Algebra - Rings and fields

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

7. Field of Quotients

  • Definition 7.1: Let $D$ be an domain. Field of quotient is defined as $\{(a,b)/ a,b \in D$ and $b\neq 0 \}$ denoted by $Q[D]$. i.e., $Q[D]= \{(a,b)/ a,b \in D$ and $b\neq 0 \}$.

  • Theorems on Field of Quotients

  • Theorem 7.1: $Q[D]$ is a field.
    Proof: Let $D$ be a domain. Then $Q[D]= \{(a,b)/ a,b \in D$ and $b\neq 0 \}$
    Let $x,y,z \in Q[D]$ $\Rightarrow$ $x=(a_1,b_1)$, $y=(a_2,b_2)$ and $z=(a_3,b_3)$ where $a_1,a_2,a_3,b_1,b_2,b_3 \in D$ and $b_1,b_2,b_3 \neq 0$ Define addition and multiplication on $Q[D]$ by $x+y=(a_1,b_1)+(a_2,b_2)=(a_1b_2+b_1a_2,b_1b_2)$ and
    $xy=(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2)$ respectively.
    Also we define $(a_1,b_1)=(a_2,b_2)$ if and only if $a_1b_2=b_1a_2$
    (1.) Closure law: Consider, $x+y$
    $=(a_1,b_1)+(a_2,b_2)$
    $=(a_1b_2+b_1a_2,b_1b_2)$ $\in$ $Q[D]$
    $\therefore$ $\forall$ $x,y \in$ $Q[D]$ $\Rightarrow$ $x+y \in Q[D]$
    (2.) Associative law: Consider, $(x+y)+z$
    $=[(a_1,b_1)+(a_2,b_2)]+(a_3,b_3)$
    $=(a_1b_2+b_1a_2,b_1b_2)+(a_3,b_3)$
    $=((a_1b_2+b_1a_2)b_3+b_1b_2a_3,b_1b_2b_3)$
    $=(a_1b_2b_3+b_1a_2b_3+b_1b_2a_3,b_1b_2b_3)$ $........(1)$
    Consider, $x+(y+z)$
    $=(a_1,b_1)+[(a_2,b_2)+(a_3,b_3)]$
    $=(a_1,b_1)+(a_2b_3+b_2a_3,b_2b_3)$
    $=(a_1b_2b_3+b_1(a_2b_3+b_2a_3),b_1b_2b_3)$
    $=(a_1b_2b_3+b_1a_2b_3+b_1b_2a_3,b_1b_2b_3)$ $........(2)$
    $\therefore$ $x+(y+z)=(x+y)+z$ $\forall$ $x,y,z \in$ $Q[D]$
    (3.) Existence of additive identity: $\exists$ an element $0=(0,1) \in$ $Q[D]$, such that \\ $x+0=(a_1,b_1)+(0,1)=(a_1,b_1)=x$ and similarly $0+x=x$
    $\therefore$ $x+0=0+x=x$ $\forall$ $x \in$ $Q[D]$
    (4.) Existence of additive inverse: For each $x=(a_1,b_1) \in$ $Q[D]$, $\exists$ an element $-x=(-a_1,b_1)$ $\in$ $Q[D]$, such that $x+(-x)=(a_1,b_1)+(-a_1,b_1)=(a_1b_1+b_1(-a_1),b_1b_1)=(0,b_1^2)=(0,1)=0$ and similarly $(-x)+x=0$
    $\therefore$ $x+(-x)=(-x)+x=0$ $\forall$ $x \in$ $Q[D]$
    Note: $(0,b_1^2)=(0,1) \Leftrightarrow 0 \times 1 = b_1^2 \times 0 \Leftrightarrow 0=0$
    (5.) Commutative law: Consider, $x+y$
    $=(a_1,b_1)+(a_2,b_2)$
    $=(a_1b_2+b_1a_2,b_1b_2)$ $........(1)$
    Consider, $y+x$
    $=(a_2,b_2)+(a_1,b_1)$
    $=(a_2b_1+b_2a_1,b_2b_1)$
    $=(a_1b_2+b_1a_2,b_1b_2)$ $........(2)$
    From $(1)$ and $(2)$ we have $x+y=y+x$
    $\therefore$ $x+y=y+x$ $\forall$ $x,y \in$ $\mathbb{Q}$
    (6.) Closure law: Consider, $xy$ $=(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2) \in$ $Q[D]$
    $\therefore$ $\forall$ $x,y \in Q[D] \Rightarrow xy \in Q[D]$
    (7.) Associative law: consider, $x(yz)$
    $=(a_1,b_1)[(a_2,b_2)(a_3,b_3)]$
    $=(a_1,b_1)[((a_2a_3,b_2b_3))]$
    $=(a_1a_2a_3,b_1b_2b_3)$
    $=(a_1a_2,b_1b_2)(a_3,b_3)$
    $=[(a_1,b_1)(a_2,b_2)](a_3,b_3)$
    $=(xy)z$
    $\therefore$ $x(yz)=(xy)z$ $\forall$ $x,y,z \in$ $Q[D]$
    (8.) Distributive law: Consider, $x(y+z)$
    $=(a_1,b_1)[(a_2,b_2)+(a_3,b_3)]$
    $=(a_1,b_1)[a_2b_3+b_2a_3,b_2b_3]$
    $=(a_1(a_2b_3+b_2a_3),b_1b_2b_3)$
    $=(a_1a_2b_3+a_1b_2a_3,b_1b_2b_3)$ $........(1)$
    and now consider, $xy+xz$
    $=(a_1,b_1)(a_2,b_2)+(a_1,b_1)(a_3,b_3)$
    $=(a_1a_2,b_1b_2)+(a_1a_3,b_1b_3)$
    $=(a_1a_2b_1b_3+b_1b_2a_1a_3,b_1b_2b_1b_3)$ $........(2)$
    From $(1)$ and $(2)$ we have $x(y+z)=xy+xz$ and similarly we can have $(x+y)z=xz+yz$
    $\therefore$ $x(y+z)=xy+xz$ and $(x+y)z=xz+yz$ $\forall$ $x,y,z \in$ $Q[D]$
    Note: $(a_1a_2b_3+a_1b_2a_3,b_1b_2b_3)=(a_1a_2b_1b_3+b_1b_2a_1a_3,b_1b_2b_1b_3)$
    $\Leftrightarrow$ $(a_1a_2b_3+a_1b_2a_3)b_1b_2b_1b_3=b_1b_2b_3(a_1a_2b_1b_3+b_1b_2a_1a_3)$
    $\Leftrightarrow$ $a_1a_2b_1^2b_2b_3^2+a_1b_2^2a_3b_1^2b_3=a_1a_2b_1^2b_2b_3+b_1^2b_2^2a_1a_3b_3$
    $\Leftrightarrow$ $a_1a_2b_2+a_1a_3b_3=a_1a_2b_2+a_1a_3b_3$
    (9.) Commuatative law: Consider, $xy$
    $=(a_1,b_1)(a_2,b_2)$
    $=(a_1a_2,b_1b_2)$
    $=(a_2a_1,b_2b_1)$
    $=(a_2,b_2)(a_1,b_1)$
    $=yx$
    $\therefore$ $xy=yx$ $\forall$ $x,y \in Q[D]$
    (10.) Existence of multiplicative identity: $\exists$ an element $1=(1,1) \in$ $Q[D]$, such that $(x)(1)=(a_1,b_1)(1,1)=(a_1,b_1)=x$ and similarly $(1)(x)=x$
    $\therefore$ $(x)(1)=(x)(1)=x$ $\forall$ $x \in$ $Q[D]$
    (11.) Existence of multiplicative inverse: For non zero $x=(a_1,b_1) \in Q[D]$ $\exists$ $x^{-1}=(b_1,a_1)$ $\in Q[D]$ such that $xx^{-1}=(a_1,b_1)(b_1,a_1)=(a_1b_1,b_1a_1)=(1,1)=1$ and $x^{-1}x=1$
    $\therefore$ $xx^{-1}=x^{-1}x=1$ $\forall$ $x \in Q[D]$
    $\therefore$ $Q[D]$ is a field. $\blacksquare$

  • Theorem 7.2: Let $D$ be an integral domain and $\phi:D\rightarrow Q[D]$. Then $\phi$ is an isomorphism.
    Proof: Let $D$ be a domain. Consider $Q[D]=\{(a,b)/a,b \in D$ and $b\neq 0\}$
    Define a map $\phi:D\rightarrow Q[D]$ by $\phi(a)=(a,1)$ $\forall$ $a \in D$
    Consider, $\phi(a+b)=(a+b,1)=(a,1)+(b,1)=\phi(a)+\phi(b)$
    Cosnider, $\phi(ab)=(ab,1)=(a,1)(b,1)=\phi(a)\phi(b)$
    $\therefore$ $\phi$ is a homomorphism.
    Let $\phi(a)=\phi(b)$
    $\Rightarrow (a,1)=(b,1)$
    $\Rightarrow a\times 1=b\times 1$
    $\Rightarrow a=b$
    $\therefore$ $\phi(a)=\phi(b) \Rightarrow a=b$ $\Rightarrow$ $\phi$ is $1-1$
    The pre image of $(a,1)$ $\in$ $Q[D]$ is $a$ $\in D$ such that $(a,1)=\phi(a)$
    $\therefore$ $\phi$ is onto.
    $\phi$ is an isomorphism $\blacksquare$

Problems 7

  1. Determine the field of quotients of an integral domain $\mathbb{Z}[\sqrt{3}]$
    Solution: Consider $\mathbb{Z}[\sqrt{3}]=\{a+b\sqrt{3}/ a,b \in \mathbb{Z}$ $ \} $
    Let $x,y \in \mathbb{Z}[\sqrt{3}] \Rightarrow x=a_1+b_1\sqrt{3}$ and $y=a_2+b_2\sqrt{3}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
    Consider, $\frac{x}{y}$
    $=\frac{a_1+b_1\sqrt{3}}{a_2+b_2\sqrt{3}}$
    $=\frac{a_1+b_1\sqrt{3}}{a_2+b_2\sqrt{3}} \times \frac{a_2-b_2\sqrt{3}}{a_2-b_2\sqrt{3}}$
    $=\frac{a_1a_2+b_1a_2\sqrt{3}-a_1b_2\sqrt{3}-3b_1b_2}{a_2^2-3b_2^2}$
    $=\frac{a_1a_2-3b_1b_2}{a_2^2-3b_2^2}+\frac{a_2b_1-a_1b_2}{a_2^2-3b_2^2}\sqrt{3}$
    $=a+b\sqrt{3}$ where $a=\frac{a_1a_2-3b_1b_2}{a_2^2-3b_2^2}$ and $b=\frac{a_2b_1-a_1b_2}{a_2^2-3b_2^2}$
    So, $a,b \in \mathbb{Q}$
    Hence, Field of quotients of $\mathbb{Z}[\sqrt{3}]$ is $\{a+b\sqrt{3}/ a,b \in \mathbb{Q}$ $ \}=\mathbb{Q}[\sqrt{3}]$ $\spadesuit$

  2. Find the field of quotients of $\mathbb{Z}_7$.
    Solution: $\mathbb{Z}_7=\{0,1,2,3,4,5,6\}$
    $Q[\mathbb{Z}_7]=\{(a,b)/ a,b \in \mathbb{Z}_7$ and $b\neq 0\}$
    $Q[\mathbb{Z}_7]=\{(a,b)/ a,b = 0,1,2,3,4,5,6$ and $b=1,2,3,4,5,6\}$
    $Q[\mathbb{Z}_7]=\{(0,1),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,5),(3,1),(3,2),$
    $(3,4),(3,5),(4,1),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,6),(6,1),(6,5)\}$ $\spadesuit$

  3. Find the field of quotients of $\mathbb{Z}_5$.
    Solution: $\mathbb{Z}_5=\{0,1,2,3,4 \}$
    $Q[\mathbb{Z}_5]=\{(a,b)/ a,b \in \mathbb{Z}_5$ and $b\neq 0 \}$
    $Q[\mathbb{Z}_5]=\{(a,b)/ a=0,1,2,3,4$ and $b=1,2,3,4 \}$
    $Q[\mathbb{Z}_5]=\{(0,1),(1,1),(1,2),(1,3),(1,4),(2,1),(2,3),(3,1),(3,2),(3,4),(4,1),(4,3) \}$ $\spadesuit$

Exercise 7

  1. Show that the set of integers can be embedded isomorphysically into the field of quotients.

  2. Determine the field of quotients of an integral domain $\mathbb{Z}[i]$

  3. Find the field of quotients of $\mathbb{Z}_3$

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