Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers

Theorem
Let $\struct {\R, +, \times}$ denote the field of real numbers.

Let $X$ be transcendental over $\R$.

Let $\R \sqbrk X$ be the ring of polynomials in $X$ over $F$.

Consider the field of quotients:
 * $\R \sqbrk X / \ideal p$

where:
 * $p = X^2 + 1$
 * $\ideal p$ denotes the ideal generated by $p$.

Then $\R \sqbrk X / \ideal p$ is the field of complex numbers.

Proof
It is taken as read that $X^2 + 1$ is irreducible in $\R \sqbrk X$.

Hence by Polynomial Forms over Field form Principal Ideal Domain: Corollary 1, $\R \sqbrk X / \ideal p$ is indeed a field.

Let $\nu$ be the quotient epimorphism from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$.

From Quotient Ring Epimorphism is Epimorphism:
 * $\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$

So $\map \nu \R$ is a monomorphism from $\R \sqbrk X$ to $\R \sqbrk X / \ideal p$.

Thus $\map \nu \R$ is an isomorphic copy of $\R$ inside $\R \sqbrk X / \ideal p$.

We identify this isomorphic copy of $\R$ with $\R$ itself, ignoring the difference between $\map \nu x$ and $x$ when $x = \R$.

Hence:


 * $(1): \quad \R \subseteq \R \sqbrk X / \ideal p$

Let $f \in \R \sqbrk X$ be arbitrary.

By Division Theorem for Polynomial Forms over Field:
 * $\exists q, r \in \R \sqbrk X: f = q p + r$

where $r =a + b X$ for some $a, b \in \R$.

Hence:

As $\nu$ is an epimorphism, it is a fortiori surjection.

Hence:
 * $(2): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed in the form:
 * $w = a + b i$
 * for some $a, b \in \R$.

Because $X^2 + 1 \in \ker \nu$, we have:

Hence we have that:


 * $(3): \quad$ In $\R \sqbrk X / \ideal p$, $i^2 = -1$

Thus $(2)$ can be improved to:


 * $(4): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed uniquely in the form:
 * $w = a + b i$
 * for some $a, b \in \R$.

From $(1)$, $(3)$ and $(4)$, the field $\R \sqbrk X / \ideal p$ is recognised as the field of complex numbers.