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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?]

H.Y Duan
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    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 $x^2+x+1$. – Gerry Myerson Nov 27 '23 at 06:47
  • @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 $;\Bbb Q(\sqrt[3]10)[x];$"? – H.Y Duan Nov 28 '23 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 $x^3-10$? – H.Y Duan Nov 28 '23 at 19:13
  • (1) The minimal polynomial for $\omega$ has to vanish at $\omega$. $x^3-1$ does; $x^3-10$ doesn't. (2) $x^2+x+1$ has no real roots. ${\bf Q}(\root3\of{10}$ has no nonreal elements. Ergo, $x^2+x+1$ remains irreducible over ${\bf Q}(\root3\of{10}$. – Gerry Myerson Nov 28 '23 at 21:47
  • @GerryMyerson Thank you! Can I ask how to explain that $Gal(L/\mathbb{Q})$ is not abelian? – H.Y Duan Nov 29 '23 at 05:17
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    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) – Gerry Myerson Nov 29 '23 at 05:52
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    (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. – Gerry Myerson Nov 29 '23 at 05:53

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What you did is fine, except the miscalculation of the minimal polynomial of $\omega$, as pointed out in the comment by Gerry Myerson.

There is a more general result along the line: [Hungerford, Chapter V, Theorem 4.12]

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 $S_3$.

Gerry Myerson
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Just a user
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  • Can I ask how to explain that $Gal(L/\mathbb{Q})$ is not abelian? – H.Y Duan Nov 29 '23 at 05:17
  • 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 $(123)(12)=(13)$. – Just a user Nov 29 '23 at 09:13