How do we explain the existence of complex conjugation?

Question

We can define the complex numbers by writing $\mathbb{C} = \mathbb{R}[i]/(i^2+1),$ where $\mathbb{R}$ is to be regarded as a commutative ring. Furthermore, since $\mathbb{R}$ happens to be a field, and since $i^2+1$ is irreducible in the PID $\mathbb{R}[i],$ thus we may deduce that $\mathbb{C}$ itself happens to be a field.

Okay so that's pretty cool, but our story isn't really complete yet, since we haven't really explained where complex conjugation comes from. I'm guessing this has something to do with Galois theory (which I know basically nothing about). Hence, question:

How do we explain the existence of complex conjugation?

Answer 1

A less technical explanation: $\mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $2$, with basis $1$ and $i$. A nontrivial field automorphism of $\mathbb{C}$ that fixes $\mathbb{R}$ must fix $1$, so it is only free to move $i$. It must send $i$ to another root of $i$'s minimal polynomial $x^2 +1$, that is: $-i$. As defined on the basis, this map extends to all of $\mathbb{C}$ as complex conjugation.

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A more technical explanation (unnecessary in this simpler context but phrased in the language of Galois theory).

The key observation:

• $\mathbb{C}$ is a degree $2$ extension of $\mathbb{R}$ (this means that $\mathbb{C}$ is an $\mathbb{R}$-vector space of dimension $2$).

Some technical points:

• Since $\mathbb{R}$ has characteristic zero, this is a separable extension.
• Since it is of degree $2$, it is normal.
• Thus $\mathbb{C}/\mathbb{R}$ is a Galois extension, and hence it makes sense to consider the Galois group.

The Galois group $G = \operatorname{Gal}(\mathbb{C}/\mathbb{R})$ encodes the symmetries (automorphisms) of $\mathbb{C}$ that fix $\mathbb{R}$. Since our extension is Galois, the order of this group is the degree of the extension, so $G$ is a group of order $2$. The nontrivial element is complex conjugation.