Chapter 6.5, Problem 11E

(a)

To determine

To find: The additive identity in the ring, the unity element, and the additive and multiplicative inverse of $\left(2,5\right)$ . For each answer a representative of the equivalence class in $P$ must be given.

Answer to Problem 11E

The additive and multiplicative inverse of $\left(2,5\right)$ is $\left\{\left(0,n\right):n\in \mathbb{Z}-\left\{0\right\}\right\}$ and $\{(5n,2n):n\in \mathbb{Z}-\{0\left\}\right\}$ .

Explanation of Solution

Given Information:

R is an equivalence relation on $\mathbb{Z}\times \left(\mathbb{Z}-\left\{0\right\}\right)$ given by $\left(x,y\right)R\left(u,v\right)$ if $xv=yu$ . Consider $P$ as the set of equivalence classes of $\mathbb{Z}\times \left(\mathbb{Z}-\left\{0\right\}\right)$ modulo $R$ . For every $\left(a,b\right)$ and $\left(c,d\right)$ in $P$ it defines the operation $\oplus$ and $\otimes$ by $\left(a,b\right)$ $\oplus$ $\left(c,d\right)$ $=$ $\left(ad+bc,bd\right)$ and $\left(a,b\right)$ $\otimes$ $\left(c,d\right)$ = $\left(ac,bd\right)$ .

It is given that Element $\left(c,d\right)$ is an additive identity if $\left(a,b\right)$ $\oplus$ $\left(c,d\right)$ $=$ $\left(a,b\right)$ $\Rightarrow$ $\left(ad+bc,bd\right)$ $=$ $\left(a,b\right)$ .

So,

  $(ad+bc)$ = $a$, $bd=b$which leads to $d=1$ and $c=0$ .

Hence, the additive identity is $\left(c,d\right)=\left(0,1\right)$ .

Now $\left(0,1\right)R\left(u,v\right)$ if $u=0$ hence the equivalence class will be $\left(o,v\right)$ where $v\ne 0$ or else it would not be present in $\mathbb{Z}\times \left(\mathbb{Z}-\left\{0\right\}\right)$ .

Therefore, additive inverse is $\left\{\left(0,n\right):n\in \mathbb{Z}-\left\{0\right\}\right\}$ .

For additive inverse of $\left(2,5\right)$ it is $(2,5)\oplus (a,b)=(2b+5a,5b)=(0,v)$

So, $2b+5a=0$ , $5b=v$ . As $a$ and $b$ must also be integers with $b\ne 0$ , then $v=5n$ , $n\in \mathbb{Z}-\left\{0\right\}$ so $b\in \mathbb{Z}-\left\{0\right\}$ .

This leads to,

  $2b+5a=0\Rightarrow a=\frac{2}{5}b$

As, $a\in \mathbb{Z}$ , $b=\{5n:n\in \mathbb{Z}-\{0\left\}\right\}$ this gives us,

  $-(2,5)=\{(-2n,5n):n\in \mathbb{Z}-\{0\left\}\right\}$

For the multiplicative inverse first we have to determine the multiplicative identity

  $(a,b)\otimes (c,d)=(a,b)\Rightarrow (ac,bd)=(a,b)$

This will give us $(c,d)=(1,1)$ and its equivalence class is of only integers $(x,y)$ such that $x=y$ that is multiplicative identity is $\{(n,n):n\in \mathbb{Z}-\{0\left\}\right\}$ .Therefore,

Multiplicative inverse of $(2,5)$ will be $(2,5)\otimes (a,b)=(n,n)\Rightarrow (2a,5b)=(n,n)\Rightarrow 2a=5b$

Hence, the multiplicative inverse of $(2,5)$ is

  ${(2,5)}^{-1}=\{(5n,2n):n\in \mathbb{Z}-\{0\left\}\right\}$

(b)

To determine

To Prove: $f$ is a ring homomorphism.

Explanation of Solution

Given Information:

Given that $f:(\mathbb{Q},+,\cdot )\to (P,\oplus ,\otimes )$is given by $f(p/q)=(p,q)$ .

Prove:

It is given that $f:(\mathbb{Q},+,\cdot )\to (P,\oplus ,\otimes )$ is given by $f(p/q)=(p,q)$ is a ring homomorphism if

  $\begin{array}{c}f(\frac{a}{b}+\frac{c}{d})=f(\frac{ad+bc}{bd})\backslash \\ =(ad+bd,bd)\\ =(a,b)\oplus (c,d)\\ =f(a/b)\oplus f(c/d)\end{array}$

So,

  $\begin{array}{c}f(\frac{a}{b}.\frac{c}{d})=f(\frac{ac}{bd})=(ac,bd)\\ =(a,b)\otimes (c,d)\\ =f(a/b)\otimes f(c/d)\end{array}$

Thus, it is a ring homomorphism

Hence, proved.