# Quotient Field 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 quotient field:

$\R \sqbrk X / \ideal p$

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 $\R \sqbrk X / \ideal p$.

$\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$

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.

$\exists q, r \in \R \sqbrk X: f = q p + r$

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

Hence:

 $\displaystyle \map \nu f$ $=$ $\displaystyle \map \nu q \, \map \nu p + \map \nu r$ as $\nu$ is a ring honomorphism $\displaystyle$ $=$ $\displaystyle \map \nu r$ as $p \in \map \ker \nu$ $\displaystyle$ $=$ $\displaystyle \map \nu a + \map \nu b \, \map \nu X$ $\displaystyle$ $=$ $\displaystyle a + b \, \map \nu X$ as we have identified $\map \nu \R$ with $\R$ $\displaystyle$ $=$ $\displaystyle a + b i$ where $i := \map \nu X$

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:

 $\displaystyle 0$ $=$ $\displaystyle \map \nu {X^2 + 1}$ $\displaystyle$ $=$ $\displaystyle \paren {\map \nu X}^2 + \map \nu 1$ $\displaystyle$ $=$ $\displaystyle i^2 + 1$

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.

$\blacksquare$