Question

Let$C$ be the field of Exercise 45 of section 3.1. Show that$C$ is isomorphic to the field$\mathrm{ℂ}$of complex numbers.

Short answer

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

Step-by-step solution

Consider the given function

Let C denote the field$\mathrm{ℝ}\times \mathrm{ℝ}$ with ordinary addition and multiplication. From exercise 45 , $C$is a field.

Define the function$f:C\to \mathrm{ℂ}$ , where$f\left(\left(a,b\right)\right)=a+bi,\forall \left(a,b\right)\in C$ . It is observed that$f$ is a bijection.

Show that the function is an isomorphism  

Let us assume two arbitrary elements of$C$as $\left(a,b\right)$and$\left(c,d\right)$. Then, we get:

$\begin{array}{rcl}f\left(\left(a,b\right)+\left(c,d\right)\right)& =& f\left(\left(a+c,b+d\right)\right)\\ & =& \left(a+c\right)+\left(b+d\right)i\\ & =& a+c+bi+di\\ & =& \left(a+bi\right)+\left(c+di\right)\\ & =& f\left(\left(a,b\right)\right)+f\left(\left(c,d\right)\right)\end{array}$

localid="1660572143132" $\begin{array}{rcl}f\left(\left(a,b\right)\left(c,d\right)\right)& =& f\left(\left(ac+bd,ad+bc\right)\right)\\ & =& \left(ac-bd\right)+\left(ad+bc\right)i\\ & =& ac+adi+bci-bd\\ & =& \left(a+bi\right)\left(c+di\right)\\ & =& f\left(\left(a,b\right)\right)f\left(\left(c,d\right)\right)\end{array}$

This implies that $f$is an isomorphism. Hence, it is proved that the map $f:C\to \mathrm{ℂ}$ is an isomorphism.