Question

Let ${\mathbf{\mathbb{Z}}}^{\mathbf{*}}$denote the ring of integers with the$\oplus$ and $⊝$operations defined in Exercise 22 of section 3.1. Prove that $\mathbf{\mathbb{Z}}$is isomorphic to${\mathbf{\mathbb{Z}}}^{*}$ .

Short answer

It is proved that $\mathrm{\mathbb{Z}}$is isomorphic to${\mathrm{\mathbb{Z}}}^{*}$ .

Step-by-step solution

Use Properties

According to properties of isomorphism, if$\varnothing :\mathrm{\mathbb{Z}}\to {\mathrm{\mathbb{Z}}}^{*}$is an isomorphism, then$\varnothing \left(0\right)=0_{Z^{*}}=1$and role="math" localid="1647911794117" $\varnothing \left(1\right)=I_{Z^{*}}=0$.

From exercise 22 of section 3.1, we have:

$$\begin{gathered} \varnothing \left(-1\right)=-\varnothing \left(1\right) \\ =2-0 \\ =2 \end{gathered}$$

$$\begin{gathered} \varnothing \left(2\right)=\varnothing \left(1+1\right) \\ =\varnothing \left(1\right)+\varnothing \left(1\right) \\ =0\oplus 0 \\ =-1 \end{gathered}$$

Proof 

Consider the function$\varnothing \left(n\right)=1-n$. Prove that the function is an isomorphism.

Consider $a,b\in \mathrm{\mathbb{Z}}$. This implies that:

$$\begin{gathered} \varnothing \left(a+b\right)=1-a-b \\ \varnothing \left(a\right)+\varnothing \left(b\right)=(1-a)\oplus \left(1-b\right) \\ =(1-a)+\left(1-b\right)-1 \\ =1-a-b \\ =\varnothing \left(a+b\right) \end{gathered}$$

Further:

$$\begin{gathered} \varnothing \left(ab\right)=1-ab \\ \varnothing \left(a\right)\otimes \varnothing \left(b\right)=\left(1-a\right)\otimes \left(1-b\right) \\ =\left(1-a\right)+\left(1-b\right)-\left(1-a\right)\left(1-b\right) \\ =2-a-b-\left(1-a-b+ab\right) \\ =1-ab \\ =\varnothing \left(ab\right) \end{gathered}$$

This implies that$\varnothing$ is ring homomorphism. Furthermore, we note that $\varnothing$is its own inverse function since,

$$\begin{gathered} \varnothing \circ \varnothing \left(n\right)=\varnothing \left(1-n\right) \\ =1-\left(1-n\right) \\ =n \end{gathered}$$

Hence,$\varnothing$ is bijective, and so, we get $\varnothing$is an isomorphism.

It is proved that$\mathrm{\mathbb{Z}}$ is isomorphic to${\mathrm{\mathbb{Z}}}^{*}$ .