Here is a homework question of mine: Exercise 24.3.12, from Shahriari, Algebra in Action.
$f\in F[x]$ is an irreducible polynomial of degree 4, and let $E$ be the splitting field of $f$ over $F$. Can Gal$(E/F)\cong S_{3}$?
Note that, the book defines Gal$(E/F)$ as Aut$(E/F)$, the collection of automorphisms of $E$ which fix $F$.
If $f$ is separable, it is easy because the Galois group has to be a transitive subgroup of $S_{4}$, so it cannot be $S_{3}$.
My problem is when $f$ is inseparable, can we get a polynomial with its Galois group $S_{3}$?
Or more generally, what kind of group can be Aut$(E/F)$ for some $f$, when $f$ is inseparable?
If $f$ is inseparable, then $f'=0$ (since $f$ is irreducible). Hence necessarily $F$ has characteristic $2$ and $f=X^4+aX^2+b$.
If $\alpha,\beta\in F_{alg}$ are roots of $f$, then $\alpha^4-\beta^4+a(\alpha^2-\beta^2)=0= (\alpha-\beta)^2((\alpha-\beta)^2+a)=0$ (because we are incharacteristic $2$). So either $\beta=\alpha$ or $\beta=\alpha+\sqrt{a}$. It follows that the splitting field $E$ of $f$ is generated by a fixed root $\alpha$ , which has degree $4$ over $F$, and $\sqrt{a}$, which has degree $1$ or $2$ over $F$, hence also degree $1$ or $2$ over $F(\alpha)$.
Thus$[E:F]=[F(\alpha)(\sqrt{a}):F(\alpha)][F(\alpha):F]=4$ or $8$. In particular, the degree cannot be $6=\vert S_3\vert$.