We know that the Galois group of an irreducible cubic polynomial is $S_3$ or $A_3$, but is every group extension whose Galois group is $S_3$ a splitting field of a cubic polynomial? If not, the extension must be the splitting field of a polynomial of degree 6. And therefore $S_3$ must be a transitive subgroup of $S_6$. Unfortunately, I found (1 2 3)(4 5 6) and (1 4)(3 5)(2 6) can generate such transitive $S_3$, but I can't find the corresponding polynomial.
Is it true that every Galois extension with Galois group $S_3$ a splitting field of a cubic irreducible polynomial? Thanks for your help!
Best Answer
If $K/\mathbb{Q}$ has Galois group $S_3$, then because $S_3$ has a subgroup of index 3, this corresponds to a subextension $F$ of degree 3. A primitive element for $F$ will be the root of a cubic polynomial and its splitting field will be the original $K$ (it can't be $F$ itself because $S_3$ has no normal subgroups of index 3).
Of course, you can also find a degree 6 polynomial for which $K$ is the splitting field (see Arturo Magidin's comment).