# Can the ''Galois group" for an inseparable irreducible polynomial of degree 4 be $S_{3}$?

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 the polynomial $f$ is inseparable, then $E$ will not be a Galois extension of $F$. Hence we also do not have a group that we call Galois group then. Do you just want to understand $\text{Aut}(E/F)$ instead?
– Con
Commented May 3, 2020 at 11:19
• @TMO As I said in my post, the book defines the Galois group in a different way, and I want to understand Aut$(E/F)$ (which is called the galois group in my textbook). Commented May 3, 2020 at 11:20
• No, that is the normal definition. But you only call it Galois group if your extension is Galois. Otherwise it is just the automorphism group of the extension.
– Con
Commented May 3, 2020 at 11:21
• $F$ is an arbitary field ? I ask because if $F$ has characteristic $0$ or is finite, $f$ must be seperable. Commented May 3, 2020 at 11:22
• @TMO: Many books call it the “Galois group” in all cases. Hungerford is one of the main books that calls it a Galois group even when the extension is not a Galois extension. And by the same token, many authors denote the automorphism groups, even when they don’t call it “the Galois group” by $G(F/K)$ or $\mathrm{Gal}(F/K)$. I seem to recall Lang does something like that... Commented May 3, 2020 at 19:20

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 $$6=\vert S_3\vert$$.