Condition on Galois Group that makes polynomial irreducible

Question

Suppose $K$ is the splitting field of a monic polynomial $f(x) \in \mathbb{Z}[x]$. What is the condition on $G = \text{Gal}(K/\mathbb{Q})$ that ensures $f$ is irreducible?

Answer 1

If you know nothing about $f$ besides that $K$ is its splitting field, then there is no such condition. For instance, the splitting field of $f$ is the same as the splitting field of $f^2$, but $f^2$ is never irreducible. On the other hand, every finite Galois extension is the splitting field of some irreducible polynomial, since you can take the minimal polynomial of a primitive element. So you cannot tell from the Galois group (or even from $K$) whether $f$ is irreducible or not: there are both reducible and irreducible polynomials with $K$ as the splitting field.

If you additionally know the roots of $f$ and how $G$ acts on them, then you can say $f$ is irreducible iff it has distinct roots in $K$ and $G$ acts transitively on them. Indeed, if $f$ is irreducible, then it is separable (since we're in characteristic $0$) so its roots are distinct. Also its roots all have the same minimal polynomial (namely $f$), so if $\alpha$ and $\beta$ are two roots, there is an isomorphism $\mathbb{Q}(\alpha)\to\mathbb{Q}(\beta)$ which sends $\alpha$ to $\beta$, which can then be extended to an automorphism of $K$ by normality. Conversely, if $G$ acts transitively on the roots of $f$ then they all have the same minimal polynomial, so $f$ is a power of the minimal polynomial. If the roots of $f$ are distinct, it must be equal to the minimal polynomial and hence irreducible.