Proving the number of real roots of a cubic polynomial corresponds to the discriminant using Galois Theory

Question

I want to prove that a cubic polynomial $f$ with real coefficients has 3 real roots iff $\Delta(f) > 0$ and $f$ has 1 real root iff $\Delta(f) <0$.

I was able to do this by just multiplying out the discriminant, but my professor says there is also a way to do this using $\operatorname{Gal}_\mathbb{R}(f)$. How would that work?

Answer 1

By hypothesis, $F\in K[X]$ is a cubic, with $K\subset \mathbf R$. The Galois group $G$ of the splitting field of $f$ over $K$ is a subgroup of $S_3$. A a priori $f$ has $3$ real roots,or $1$ real and $2$ complex conjugate roots.

• if $f$ is irreducible over $K$, the order of $G$ is a multiple of $3$, hence $G=S_3$ or $A_3$ (which is $C_3$), and $G=C_3$ iff $G$ is generated by a circular permutation of the $3$ roots. This means, writing $\Delta=D^2$ (ovious notation), that $G$ fixes $D$, and the discriminant is a square in $K$, or equivalently $\Delta>0$
• if $f$ is reducible over $K$, either $f=(X-a)g$, where $a\in K$ and $g$ is irreducible, hence has no root in $K$; or $f$ splits in $K$, and again $G$ fixes $D$ and $\Delta$ is a square ./.