*(This is part of my series *Field theory for high-schoolers*.)*

The previous post shows how to construct $\mathbf{C}$ as a simple algebraic extension of $\mathbf{R}$. This came very naturally from all our work, and was given only as an example. In this final post, we summarise how all the ideas of the previous posts lead to that result.

It starts with the polynomial $f(x) = x^2 + 1 \in \mathbf{R}[x]$. Since we know that $x^2 \ge 0$ for any real number $x$, we see that $f(x)$ can have no root in $\mathbf{R}$.

Kronecker’s theorem states that there must exist some extension $E$ of $\mathbf{R}$ and some element $\alpha$ of $E$ such that $f(\alpha) = \alpha^2+1 = 0$. In other words, $\alpha^2 = -1$. This number had already been studied long before Kronecker’s birth, and it already had a name: $i$.

The next step is to construct the extension $\mathbf{R}(i)$ as described in post 6. For that, we need to find the degree of $i$ over $\mathbf{R}$ (post 4), and thus in turn we need to find the minimal polynomial of $i$ over $\mathbf{R}$. Recall that the minimal polynomial of $i$ over $\mathbf{R}$ is a polynomial with coefficients in $\mathbf{R}$ which is monic, is irreducible, and has $i$ as a root. We are lucky: it is precisely $f(x)$: $f(x)$ is certainly monic and has $i$ as a root, and since it has no root in $\mathbf{R}$ and is of degree $2$, it is irreducible (post 2).

Thus $f(x)$ has degree $2$, and so $i$ has degree $2$ over $\mathbf{R}$, and our field $\mathbf{R}(i)$ consists of all elements of the form $a+bi$, with $a$ and $b$ real. We call this field $\mathbf{C}$, and we are done.