ $b=c$
Thus $ba=ca$ $\Rightarrow$ $b=c$ i.e., right cancellation holds in $R$.
Conversely: Let us suppose cancellation law holds in $R$. Let $a,b \in R$ such that $ab=0$
Suppose $ab=0$ $\Rightarrow$ $a\neq 0$ and $b\neq 0$
Then $ab=0$
$\Rightarrow$ $ab=a0$
$\Rightarrow$ $b=0$ [by left cancellation in $R$] which is a contradiction to the fact that $b\neq 0$
Thus $ab=0$ $\Rightarrow$ $a=0$ or $b=0$ i.e., zero divisors are absent.
$\therefore$ $R$ is without zero divisors. $\blacksquare$

• Theorem 1.2: Every field is an integral doamin.
Proof: Let $F$ be a field. To prove: $F$ has no zero divisors. Let $a,b \in F$ such that $ab=0$ with $a \neq 0$. Since $a\neq 0 \in F$ $\Rightarrow$ $a^{-1} \in F$
Consider $ab=0$
$\Rightarrow$ $a^{-1}(ab)=a^{-1}0$ [by pre multiplying $a^{-1}$]
$\Rightarrow$ $(a^{-1}a)b=0$ [by associative law in $F$]
$\Rightarrow$ $b=0$
Similarly if $ab=0$ with $b\neq 0$ we get $a=0$
Thus if $ab=0$ $\Rightarrow$ $a=0$ or $b=0$ i.e., zero divisors are absent.
$F$ is an integral domain. $\blacksquare$
The converse of the above theorem need not be true, i.e., every integral domain may not be a field.
Example: $(\mathbb{Z},+,\cdot)$ is an integral domain but not a field.

• Theorem 1.3: A finite integral domain is a field.
Prooof: Let $D$ be a finite integral domain with $n$ distinct elements say $x_1,x_2,...,x_n$. If $a\in D$ with $a \neq 0$, then $x_1a,x_2a,.....,x_na \in D$
Claim: $x_1a,x_2a,....,x_na$ are $n$ distinct elements.
uppose if $x_ia=x_ja$ when $i\neq j$
$\Rightarrow$ $x_ia-x_ja=0$
$\Rightarrow$ $(x_i-x_j)a=0$
$\Rightarrow$ $x_i-x_j=0$ [$\because$ $a\neq 0$]
$\Rightarrow$ $x_i=x_j$ which is a contradiction to the fact that $x_1,x_2,...,x_n$ are distinct.
So $x_1a,x_2a,....,x_na$ are $n$ distinct elements in $D$.
Since $D$ is a domain, $1\in D$ $\Rightarrow$ $1=x_ia$ for some $i$, $1\leq i\leq n$
$\Rightarrow$ $a=x^{-1}_i$ $\in D$0
$\therefore$ Every non zero element in $D$ has a multiplictive inverse in $D$ and so $D$ is a field. $\blacksquare$

• Theorem 1.4: $\mathbb{Z}_n$ is an integral domain if and only if $n$ is prime.
Proof: Suppose $\mathbb{Z}_n$ is an integral domain. To prove: $n$ is prime.
Suppose if possible $n$ is not a prime. Then $n=ab$ where $a\neq 0, b \neq 0$ and $a<n, b<n$
$\Rightarrow$ $0=ab$ where $a\neq 0$ and $b\neq 0$ [$\because$ In $\mathbb{Z}_n$, $n\equiv 0$]
$\Rightarrow$ $a\neq 0$ and $b\neq 0$ $\Rightarrow$ $ab=0$
$\Rightarrow$ $a$ and $b$ are zero divisors in $\mathbb{Z}_n$
$\Rightarrow$ $\mathbb{Z}_n$ is not an integral domain, which is a contradiction to the fact that $\mathbb{Z}_n$ is an integral domain. Thus $n$ must be prime.
Conversely: Suppose $n$ is prime. To prove: $\mathbb{Z}_n$ is an integral domain. Let $a,b \in \mathbb{Z}_n$ such that $ab=0$
$\Rightarrow$ $ab=nk$ where $k$ is an integer.
$\Rightarrow$ $\frac{ab}{n}=k$
$\Rightarrow$ $n|ab$
$\Rightarrow$ $n|a$ or $n|b$ [$\because$ $n$ is prime]
$\Rightarrow$ $\frac{a}{n}=k_1$ or $\frac{b}{n}=k_2$ where $k_1,k_2$ are integers
$\Rightarrow$ $a=k_1n$ or $b=k_2n$
$\Rightarrow$ $a=0$ or $b=0$
$\therefore$ $ab=0$ $\Rightarrow$ $a=0$ or $b=0$ i.e., zero divisors are absent in $\mathbb{Z}_n$ and so $\mathbb{Z}_n$ is an integral domian. $\blacksquare$

• Theorem 1.5: $\mathbb{Z}_n$ is a field if and only if $n$ is prime.
Proof: Follows from theorem 1.2, 1.3 and 1.4 $\square$

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