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I'm working with Galois group $Gal \Big(\frac{\mathbb{Q}(\sqrt[3]{3},\sqrt{-3})}{\mathbb{Q}}\Big)$ and I'm able to determine that all automorphism work in the following way:

$\sigma(\sqrt[3]{3}) = \sqrt[3]{3}, \omega \sqrt[3]{3}, \omega^2\sqrt[3]{3}$ and $\sigma(\sqrt{-3}) = \sqrt{-3},-\sqrt{-3}$ I'm trying to calculate all the orders of the automorphisms.

However, say I have the particular automorphism of the form $\sigma(\sqrt[3]{3}) = \omega\sqrt[3]{3}$ and $\sigma(\sqrt{-3}) = -\sqrt{-3}$ my problem is that if I want to compute the order i need to know the image of $\omega$.

$\sqrt[3]{3} \to \omega\sqrt[3]{3} \to \cdots$

How do I know how to continue this sequence?

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  • $\begingroup$ The problem is solved observing $\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$ $\endgroup$ – Rodrigo Nov 8 '17 at 1:48
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You know that $\sqrt[3]{3}$ is a root of the irreducible polynomial $x^3 - 3$ and $\sqrt{-3}$ is a root of the polynomial $x^2+3$.

Thus if $\omega$ is a cube root of unity, then $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, and $\omega^2 \sqrt[3]{3}$ are the three roots of $x^3 - 3$ and an automorphism must permute them (automorphisms can't take an element of degree 3 to one of degree 2).

Further, using the tower theorem you have: $[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}(\sqrt{-3})][\mathbb{Q}(\sqrt{-3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{3},\sqrt{-3}): \mathbb{Q}] = 6$ by looking at the degrees of the polynomials.

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  • $\begingroup$ But how do i know the image of $\omega 3^{1/3}$ by the automorphism above? $\endgroup$ – Rodrigo Nov 6 '17 at 2:50
  • $\begingroup$ It can be any of the roots of $x^3-3$, depending on the automorphism. The group acts like the permutation group on three elements. $\endgroup$ – Chris C Nov 6 '17 at 2:54
  • $\begingroup$ Thus the part acting on the degree two part has order two, but I don't think you have enough info on the degree three portion to see if it has order 2 or 3. $\endgroup$ – Chris C Nov 6 '17 at 2:57
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    $\begingroup$ You haven't actually done the "meat" of the problem, which is showing that $\sqrt[3]{3}$ still has degree $3$ over $\mathbb{Q}(\sqrt{-3})$. $\endgroup$ – Steve D Nov 6 '17 at 4:25
  • $\begingroup$ @ChrisC if this is the case all the elements of the group would be of order 2 and this is impossible for order 6, i believe... $\endgroup$ – Rodrigo Nov 8 '17 at 2:42

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