Prologue~
“What I have done... will not be lost. It is a study which must lie in wait for its time, perhaps for many years. The mathematical language I have created will someday be common property.”
— From Galois' final letter before his death
Verse I: The Ground~
Even before a formal proof was established for a very fundamental result that we so easily take for granted, algebraists observed empirically that polynomials with real coefficients had non-real roots occurring in pairs. This was a much under-appreciated founding listed in the works of Cardano, Euler and Lagrange.
The rigorous and structural explanation — that conjugate roots arise because of the action of a Galois automorphism — was made possible through the revolutionary framework introduced by Évariste Galois, whose ideas provided a structural approach to the roots of polynomials. These insights were later expanded and rigorously developed by modern algebraists like Dedekind, Kronecker, Artin, and Zariski.
What we encounter here is a proof of the statement.
Verse II: The statement and the proof~
Theorema
Let $f(x) \in \mathbb{R[x]}$. If $ \alpha \in \mathbb{C}$ is a root of $f$, then $\overline{\alpha}$ is also a root of $f$.
Probationem
To prove the above statement, let us rely a little on our morphism skills and try defining an automorphism from $\mathbb{C}$ to $\mathbb{C}$ such that: $$\phi : \mathbb{C} \rightarrow \mathbb{C}$$ $$ \phi(z)= \overline{z}$$ A few more properties of this map would then be:
1) $\phi(z+ w)= \phi(z) + \phi(w)$
2) $\phi(zw)=\phi(z) \phi(w)$
3) $\phi(1/z)= \frac{1}{ \phi(z)}$, for $z \neq 0$
4) $\phi(a)= a, \ \forall \ a \in \mathbb{R}$
Hence, for every $a$ in $\mathbb{R}$, the automorphism map $\phi$ fixes $a$.
Therefore, complex conjugation is a field automorphism of the field extension $\mathbb{C} / \mathbb{R}$ that fixes every element of $\mathbb{R}$.
We now take into consideration $\mathbb{R} \ \subset \mathbb{C}$ and a polynomial $P(z) \in \mathbb{R[x]}$ such that the coefficients are real.
$\Rightarrow P(z)= a_n z^n + a_{n-1} z^{n-1}+ \cdots + a_1 z + a_0$ , where $a_i \in \mathbb{R}$.
Now, let us consider some $\alpha \in \mathbb{C}$ as a root of $P(z)$, such that $P(\alpha)=0$.
If we attempt to apply the field automorphim to the polynomial $P(z)$ at $\alpha$, we obtain:
$$ \phi(P(\alpha))= \overline{P(\alpha)}= \overline{0}= 0$$
This implies that for any $\alpha \in \mathbb{C}, \ \overline{\alpha} \in \mathbb{C}$ is also a root of the polynomial $P(z)$.
Epilogue~
Upon deeper reflection, we realise that the Galois group of the field extension $\mathbb{C}/ \mathbb{R}$ is remarkably simple — it contains only two automorphisms: the identity map, which leaves every complex number unchanged, and the complex conjugation map $\phi$, which reflects each number across the real axis. These are the only field automorphisms of $\mathbb{C}$ that fix every element of $\mathbb{R}$. $$ Gal (\mathbb{C} / \mathbb{R}) = \{ id, \phi \}$$ This Galois group, though minimal in size, captures a powerful symmetry: the necessity that every non-real root of a real polynomial must be accompanied by its mirror image — its complex conjugate. The structure of the field dictates the symmetry of the solutions.
Thus, the pairing of complex roots in real polynomials is no coincidence — it is instead a reflection of deep algebraic symmetry. A symmetry so deeply etched into the soul of mathematics that it whispers through every structure it touches.