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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.

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

1 Answer 1

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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 $3\times 3=9$.

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