Theorem: Complex Numbers as Quotient Ring of Real Polynomial

Let $\C$ be the set of complex numbers.
Let $P \sqbrk x$ be the set of polynomials over real numbers, where the coefficients of the polynomials are real.
Let $\ideal {x^2 + 1} = \set {\map Q x \paren {x^2 + 1}: \map Q x \in P \sqbrk x}$ be the ideal generated by $x^2 + 1$ in $P \sqbrk x$.
Let $D = P \sqbrk x / \ideal {x^2 + 1}$ be the quotient of $P \sqbrk x$ modulo $\ideal {x^2 + 1}$.
Then:
:$\struct {\C, +, \times} \cong \struct {D, +, \times}$

Proof:
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:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = a + b i$
We have that:
:$\forall z = a + b i \in \C : \exists \eqclass {a + b x} {x^2 + 1} \in D$
such that:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = a + b i = z$
So $\phi$ is a surjection.
To prove that it is a injection, we let:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = \map \phi {\eqclass {c + d x} {x^2 + 1} }$
So:
{{begin-eqn}}
{{eqn | l = \map \phi {\eqclass {a + b x} {x^2 + 1} }
      | r = \map \phi {\eqclass {c + d x} {x^2 + 1} }
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = a + b i
      | r = c + d i
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadstoandfrom
      | l = a = c
      | o = \land
      | r = b = d
      | c = Equality of Complex Numbers
}}
{{eqn | ll= \leadstoandfrom
      | l = a + b x
      | r = c + d x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \eqclass {a + b x} {x^2 + 1}
      | r = \eqclass {c + d x} {x^2 + 1}
      | 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 = \map \phi {\eqclass {a + b x} {x^2 + 1} + \eqclass {c + d x} {x^2 + 1} }
      | r = \map \phi {\eqclass {\paren {a + c} + \paren {b + d} x} {x^2 + 1} }
      | c = 
}}
{{eqn | r = \paren {a + c} + \paren {b + d} i
      | c = Definition of $\phi$
}}
{{eqn | r = \paren {a + b i} + \paren {c + d i}
      | c = {{Defof|Complex Addition}}
}}
{{eqn | r = \map \phi {\eqclass {a + b x} {x^2 + 1} } + \map \phi {\eqclass {c + d x} {x^2 + 1} }
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {\eqclass {a + b x} {x^2 + 1} \times \eqclass {c + d x} {x^2 + 1} }
      | r = \map \phi {\eqclass {\paren {a + b x} \times \paren {c + d x} } {x^2 + 1} }
      | c = 
}}
{{eqn | r = \map \phi {\eqclass {a \times c + \paren {a \times d + b \times c} x + b \times d \, x^2} {x^2 + 1} }
      | c = 
}}
{{eqn | r = \map \phi {\eqclass {a \times c + \paren {a \times d + b \times c} x + b \times d \, x^2 - b \times d \paren {x^2 + 1} } {x^2 + 1} }
      | c = Definition of $D$ as a quotient ring modulo $\ideal {x^2 + 1}$
}}
{{eqn | r = \map \phi {\eqclass {\paren {a \times c - b \times d} + \paren {a \times d + b \times c} x} {x^2 + 1} }
      | c = 
}}
{{eqn | r = \paren {a \times c - b \times d} + \paren {a \times d + b \times c} i
      | c = Definition of $\phi$
}}
{{eqn | r = \paren {a + b i} \times \paren {c + d i}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = \map \phi {\eqclass {a + b x} {x^2 + 1} } \times \map \phi {\eqclass {c + d x} {x^2 + 1} }
      | 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
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