# Knowing the Galois group of the splitting field of a polynomial $f$, how can I show that $f$ is irreducible in the ground field?

So I'm given $f(x) = \sum_{k=0}^{8}\frac{x^k}{k!} \in \mathbb{Q}[x]$. Denote its splitting field by $E$, then I'm also given that ${\rm Gal}(E/\mathbb{Q}) \cong A_8$. The task is to prove that $f(x)$ is irreducible over $\mathbb{Q}$.

So $A_8$ is known to be simple, so by the fundamental theorem there cannot exist intermediate fields $\mathbb{Q} \subset K \subset E$ such that $\mathbb{Q} \subset K$ is normal. So if $f$ is reducible over $\mathbb{Q}$, but not completely, then the splitting field of the remaining irreducible polynomial will also be $E$. Any irreducible polynomial in $\mathbb{Q}[x]$ which splits in $E$ will not split in any subfield of $E$ containing $\mathbb{Q}$.

Assuming that my reasoning so far is correct, I cannot see how to proceed from here. I don't quite see for instance how it can't be the case that one root of $f$ lies in $\mathbb{Q}$ and the rest in $E$. Any help is appreciated.

• Note that if one root was in $\Bbb Q$ and the rest were in $E$, then the action of a Galois automorphism would be determined by its action on the seven roots not in $\Bbb Q$. Thus, the Galois group would be isomorphic to a subgroup of $S_7$. – Rolf Hoyer Jun 8 '16 at 16:39
• @RolfHoyer Oh right! It's only true that the Galois group is isomorphic to a subgroup of $S_r$ if $f$ has $r$ distinct roots in the splitting field, right? Of course, this is the case here since $\mathbb{Q}$ has characteristic 0. – Auclair Jun 8 '16 at 16:42

You can appeal to the fact that $f$ is irreducible iff the Galois group acts transitively on the set of roots. Here, we expect that $\text{Gal}(E/\mathbb{Q}) \cong A_8$ under the action of the Galois group on the roots, but let's ensure that is true:
Consider the action of the Galois group on the $n$ roots of $f$ in $\mathbb{C}$. This induces a homomorphism $f:A_8 \to S_n$. Since $A_8$ is simple, this must be injective, and so $n \geq 8$. Since $n\leq 8$ anyway, $n=8$. So $[S_8:f(A_8)] = 2$, so $f(A_8) \vartriangleleft S_8$, whence $f(A_8) = A_8$. In particular, the Galois group acts transitively on the set of roots of $f$.
So, $f$ is irreducible.