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Been reviewing some Galois theory, and computing Galois groups has been relatively routine, but one has me stumped.

Suppose $\omega$ is a primitive 37th root of unity. Let's set $\alpha=\omega+\omega^{10}+\omega^{26}$. I want to compute the Galois group of $\mathbf{Q}(\alpha)/\mathbf{Q}$.

I realize that $\mathbf{Q}(\alpha)\subseteq\mathbf{Q}(\omega)$, so $[\mathbf{Q}(\alpha):\mathbf{Q}]$ must divide $[\mathbf{Q}(\omega):\mathbf{Q}]=\varphi(37)=36$. So the order of the Galois group must be a divisor of 36, but that's about all I've managed.

What's the trick to finding this Galois group? Thanks!

Edit: Any automorphism has to map $\omega$ to $\omega^k$ where $(k,37)=1$. I noticed that the maps sending $\omega\mapsto\omega^{10}$ and $\omega\mapsto\omega^{26}$ both fix $\alpha$. Where does one go from there?

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    $\begingroup$ The elements of the Galois group of $\mathbf{Q}(\omega)/\mathbf{Q}$ must send $\omega$ to another primitive 37th root of unity, that is, to $\omega^k$ with $\gcd(k,37)=1$. What does one of those automorphisms do to $\alpha$? $\endgroup$ – lhf Dec 1 '11 at 0:54
  • $\begingroup$ Oh, I see that the maps sending $\omega\mapsto\omega^{10}$ and $\omega\mapsto\omega^{26}$ both fix $\alpha$. $\endgroup$ – Nastassja Dec 1 '11 at 1:19
  • $\begingroup$ The idea is to try to find easy automorphism that fix $\alpha$ to try to increase the lower bound on the order of the Galois group you're seeking. $\endgroup$ – lhf Dec 1 '11 at 2:03
  • $\begingroup$ @lhf Finally got it, thanks very much for your kind help. $\endgroup$ – Nastassja Dec 1 '11 at 3:03
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You have pretty much answered your own question! Indeed, suppose $\sigma \in \text{Gal}(\mathbf{Q}(\omega)/\mathbf{Q})$ is such that $\sigma \alpha = \alpha$. As you noted, $\sigma \omega = \omega^k$ for some $k$ with $(k, 37)=1$. Then we have $\omega^k + \omega^{10k} + \omega^{26k} = \omega+\omega^{10}+\omega^{26}$. This implies $\{k,10k,26k\}=\{1,10,26\}$ mod $37$, by the linear independence of $\{1, \omega, \dots, \omega^{36}\}$. In particular, we have either $k=1, 10k=1$ or $26k=1$. Hence $k=1, k=10$ or $k=26$, and a quick check shows that all of these work. Therefore...

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  • $\begingroup$ Thanks Bruno. I honestly don't understand why I'm looking at automorphisms in $\mathrm{Gal}(\mathbf{Q}(\omega)/\mathbf{Q})$ that happen to fix $\alpha$ when I want to compute $\mathrm{Gal}(\mathbf{Q}(\alpha)/\mathbf{Q})$ instead. So there are 3 automorphisms of $\mathbf{Q}(\omega)$ that fix $\mathbf{Q}$ that restrict to the identity on $\mathbf{Q}(\alpha)$? I don't see how I'm close to answering the question. $\endgroup$ – Nastassja Dec 1 '11 at 2:38
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    $\begingroup$ Remember that if $H$ is a normal subgroup of the Galois group $G$ of the Galois extension $\text{Gal}(F/\mathbb{Q})$, with fixed field $K$, then $\text{Gal}(K/\mathbb{Q})$ is Galois with Galois group $G/H$. $\endgroup$ – Bruno Joyal Dec 1 '11 at 2:41
  • $\begingroup$ Ok, so $\{1,\varphi_{10},\varphi_{26}\}$ is a normal subgroup of the Galois group of $\mathrm{Gal}(\mathbf{Q}(\omega)/\mathbf{Q})$ with fixed field $\mathbf{Q}(\alpha)$, and hence $\mathrm{Gal}(\mathbf{Q}(\alpha)/\mathbf{Q})$ has Galois group $G/\{1,\varphi_{10},\varphi_{26}\}$? $\endgroup$ – Nastassja Dec 1 '11 at 2:53
  • $\begingroup$ Yup! It's normal because the Galois group of the big extension is abelian. The Galois group of your extension is cyclic of order 12. $\endgroup$ – Bruno Joyal Dec 1 '11 at 2:54
  • $\begingroup$ Not quite. $\text{Gal}(\mathbf{Q}(\omega)/\mathbf{Q}) = \mathbb{Z}/36\mathbb{Z} = (\mathbb{Z}/37\mathbb{Z})^\times$ $\endgroup$ – Bruno Joyal Dec 1 '11 at 2:57

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