# 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:

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:

\(\displaystyle \map \nu f\) | \(=\) | \(\displaystyle \map \nu q \, \map \nu p + \map \nu r\) | 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.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 65$. Some properties of $F \sqbrk X$, where $F$ is a field