By Division Algorithm of Polynomial, any set in $D$ has an element in the form $a + b x$. Define $\phi: D \to \C$ as a mapping: :$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = a + b i$ We have that: :$\forall z = a + b i \in \C : \exists \left[\!\left[{a + b x}\right]\!\right] \in D$ such that: :$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = a + b i = z$ So $\phi$ is a surjection. To prove that it is a injection, we let: :$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)$ So: {{begin-eqn}} {{eqn | l = \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right)       | r = \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)       | c =  }} {{eqn | ll= \iff       | l = a + b i       | r = c + d i       | c = Definition of $\phi$ }} {{eqn | ll= \iff       | l = a = c       | o = \land       | r = b = d       | c = Equating both real part and imaginary part }} {{eqn | ll= \iff       | l = a + b x       | r = c + d x       | c =  }} {{eqn | ll= \iff       | l = \left[\!\left[{a + b x}\right]\!\right]       | r = \left[\!\left[{c + d x}\right]\!\right]       | c =  }} {{end-eqn}} So $\phi$ is an injection and thus a bijection. It remains to show that $\phi$ is a homomorphism for the operation $+$ and $\times$. {{begin-eqn}} {{eqn | l = \phi \left({\left[\!\left[{a + b x}\right]\!\right] + \left[\!\left[{c + d x}\right]\!\right]}\right)       | r = \phi \left({\left[\!\left[{\left({a + c}\right) + \left({b + d}\right) x}\right]\!\right]}\right)       | c =  }} {{eqn | r = \left({a + c}\right) + \left({b + d}\right) i       | c = Definition of $\phi$ }} {{eqn | r = \left({a + b i}\right) + \left({c + d i}\right)       | c = {{Defof|Complex Addition}} }} {{eqn | r = \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) + \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \phi \left({\left[\!\left[{a + b x}\right]\!\right] \times \left[\!\left[{c + d x}\right]\!\right]}\right)       | r = \phi \left({\left[\!\left[{\left({a + b x}\right) \times \left({c + d x}\right)}\right]\!\right]}\right)       | c =  }} {{eqn | r = \phi \left({\left[\!\left[{a \times c + \left({a \times d + b \times c}\right) x + b \times d \, x^2}\right]\!\right]}\right)       | c =  }} {{eqn | r = \phi \left({\left[\!\left[{a \times c + \left({a \times d + b \times c}\right) x + b \times d \, x^2 - b \times d \left({x^2 + 1}\right)}\right]\!\right]}\right)       | c = Definition of $D$ as a quotient ring modulo $\left\langle{x^2 + 1}\right\rangle$ }} {{eqn | r = \phi \left({\left[\!\left[{\left({a \times c - b \times d}\right) + \left({a \times d + b \times c}\right) x}\right]\!\right]}\right)       | c =  }} {{eqn | r = \left({a \times c - b \times d}\right) + \left({a \times d + b \times c}\right) i       | c = Definition of $\phi$ }} {{eqn | r = \left({a + b i}\right) \times \left({c + d i}\right)       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) \times \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)       | c = Definition of $\phi$ }} {{end-eqn}} Thus $\phi$ has been demonstrated to be a bijective ring homomorphism and thus by definition a ring isomorphism. {{qed}} Category:Complex Numbers Category:Quotient Rings ly358o0xbn7fa4eepsu7ghe353n3763