📂Abstract Algebra

An Algebraic Way to Construct the Complex Field from the Real Field

Theorem¹

$$ \mathbb{R} [x ] / \left< x^2 + 1 \right> \simeq \mathbb{C} $$

Explanation

Taken merely as a fact it seems obvious, but the process of constructing the complex field from the real field is quite beautiful.

Whether we cut $\mathbb{R} [x ]$ by $\left< x^2 \right>$ or by $\left< x^2 + x \right>$, the elements would come out in the form $ax + b$ either way, yet there is a reason for cutting by $\left< x^2 + 1 \right>$ in particular. Try proving it yourself at least once and savor this beauty.

Proof

Since $\left( x^2 + 1 \right)$ is an irreducible element over $F$, $\left< x^2 + 1 \right>$ is a maximal ideal of $\mathbb{R} [ x ]$, and therefore $\mathbb{R} [x ] / \left< x^2 + 1 \right>$ is a field.

The extension field $\mathbb{R} [x ] / \left< x^2 + 1 \right>$ of $\mathbb{R}$ has as its elements cosets such as $(ax + b) + \left< x^2 + 1 \right>$. Since every such element can be expressed as $a + b \alpha$ for $a,b \in \mathbb{R}$ and some $\alpha$, $$ \mathbb{R} [x ] / \left< x^2 + 1 \right> = \mathbb{R} ( \alpha ) $$ is a simple extension field.

Concretely, if we set $\alpha := x + \left< x^2 + 1 \right>$, then $$ \begin{align*} & \alpha^2 = \left( x + \left< x^2 + 1 \right> \right)^2 \\ \implies& \alpha^2 + 1 = \left( x^2 + \left< x^2 + 1 \right> \right) + \left( 1 + \left< x^2 + 1 \right> \right) \\ \implies& \alpha^2 + 1 = \left( x^2 + 1 \right) + \left< x^2 + 1 \right> = \left< x^2 + 1 \right> = 0 + \left< x^2 + 1 \right> \end{align*} $$ Since $\alpha$ is a zero of $\left( x^2 + 1 \right)$, $\alpha$ effectively plays the same role as the imaginary unit $i$, and the following holds. $$ \mathbb{R} ( \alpha ) \simeq \mathbb{C} $$

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