# Identify the property of Galois group

After reading this question Splitting field of $x^3-5 \in \mathbb{Q}[X]$. Galois group and fields?, I tried to think about the following question: Find the splitting field $$L$$ of the polynomial $$X^3-10\in \mathbb{Q}[X]$$. And I try to figure out if the Galois extension $$L$$ over $$\mathbb{Q}$$ is abelian.

My idea is that

(1) splitting field $$L$$:

Note that all roots are $$10^{1/3}\omega^i$$ for $$i=0,1,2$$ where $$\omega^3=1$$. So the splitting field is $$L=\mathbb{Q}(10^{1/3}, \omega)$$.

(2) order of extension $$[L:\mathbb{Q}]$$:

since $$\;x^3-1=(x-1)(x^2+x+1)\;$$ , we have that the minimal polynomial of $$\;\omega\;$$ over the rationals is $$\;x^2+x+1\;$$

This polynomial remains irreducible in $$\;\Bbb Q(\sqrt[3]10)[x]\;$$ since $$\;\Bbb Q(\sqrt[3]10)\subset\Bbb R\;$$ , whereas $$\;\omega\in\Bbb C\setminus\Bbb R\;$$, and from here $$\;[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q(\sqrt[3]10)]=2\;$$ , [Question: Am I right in saying this?]

Also, $$\{1, 10^{1/3}, 10^{2/3}\}$$ is a basis for $$\Bbb{Q}(10^{1/3})$$ over $$\Bbb{Q}$$, so $$[\Bbb{Q}(10^{1/3}):\Bbb{Q}]=3$$.

So altogether:

$$[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q]=[\Bbb Q(\sqrt[3]10,\,\omega):\Bbb Q(\sqrt[3]10)][\Bbb Q(\sqrt[3]10):\Bbb Q]=2\cdot3=6$$

(3) Because this is a splitting field over $$\mathbb{Q}$$, the extension is normal and separable, hence Galois, so the Galois group has order 6. The splitting field of a degree $$n$$ polynomial is a subgroup of $$S_n$$, and $$\vert S_3\vert=6$$, so has to be it. So $$Gal(L/\mathbb{Q})\cong S_3$$ which is not abelian. [Question: Am I right in saying this?]

• The minimal polynomial of any $\alpha$ over the rationals must, by definition, be a polynomial with rational coefficients, so what you have written about $\omega$ can't be right. The minimal polynomial for $\omega$ is $x^2+x+1$. Nov 27, 2023 at 6:47
• @GerryMyerson Thank you! Do you mean that "since $\;x^3-1=(x-1)(x^2+x+1)\;$ , we have that the minimal polynomial of $\;\omega\;$ over the rationals is $\;x^2+x+1\;$ ...but this polynomial remains irreducible in $\;\Bbb Q(\sqrt[3]10)[x]\;$"? Nov 28, 2023 at 19:07
• @GerryMyerson One thing I am confused is that we consider polynomial $x^3-10$. But why for minimal polynomial for $\omega$ considering $x^3-1$ but not $x^3-10$? Nov 28, 2023 at 19:13
• When you write, "the splitting field ... is a subgroup of $S_n$", what you mean is the Galois group of the splitting field ... is a subgroup of $S_n$. So the Galois group is a subgroup of $S_3$, and it's a group of order six, but $S_3$ is of order six, so the Galois group is all of $S_3$. Now, $S_3$ is not abelian, which says the Galois group is not abelian. If you want to see that the Galois group is not abelian without facts about $S_3$, you just have to find two automorphisms that don't commute. There's one automorphism that fixes $\root3\of{10}$ while taking $\omega$ to $1/\omega$ (cont) Nov 29, 2023 at 5:52
• (continued) and there's one that fixes $\omega$ while taking $\root3\of{10}$ to $\omega\root3\of{10}$, and you can show these two automorphisms don't commute. Nov 29, 2023 at 5:53

What you did is fine, except the miscalculation of the minimal polynomial of $$\omega$$, as pointed out in the comment by Gerry Myerson.
If $$p$$ is prime and $$f$$ is an irreducible polynomial of degree $$p$$ over $$\mathbb Q$$ which has precisely two nonreal roots in $$\mathbb C$$, then the Galois group of $$f$$ is isomorphic to $$S_p$$.
The proof is very simple: Let $$\alpha$$ be (any) root of $$f$$, then $$[\mathbb Q[\alpha]:\mathbb Q]=p$$, and by Galois correspondence, the Galois group $$G$$ contains a subgroup of index $$p$$, hence $$p\mid |G|$$, and by Cauchy's theorem, there exists an element of order $$p$$ in $$G$$. As $$G$$ can be regarded as a subgroup of $$S_p$$, and in $$S_p$$, an element of order $$p$$ must be a $$p$$-cycle. The complex conjugate will keep all the real roots fixed and interchange the two imaginary roots, therefore is a transposition in $$S_p$$. Now it's a group theory exercise to show that a $$p$$-cycle and a transposition generate $$S_p$$ (In the special case of $$p=3$$, this is easy by both $$2$$ and $$3$$ divide $$|G|$$, as you have done).
In particular, for any polynomial $$x^3-n$$ where $$n\in\mathbb Z$$ is not a perfect cubic number, the Galois group is isomorphic to $$S_3$$.
• Can I ask how to explain that $Gal(L/\mathbb{Q})$ is not abelian? Nov 29, 2023 at 5:17
• Is $L$ the splitting field of $x^3-10$? Then the Galois group is isomorphic to $S_3$ as you have shown, and $S_3$ is not abelian. For example, $(12)(123)=(23)$ but $(123)(12)=(13)$. Nov 29, 2023 at 9:13