2
$\begingroup$

I want to find the Galois group over $\mathbb{Q}$ of $(x^3-2)(x^2-2).$


I already know that the Galois group over $\mathbb{Q}$ of $f(x) = x^3-2$ is isomorphic to $S_3$ and there are four intermediate fields between the splitting field of $f$ and $\mathbb{Q}$: $\mathbb{Q}(\sqrt[3]{2}),\,\mathbb{Q}(\omega),\,\mathbb{Q}(\omega\sqrt[3]{2}),\,\mathbb{Q}(\omega^2\sqrt[3]{2}),$ where $\omega$ is a primitive $3^\mathrm{rd}$ root of unity. It's easy to verify that $\sqrt{2}$ does not belong in one of these intermediate fields, so by the fundamental theorem, $\mathbb{Q}(\sqrt{2})\not\subset\mathbb{Q}(\sqrt[3]{2},\omega).$ Now we have that $$[\mathbb{Q}(\sqrt[3]{2},\omega,\sqrt{2}) : \mathbb{Q}]=12.$$ This is where I'm stuck. I don't know how to conclude what the Galois group of the polynomial is, other than it is some subgroup of $S_6$ of order $12$.

$\endgroup$
  • 2
    $\begingroup$ First, show that your extension field is a splitting field of $x^6-2$ over $\mathbb{Q}$. Then see this: math.stackexchange.com/questions/1575214/… $\endgroup$ – Sungjin Kim Aug 5 '16 at 22:50
  • 2
    $\begingroup$ Actually you should be able to realize this as a subgroup of $S_5$ because your polynomial only has five zeros. Anyway, the imaginary part of the eighth root of unity found a nice way of doing it. Also, this question is quite similar to this. It doesn't seem to matter, which square root you adjoin in addition to the cube roots (ok, $\sqrt{-3}$ would be an exception). $\endgroup$ – Jyrki Lahtonen Aug 5 '16 at 23:11
4
$\begingroup$

Well, you know what the generating automorphisms of the Galois group are, namely $\sigma, \tau, \rho$ which act as follows on the generators of the splitting field: $$\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}; \sigma(\omega) = \omega; \sigma(\sqrt{2}) = \sqrt{2}$$ $$\tau(\sqrt[3]{2}) = \sqrt[3]{2}; \tau(\omega) = \omega^{2}; \tau(\sqrt{2}) = \sqrt{2}$$ $$\rho(\sqrt[3]{2}) = \sqrt[3]{2}; \rho(\omega) = \omega; \rho(\sqrt{2}) = -\sqrt{2}$$ Now, we compute $$(\sigma\rho)(\sqrt[3]{2}) = \omega\sqrt[3]{2}; (\sigma\rho)(\omega) = \omega; (\sigma\rho)(\sqrt{2}) = -\sqrt{2}$$ It is easy to see that $\sigma\rho$ has order $6$ (why?). Further, $(\sigma\rho)^{-1} = \rho^{-1}\sigma^{-1} = \rho\sigma^{2}$. Finally, note that $\sigma$ and $\rho$ commute, as well as $\tau$ and $\rho$, and $\sigma, \tau$ obey the relations $\sigma\tau = \tau\sigma^{2}$. Now, your group $G$ cannot be abelian, and must contain $S_{3}$. A good candidate for $G$ is thus $D_{6}$, which is nonabelian of order $12$ and contains an isomorphic copy of $D_{3} \cong S_{3}$. Let $\beta = \sigma\rho$. Then $$\beta\tau = \sigma\rho\tau = \sigma\tau\rho = \tau\sigma^{2}\rho = \tau\rho\sigma^{2} = \tau\beta^{-1}$$ Furthermore, $\beta^{4} = \sigma$, $\beta^{3} = \rho$, so $G$ has the presentation $\langle \beta, \tau \mid \beta^{6} = \tau^{2} = e, \beta\tau = \tau\beta^{-1}\rangle \cong D_{6}$.

$\endgroup$
  • $\begingroup$ user346096: feel free to let me know if there is something in my answer that needs further clarification/explanation! $\endgroup$ – Alex Wertheim Aug 8 '16 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.