Casus irreducibilis

Remarks

The three roots are distinct, since an irreducible polynomial over a field of characteristic $0$ has no repeated roots. For the idea of the proof, adjoin first the square root of the discriminant, which is real. Then over every real field $F$ that contains this square root and over which $f$ is irreducible, adjoining one root of $f$ adjoins all three roots. Take a real tower of steps adjoining $q$-th roots, $q$ prime, whose last step $E \subseteq E(\alpha)$ with $\alpha^q \in E$ first makes $f$ reducible. Then $3$ divides $[E(\alpha) : E]$, which is $1$ or $q$, so $q = 3$ and $E(\alpha)/E$ is normal of degree $3$. Hence $x^3 - \alpha^3$ splits in $E(\alpha)$, so this real field contains a nonreal cube root of unity, which is impossible. Example: $f = x^3 - 3x + 1$ over $\Q$ is irreducible, since $\pm 1$ are not roots, and its roots $2\cos(2\pi/9)$, $2\cos(4\pi/9)$, $2\cos(8\pi/9)$ are real. So every expression of these roots by radicals passes through nonreal numbers, as in the cubic formula. Irreducibility is necessary: $x^3 - x = x(x - 1)(x + 1)$ has three real roots, all in $\Q$.

Going deeper

For a student

For a real cubic with three distinct real roots, the discriminant is positive, so the square root in the cubic formula is of a negative number. The theorem shows that the problem is not the formula but the numbers themselves. The roots can still be written with real numbers using trigonometry, for example as $2\sqrt{-p/3}\cos\theta$ for suitable angles $\theta$ when the cubic is $t^3 + pt + q$.