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

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

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