(related to Proposition: Complex Numbers are a Field Extension of Real Numbers)
Let \((\mathbb R,\oplus,\odot)\) be the field of real numbers and let \((\mathbb C, + ,\cdot)\) be the field of complex numbers. We want to show that the function. \[f:=\cases{\mathbb R\to\mathbb C,\\ x\to (x,0),}\] is an bijective field homomorphism. Obviously, if the real numbers \(a,b\) are unequal \(a\neq 0\), then \[f(a)=(a,0)\neq (b,0)=f(b).\]
Thus, \(f\) is injective. Furthermore, and trivially, for all \(f(a)=(a,0)\in\mathbb C\), there exists \(a\in\mathbb R\). Thus \(f\) is surjective. Because \(f\) is injective as well as surjective, \(f\) is also bijective.
Moreover, for all real numbers \(a,b\) it follows from the existence of real zero and the definition of complex addition:
\[\begin{array}{rcll} f(a\oplus b)&=&f(a\oplus b,0)&\text{by definition of }f\\ &=&(a\oplus b,0\oplus 0)&\text{because real zero is neutral with respect to the addition or real numbers}\\ &=&(a,0)+(b,0)&\text{by definition of complex addition}\\ &=&f(a)+f(b)&\text{by definition of }f \end{array}\]
We also have for all real numbers \(a,b\), by virtue of the multiplication of real numbers by real zero and the definition of complex multiplication. \[\begin{array}{rcll} f(a\odot b)&=&f(a\odot b,0)&\text{by definition of }f\\ &=&(a\odot b\oplus 0,0\oplus 0)&\text{because real zero is neutral with respect to the addition or real numbers}\\ &=&(a\odot b\oplus 0\odot 0,a\odot 0\oplus 0\odot b)&\text{multiplication of real numbers by real zero}\\ &=&(a,0)\cdot(b,0)&\text{by definition of complex multiplication}\\ &=&f(a)\cdot f(b)&\text{by definition of }f \end{array}\]
Thus, \(f\) is field homomorphism.