Question

Let $\overline{Z}$denote the ring of integers with the $\oplus$and $\odot$operations defined in Exercise 24 of section 3.1. Prove that $\overline{\mathrm{\mathbb{Z}}}$is isomorphic to $\mathrm{\mathbb{Z}}$

Short answer

It is proved that the map$f:E\to \mathrm{\mathbb{Z}}$ is an isomorphism.

Step-by-step solution

Consider the given function

We note that by the properties of an isomorphism, if$\varnothing :\mathrm{\mathbb{Z}}\to \overline{\mathrm{\mathbb{Z}}}$ is an isomorphism, thenrole="math" localid="1648463060458" $\varnothing \left(0\right)=0_{\overline{\mathrm{\mathbb{Z}}}}=1$ and$\varnothing \left(1\right)=I_{\overline{\mathrm{\mathbb{Z}}}}=0$ . Furthermore, using the information from Exercise 24 in Section 3.1,

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

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

Therefore, a reasonable function to try would be$\varnothing \left(n\right)=1-n$ .

Show that the function is isomorphism  

We prove this is an isomorphism. Let$a,b\in \mathrm{\mathbb{Z}}$. Then$\varnothing \left(a+b\right)=1-a-b$ and,
$\begin{array}{rcl}\varnothing \left(a\right)\oplus \varnothing \left(b\right)& =& \left(1-a\right)\oplus \left(1-b\right)\\ & =& \left(1-a\right)+\left(1-b\right)-1\\ & =& 1-a-b\\ & =& \varnothing \left(a+b\right)\end{array}$

Also, $\varnothing \left(ab\right)=1-ab$while

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

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

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

Hence,$\varnothing$ is bijective, and so, we get $\varnothing$is a ringisomorphism.