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 $\left\langle x^2+1 \right\rangle = \left\{ { Q\left({x}\right) \left({x^2+1}\right) \vert Q\left({x}\right) \in P\left[{x}\right]}\right\}$ 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:
 * $\phi\left({\left[\!\left[a+bx\right]\!\right]}\right) = a+bi$

Obviously it is a surjection.

To prove that it is a injection, we let $\phi\left(\left[\!\left[a+bx\right]\!\right]\right)=\phi\left(\left[\!\left[c+dx\right]\!\right]\right)$

So it is an injection and thus a bijection.

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