Let $f$ be an irreducible cubic polynomial over $\mathbb Q$ with exactly one real root and $K$ the splitting field of $f$.

Question

Let $f$ be an irreducible cubic polynomial over $\mathbb Q$ with exactly one real root and $K$ the smallest subfield of $\mathbb C$ containing the roots of $f$.

My question is

1. $\sigma (K) \subset K$, where $\sigma $ denotes complex conjugation.

2. [$K$ : $\mathbb Q$ ] is an even number.

Now what is complex conjugation here and how can I prove these statements?

Any hint will be appreciated.

Answer 1

Let assume $[K:\mathbb{Q}]$ is an odd number then we shall prove that all roots of $f$ are real numbers. If $f$ has only one real root ($f$ is a cubic so always has at least one real root), denote $\alpha_1,\alpha_2,\alpha_3$ three roots $f$ in which $\alpha_1 \in \mathbb{R}$. The complex conjugation $\sigma: K \longrightarrow K, x + iy \mapsto x - iy=\overline{x+iy}$ satisfies that $g(\overline{z}) = \overline{g(z)}$ for all polynomials so in particular $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ and further more $\sigma(\alpha_1)=\alpha_1,\sigma(\alpha_2)=\alpha_3,\sigma(\alpha_3)=\alpha_2$ which implies that $\sigma^2 = 1$. Consequently, $2 =\mathrm{ord}(\sigma) \mid [K: \mathbb{Q}] = \left |\mathrm{Gal}(K/\mathbb{Q}) \right|$ (since $\mathbb{Q}$ has characteristic $0$) which contradicts to our assumption.