Finding the Galois group of over $\mathbb{Q}$ using the Galois groups over finite fields
问题内容
One question in a past qualifying exam asked us to find the Galois group of $x^6+3$ over $\mathbb{F}_5$, $\mathbb{F}_7$, and $\mathbb{Q}$.
Each part of this question has been answered individually on this website: over $\mathbb{F}_5$, $\mathbb{F}_7$, and $\mathbb{Q}$. I asked this question again because I think we need to take advantage of the Galois groups over $\mathbb{F}_5$ and $\mathbb{F}_7$ to find the Galois group over $\mathbb{Q}$ in our exam. A theorem in Lang's algebra states that the Galois groups over $\mathbb{F}_5$ and $\mathbb{F}_7$ are embedded into the Galois group over $\mathbb{Q}$.
$x^6+3=(x^2+2)(x^2-2x+2)(x^2+2x+2)$ over $\mathbb{F}_5$, so its Galois group as a subgroup of $S_6$ is generated by a product of three disjoint 2-cyles. $x^6+3=(x^3-2)(x^3+2)$ over $\mathbb{F}_7$, so its Galois group as a subgroup of $S_6$ is generated by a product of two disjoint 3-cyles. The Galois group $G$ over $\mathbb{Q}$ has degree six, because $x^6+3$ is irreducible and the sixth-root of unity exists in $\mathbb{Q}(\alpha); \alpha=\sqrt[6]{-3}$.
Now, I want to identify whether the Galois group $G$ is cyclic or isomorphic to $S_3$. We know $G$ contains a product of three disjoint 2-cycles and a product of two disjoint 3-cycles. However, I am not able to identify these cycles. I choose an ordering for the roots of $x^6+3$ in a way that the Galois group over $\mathbb{F}_5$ corresponding to $(x^2+2)(x^2-2x+2)(x^2+2x+2)$ is generated by $(12)(34)(56)$. Now based on this choice of ordering, I only need to specify the Galois group over $\mathbb{F}_7$ corresponding to $(x^3-2)(x^3+2)$ is generated by which product of two disjoint 3-cycles $(abc)(def)$. If I specify $(abc)(def)$, then I am able to say $G$ is cyclic or not.
Do you know, how I can identify $(abc)(def)$ when I choose $(12)(34)(56)$ to generate the Galois group corresponding to $(x^2+2)(x^2-2x+2)(x^2+2x+2)$?
回答 (1)
Apologies for not directly answering the question with the method you are attempting to use: I am doing so because, while I could be wrong, speaking from experience I highly doubt that invoking this theorem of Lang is what the author of this qualifying exam had in mind.
You've already noted that the group is isomorphic to $\mathbb Z/6\mathbb Z$ or $S_6$. But since your field has $\alpha=\sqrt[6]{-3}$ as you've noted, it also has $\alpha^2$, which is $\sqrt[3]3$ times a sixth root of unity, and hence it has $\sqrt[3]3$. So $\mathbb Q(\sqrt[3]3)$ is a subfield of your field which is not Galois over $\mathbb Q$, which proves that the Galois group cannot be abelian. So the Galois group is $S_6$.