Algebra - Rings and fields

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: nstejas@gmail.com Mobile: +91 9845410469

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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