## Algebra - Rings and fields

### 10. Quotient rings

• Definition 10.1: Let $I$ be an ideal of ring $R$ then $R/I=\{I+a / a \in R\}$ forms a ring with respect to addition and multiplication defined by $(I+a)+(I+b)=(I+a+b)$ and $(I+a)(I+b)=(I+ab)$ respectively called Quotient ring.

Theorems on Quotient rings

• Theorem 10.1: Quotient ring is a ring.
Proof: Let $I$ be an ideal of a ring $R$. Then quotient ring of $R$ is $R/I=\{I+a/a \in R\}$
To prove: $R/I$ is a ring w.r.t addition and multiplication defined by $(I+a)+(I+b)=(I+a+b)$ and $(I+a)(I+b)=(I+ab)$ respectively.
Let $x,y,z \in R/I$ $\Rightarrow$ $x=I+a$ and $y=I+b$ and $z=I+c$ where $a,b,c \in R$
(1.) Closure law: Consider $x+y$
$=(I+a)+(I+b)$
$=(I+a+b) \in R/I$
$\therefore$ $\forall$ $x,y \in R/I$ $\Rightarrow$ $x+y \in R/I$
(2.) Associative law: Consider $(x+y)+z$
$=[(I+a)+(I+b)]+(I+c)$
$=[I+a+b]+(I+c)$
$=(I+a+b+c)$
$=(I+a)+[I+b+c]$
$=(I+a)+[(I+b)+(I+c)]$
$=x+(y+z)$
$\therefore$ $\forall$ $x,y,z \in R/I$, we have $(x+y)+z=x+(y+z)$
(3.) Existence of additive identity: $\exists$ an element $0=(I+0) \in$ $R/I$, such that $x+0=(I+a)+(I+0)=(I+a+0)=(I+a)=x$ and $0+x=(I+0)+(I+a)=(I+0+a)=(I+a)=x$
$\therefore$ $x+0=0+x=x$ $\forall$ $x \in R/I$
(4.) Existence of additive inverse: For each $x=I+a \in$ $R/I$, $\exists$ an element $-x=I+(-a)=I-a$ $\in$ $R/I$, such that $x+(-x)=(I+a)+(I-a)=(I+a-a)=(I+0)=0$ and $(-x)+x=(I+(-a)+(I+a)=(I-a+a)=(I+0)=0$.
$\therefore$ $x+(-x)=(-x)+x=0 \in R/I$
(5.) Commutative law: Consider $x+y$
$=(I+a)+(I+b)$
$=(I+a+b)$
$=(I+b+a)$
$=(I+b)+(I+a)=y+x$
$\therefore$ $\forall x,y \in$ $R/I$, we have $x+y=y+x$.
(6.) Closure law: Consider, $xy$
$=(I+a)(I+b)$
$=(I+ab) \in R/I$
$\therefore$ $\forall$ $x,y \in R/I$, we have $xy \in$ $R/I$
(7.) Associative law: Consider $x(yz)$
$=(I+a)[(I+b)(I+c)]$
$=(I+a)(I+bc)$
$=(I+abc)$
$=(I+ab)(I+c)$
$=[(I+a)(I+b)](I+c)$
$=(xy)z$
$\therefore$ $\forall$ $x,y,z \in$ $R/I$, we have $x(yz)=(xy)z$
(8.) Distributive law: Consider, $x(y+z)$
$=(I+a)[(I+b)+(I+c)]$
$=(I+a)(I+b+c)$
$=(I+a(b+c))$
$=(I+ab+ac)$
$=(I+ab)+(I+ac)$
$=(I+a)(I+b)+(I+a)(I+c)$
$=xy+xz$
Similarly $(x+y)z=xz+yz$
$\therefore$ $x(y+z)=xy+xz$ and $(x+y)z=xz+yz$, $\forall$ $x,y,z \in$ $R/I$
$\therefore$ $R/I$ is a ring. $\blacksquare$

• Corollory 10.1.1: If $R$ is commutative a ring then so is $R/I$
Proof: $R$ is commutative $\Rightarrow$ $ab=ba$ $\forall$ $a,b \in R$
Consider $xy=I+ab=I+ba=yx$ $\Rightarrow$ $R/I$ is commutative $\square$

• Corollory 10.1.2: If $R$ is a ring with identity then so is $R/I$
Proof: $R$ is a ring with identity $\Rightarrow$ $1 \in R$
So $\exists$ $1=I+1 \in R/I$ such that $x(1)=(1)x=x$ $\forall$ $x \in R/I$ $\Rightarrow$ $R/I$ is a ring with identity. $\square$

• Theorem 10.2: (Fundamental theorem of ring homomorphism) If $f$ is a onto homomorphism from a ring $R$ to a ring $R'$ with kernel $K_f$ then, $f(R) \cong R/K_f$.
Proof: Let $f$ is a onto homomorphism from a ring $R$ to a ring $R'$ with kernel $K_f$. We know that every kernel is an ideal and so $R/K_f$ is a quotient ring.
Define $\phi:R/K_f \rightarrow f(R)$ by $\phi(K_f+a)=f(a)$ $\forall$ $a \in R$
(1.) $\phi$ is well defined.
Consider $K_f+a=K_f+b$ where $a,b \in R$
$\Rightarrow$ $a-b \in K_f$
$\Rightarrow$ $f(a-b)=0$
$\Rightarrow$ $f(a)-f(b)=0$ [$\because$ $f$ is an homomorphism]
$\Rightarrow$ $f(a)=f(b)$
$\Rightarrow$ $\phi(K_f+a)=\phi(K_f+b)$
(2.) $\phi$ is an homomorphism
Consider, $\phi[(K_f+a)+(k_f+b)]$
$=\phi(K_f+a+b)$
$=f(a+b)$
$=f(a)+f(b)$ [$\because$ $f$ is an homomorphism]
$=\phi(K_f+a)+\phi(K_f+b)$
Now consider, $\phi[(K_f+a)(K_f+b)]$
$=\phi(K_f+ab)$
$=f(ab)$
$=f(a)f(b)$ [$\because$ $f$ is an homomorphism]
$=\phi(K_f+a)\phi(K_f+b)$
(3.) $\phi$ is $1-1$
Let $\phi(K_f+a)=\phi(K_f+b)$
$\Rightarrow$ $f(a)=f(b)$
$\Rightarrow$ $f(a)-f(b)=0$
$\Rightarrow$ $f(a-b)=0$ [$\because$ $f$ is an homomorphism]
$\Rightarrow$ $a-b \in K_f$
Hence $K_f+a-b=k_f$ $\Rightarrow$ $K_f+a=K_f+b$
(4.) $\phi$ is onto
Let $x \in f(R)$ then $\exists$ $a \in R$ such that $x=f(a)$ $\Rightarrow$ $x=\phi(K_f+a)$
$\therefore$ The pre image of $x$ is $K_f+a$
$\therefore$ $\phi$ is an isomorphism. i.e., $R/K_f\cong f(R)$ $\blacksquare$

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