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My question is part of a larger problem: I'm supposed to find the minimal polynomial over $\mathbb{Q}$ of $1 + \sqrt[3]{2} + \sqrt[3]{4}$ "using the automorphisms of the corresponding Galois extension."

After talking to my professor, I know that the Galois group I'm looking for is $S_3$. However, I don't know how to ascertain what the Galois extension is with the information I've been given.

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    $\begingroup$ I think, @астонвіллаолофмэллбэрг, that you get the Galois extension by adding $\omega=-\frac12+\frac{\sqrt{-3}}2$, not $i$. But in broad outline, your suggestions seem on the mark. $\endgroup$
    – Lubin
    Commented Nov 23, 2019 at 4:16
  • $\begingroup$ @Lubin Yes, sorry, I meant $\omega$ instead of $i$ above, but got carried away with the rest of the explanation, which thankfully is right. $\endgroup$ Commented Nov 23, 2019 at 4:22
  • $\begingroup$ Indeed, @астонвіллаолофмэллбэрг, it was hardly more than a slip of the pen (keyboard?), but would have been debilitating for OP if allowed to stand. $\endgroup$
    – Lubin
    Commented Nov 23, 2019 at 4:27
  • $\begingroup$ Thank you both; the problem makes sense now. $\endgroup$
    – Annapox
    Commented Nov 23, 2019 at 4:57
  • $\begingroup$ @Sarah Once you are done, kindly write an answer yourself and close the question. $\endgroup$ Commented Nov 23, 2019 at 4:59

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Let $\alpha = \sqrt[3]{2}$. We are looking for the minimal polynomial of $1 + \alpha + \alpha^2$. To find the Galois extension of $\mathbb{Q}$ that contains $1 + \alpha + \alpha^2$, note that $1, \alpha, \alpha^2 \in \mathbb{Q}(\alpha)$. However, $\mathbb{Q}(\alpha)$ is not Galois over $\mathbb{Q}$ because the polynomial $p(x) = x^3 - 2$ is irreducible in $\mathbb{Q}$, has a root in $\mathbb{Q}(\alpha)$, and does not split into linear factors in $\mathbb{Q}(\alpha)$. However, $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\alpha, \omega)$, where $\omega = e^{2i\pi/3}$. Furthermore, $p(x)$ splits into linear factors in $\mathbb{Q}(\alpha, \omega)$, so $\mathbb{Q}(\alpha, \omega)$ is Galois over $\mathbb{Q}$.

Next we must find the Galois group of $\mathbb{Q}(\alpha, \omega)/\mathbb{Q}$. We know that $[\mathbb{Q}(\alpha, \omega): \mathbb{Q}] \leq 3! = 6$. We also know that $[\mathbb{Q}(\alpha): \mathbb{Q}] = 3$ since $p(x)$ is the minimal polynomial over $\mathbb{Q}$ of $\alpha$. Since $[\mathbb{Q}(\alpha, \omega): \mathbb{Q}] = [\mathbb{Q}(\alpha, \omega): \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}]$, it must be that $3 < [\mathbb{Q}(\alpha, \omega): \mathbb{Q}] \leq 6$ and 3 divides $[\mathbb{Q}(\alpha, \omega): \mathbb{Q}]$. So $[\mathbb{Q}(\alpha, \omega): \mathbb{Q}] = 6$; hence $\operatorname{Gal}(\mathbb{Q}(\alpha, \omega)/\mathbb{Q})$ is isomorphic to $S_3$. In particular, consider the set of permutations in $S_3$ that map $\alpha$ to roots of $p(x)$. If we define $\sigma_1, \sigma_2, \sigma_3 \in S_3$ by \begin{align*} \sigma_1(\alpha) &= \alpha\\ \sigma_2(\alpha) &= \alpha \omega\\ \sigma_3(\alpha) &= \alpha \omega^2 \end{align*} then the images under $\sigma_1, \sigma_2, \sigma_3$ of $1 + \alpha + \alpha^2$ are the roots of the minimal polynomial of $1 + \alpha + \alpha^2$. We have \begin{align*} \sigma_1\left(1 + \alpha + \alpha^2\right) = 1 + \alpha + \alpha^2\\ \sigma_2\left(1 + \alpha + \alpha^2\right) = 1 + \alpha \omega + \alpha^2 \omega^2\\ \sigma_3\left(1 + \alpha + \alpha^2\right) = 1 + \alpha \omega^2 + \alpha^2 \omega^4 \end{align*} So the minimal polynomial of $1 + \sqrt[3]{2} + \sqrt[3]{4}$ is $$q(x) = \left(x - 1 - 2^{1/3} - 2^{2/3}\right)\left(x - 2^{1/3}e^{2i\pi/3} - 2^{2/3}e^{-2i\pi/3}\right)\left(x - 2^{1/3}e^{-2i\pi/3} - 2^{2/3}e^{2i\pi/3}\right).$$ After doing a great deal of algebra, we find that $q(x) = x^3 - 3x^2 - 3x - 1$.

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