Complex Numbers as Quotient Ring of Real Polynomial

Theorem
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:

So $\phi$ is an injection and thus a bijection.

It remains to show that $\phi$ is a homomorphism for the operation $+$ and $\times$.

Thus $\phi$ has been demonstrated to be a bijective ring homomorphism and thus by definition a ring isomorphism.