Complex Numbers as Quotient Ring of Real Polynomial

Theorem
Let $\C$ be the complex number set.

Let $P \left[{x}\right]$ be the set of polynomial over real numbers, where the coefficient of the polynomial is real.

Let $\left\langle{x^2 + 1}\right\rangle = \left\{ {Q \left({x}\right) \left({x^2 + 1}\right): Q \left({x}\right) \in P \left[{x}\right]}\right\}$ be the ideal generated by $x^2 + 1$ in $P \left[{x}\right]$.

Let $D = P \left[{x}\right] / \left\langle{x^2 + 1}\right\rangle$ be the quotient of $P \left[{x}\right]$ modulo $\left\langle{x^2 + 1}\right\rangle$.

Then:
 * $\left({\C, +, \times}\right) \cong \left({D, +, \times}\right)$

Proof
By Division Algorithm of Polynomial, any set in $D$ has an element in the form $a+bx$.

Define $\phi: D \to \C$ as a mapping:
 * $\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = a + b i$

$\forall z = a+bi \in \C : \exists \left[\!\left[{a + b x}\right]\!\right] \in D$ 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 it is an injection and thus a bijection.

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