# Galois group of $x^3-2$

In Abstract Algebra: 3rd Edition by Dummit and Foote, page 564, example (5), the following is stated:

The splitting field of $$x^3-2$$ over $$\mathbb{Q}$$ is Galois of degree 6. The roots of this equation are $$\sqrt[3]{2}, \rho\sqrt[3]{2}, \rho^2\sqrt[3]{2}$$ where $$\rho=\zeta_3=\frac{-1+\sqrt{-3}}{2}$$ is a primitive cube root of unity. Hence the splitting field can be written $$\mathbb{Q}(\sqrt[3]{2}, \rho\sqrt[3]{2})$$. Any automorphism maps each of these two elements to one of the roots of $$x^3-2$$, giving 9 possibilities, but since the Galois group has order 6 not every such map is an automorphism of the field.

My question:

What exactly is the author seeing here to give 9 possibilities? I can only see $$\sqrt[3]{2}\mapsto \sqrt[3]{2}, \rho\sqrt[3]{2}, \rho^2\sqrt[3]{2}$$ and $$\rho\sqrt[3]{2}\mapsto \sqrt[3]{2}, \rho\sqrt[3]{2}, \rho^2\sqrt[3]{2}$$, giving 6 possibilities.

• There are three choices for the first element, and three choices for the second element. Mar 26, 2023 at 22:35
• What are the three choices for the first element? "An automorphism maps each of these two elements to one of the roots of $x^3-2$..."
– IAAW
Mar 26, 2023 at 22:48
• You wrote down three choices for each basis element. So the total number of choices is $3\times 3 = 9$. Apr 4, 2023 at 16:56

An automorphism of $$\mathbb{Q}(2^{1/3},\rho 2^{1/3})$$ fixing $$\mathbb{Q}$$ is determined by both the image of $$2^{1/3}$$ and the image of $$\rho 2^{1/3}$$. Only listing out $$2^{1/3}\to2^{1/3},\rho2^{1/3},\rho^22^{1/3}$$ does not give you $$3$$ maps. Instead, the combination $$2^{1/3}\to \rho2^{1/3}$$ and $$\rho2^{1/3}\to\rho^22^{1/3}$$ (for example) gives you one candidate.
The problem becomes finding the number of possible combinations of the images of $$2^{1/3}$$ and $$\rho 2^{1/3}$$. Of course, it is $$3\times 3=9$$.