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Let $L$ be the splitting field of the polynomial $f = T^3 - 2 \in \mathbb{Q}[T]$.

$\textbf{a)}$ Determine $[L:\mathbb{Q}]$ and a $\mathbb{Q}-$basis of $L$.
$\textbf{b)}$ Let $\zeta$ be a primitive $3$rd unit root and $\sigma: \mathbb{Q}(\zeta) \to \mathbb{Q}(\zeta)$ be a morphism such that $\sigma(\zeta) = \zeta^{2}$. Determine all extensions $\tilde{\sigma}: L \to \mathbb{Q}(\zeta)$ of $\sigma$
$\textbf{c)}$ Determine all $\mathbb{Q}-$automorphisms $\sigma: L \to L$, i.e. determine $\text{Gal}(L/\mathbb{Q})$.
$\textbf{d)}$ Determine two (canonical) generators for $\text{Gal}(L/\mathbb{Q})$
$\textbf{e)}$ You may use that we have $\text{Gal}(L/\mathbb{Q}) \cong S_3$ in order to determine all $M$ with $\mathbb{Q} \leq M \leq L$ and $[M:\mathbb{Q}] = 3$

First of all, I observed that the splitting field for $f$ is $L = \mathbb{Q}(\sqrt[3]{2}, \zeta)$ where $\zeta$ is a third primitive root. Consequently, I got for a) that $[L:\mathbb{Q}] = 6$ and as a basis I got $B = \{1, \sqrt[3]{2}, \sqrt[3]{4}, \zeta, \zeta\sqrt[3]{2}, \zeta\sqrt[3]{4}\}$

However, my problems start with the second point. In order to get all extensions I know that I may look at the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}(\zeta)$ (which is $m(\sqrt[3]{2}, \mathbb{Q}(\zeta)) = T^{3} - 2$ because $\sqrt[3]{2} \notin \mathbb{Q}(\zeta)$). Then I apply the morphism $\sigma$ on the minimal polynomial and look for all zeros of the minimal polyomial in $\mathbb{Q}(\zeta)$. The number of zeros in $\mathbb{Q}(\zeta)$ is equal to the number of extensions for $\sigma$.

So, I want to apply $\sigma$ on $m(\sqrt[3]{2}, \mathbb{Q}(\zeta))$ which gives me: $\sigma(m(\sqrt[3]{2}, \mathbb{Q}(\zeta))) = \sigma(T^{3} - 2) = \sigma(T)^3 - \sigma(2)$.

Now I don't really know how to apply that we have $\sigma(\zeta) = \zeta^2$. We have basically determined the image for $\zeta$ for $\sigma$ but the mapping doesn't tell me anything about the image of $\sigma(T)$ and $\sigma(2)$.

And without determining the extensions $\tilde{\sigma}$, I have some problems to determine the automorphisms, i.e. determine $\text{Gal}(L/\mathbb{Q})$. Could anybody help me with that exercise?

MathGeek
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    Surely this example has been discussed on this site already, probably several times. – Gerry Myerson Feb 05 '24 at 06:30
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    See https://math.stackexchange.com/questions/564693/determine-the-galois-groups-of-each-of-these-polynomials-in-mathbbq and https://math.stackexchange.com/questions/643765/permutation-of-roots-for-galois-group-with-six-elements and https://math.stackexchange.com/questions/345097/various-definitions-of-galois-groups and https://math.stackexchange.com/questions/2702662/finding-the-galois-group-of-x3-2-over-mathbbq and https://math.stackexchange.com/questions/1753097/determining-automorphisms-of-this-extension and surely many more. – Gerry Myerson Feb 05 '24 at 06:31
  • Thank you! However, none of those questions gives me answers about b) and e). Especially my question about how to find the extensions with this specific morphism $\sigma$ is not answered there. – MathGeek Feb 05 '24 at 14:11
  • Ok, for the part e) you just use that there is a 1-to-1 correspondence between the subgroups of a Galois group and the intermediate fields $M$. Since we have already given that $Gal(L/\mathbb{Q}) \cong S_3$ and we know that there is only one subgroup of $S_3$ with three elements (which is $A_3$), we can say that the only possible choice for $M$ is $A_3$. So, there is no need to answer this question. However, I am still struggling with part b)... – MathGeek Feb 05 '24 at 15:34
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    Part (b) is impossible. I will bet (a beer, may be) that the intention was to ask for all the extensions $\tilde{\sigma}:L\to L$. Surely the image has to be (at least) a six-dimensional extension field also! – Jyrki Lahtonen Feb 06 '24 at 05:56

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