Prove (or disprove by example) the following: If $F$ is a field extension of $K$, not necessarily a Galois extension, with $[F:K]=6$ and $\mathrm{Aut}_K F \simeq S_3$, then $F$ must be the splitting field of an irreducible cubic in $K[x]$.
This is a question from a qualifying exam at my university. I think the requirement that the extension be Galois is necessary for this to be true. Like, if $F$ doesn't have to be Galois over $K$, then there must be an example where $F$ isn't a splitting field at all, right? I've yet to come up with such an example though.
Let $G=\textrm{Aut}_KF\cong S_3$. Then the fixed field of $G$, $K'$ say, contains $K$. Moreover, from a standard theorem in Galois theory, $F/K'$ is a Galois extension with degree $|G|=6$ and Galois group $G$. From the product formula for degrees $K'=K$, so $F/K$ is Galois.
Let $L$ be the fixed field of $(1\,2)\in S_3$. Then $|L:K|=3$. Let $a\in L$, $a\notin K$. Since $3$ is prime, $K(a)$ cannot be an intermediate extension of $L/K$ by the Galois correspondence, so $K(a)=L$ and $a$ is a zero of an irreducible cubic $g$ over $K$. But $\langle(1\,2)\rangle$ is not a normal subgroup of $S_3$, so $L$ cannot contain all the roots of $g$ and so $F$ must be the splitting field of $g$.