# Finding a Galois Extension

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.

• 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. Commented Nov 23, 2019 at 4:16
• @Lubin Yes, sorry, I meant $\omega$ instead of $i$ above, but got carried away with the rest of the explanation, which thankfully is right. Commented Nov 23, 2019 at 4:22
• Indeed, @астонвіллаолофмэллбэрг, it was hardly more than a slip of the pen (keyboard?), but would have been debilitating for OP if allowed to stand. Commented Nov 23, 2019 at 4:27
• Thank you both; the problem makes sense now. Commented Nov 23, 2019 at 4:57
• @Sarah Once you are done, kindly write an answer yourself and close the question. Commented Nov 23, 2019 at 4:59

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 $$q(x) = x^3 - 3x^2 - 3x - 1$$.