## Cubic irreducibles over QQ are irreducible over extensions of degree 4

Let $p(x)$ be an irreducible cubic polynomial over $\mathbb{Q}$. Suppose $K$ is an extension of $\mathbb{Q}$ of degree 4; prove that $p(x)$ is irreducible over $K$.

We may assume without loss of generality that $p(x)$ is monic.

Now suppose to the contrary that $p(x)$ is reducible over $K$. Since $p(x)$ is cubic, it must have a linear factor, and thus must have a root $\alpha$ in $K$. Since $\alpha$ is a root of the monic irreducible $p(x)$ over $\mathbb{Q}$, $p(x)$ is in fact the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and thus $\mathsf{dim}_\mathbb{Q} \mathbb{Q}(\alpha) = 3$. We thus have the following chain of fields: $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq K$. By Theorem 5.7 in TAN, we have that 3 divides 4, a contradiction.

So $p(x)$ is irreducible over $K$.