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:
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:
\(\ds \map \nu f\) | \(=\) | \(\ds \map \nu q \, \map \nu p + \map \nu r\) | as $\nu$ is a ring honomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \nu r\) | as $p \in \map \ker \nu$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \nu a + \map \nu b \, \map \nu X\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a + b \, \map \nu X\) | as we have identified $\map \nu \R$ with $\R$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a + b i\) | 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:
\(\ds 0\) | \(=\) | \(\ds \map \nu {X^2 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \nu X}^2 + \map \nu 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds i^2 + 1\) |
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