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Let $F$ be the splitting field of $x^8-5$, we have $F=\mathbb{Q}(5^{1/8},i,\sqrt{2})$. I need to calculate the galois group Gal$(F/\mathbb{Q})$.

I know the three generators are \begin{align*} &\sigma_1:5^{1/8}\mapsto 5^{1/8}\zeta_8,\;\sqrt{2}\mapsto\sqrt{2},\;i\mapsto i\\ &\sigma_2:5^{1/8}\mapsto 5^{1/8},\;\sqrt{2}\mapsto-\sqrt{2},\;i\mapsto i\\ &\sigma_3:5^{1/8}\mapsto 5^{1/8},\;\sqrt{2}\mapsto\sqrt{2},\;i\mapsto -i\\ \end{align*}

Then Gal$(F/\mathbb{Q})=<\sigma_1,\sigma_2,\sigma_3>$

So I got (updated: adding $\beta\gamma=\gamma\beta$) $$ Gal(F/\mathbb{Q})=<\alpha,\beta,\gamma:\alpha^8=1,\beta^2=\gamma^2=1,\gamma\beta=\beta\gamma, \beta\alpha=\alpha^3\beta,\gamma\alpha=\alpha^7\gamma> $$

But I am not sure does such result count as working out the galois group? Is there any way to work out an explicit expression like $D_8\times K$ or something?

I know $|Gal(F/\mathbb{Q})|=32$, and $<\sigma_2,\sigma_3>=\mathbb{Z}_2\times\mathbb{Z}_2$, and $<\sigma_1>=\mathbb{Z}_8$, but how to combine them?

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    $\begingroup$ A LaTeX tip: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use \langle and \rangle. Also, you can use \mathrm{Gal} or \operatorname{Gal} to make the text upright (as it should be). $\endgroup$ – Zev Chonoles Jun 22 '13 at 5:10
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    $\begingroup$ Where did the sqrt of two come in? $\endgroup$ – Mariano Suárez-Álvarez Jun 22 '13 at 5:12
  • $\begingroup$ @ZevChonoles, thanks! $\endgroup$ – hxhxhx88 Jun 22 '13 at 5:19
  • $\begingroup$ @MarianoSuárez-Alvarez, $\zeta_8^2=e^{i\pi/4}$ $\endgroup$ – hxhxhx88 Jun 22 '13 at 5:19
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    $\begingroup$ Why not just let the field be generated by $5^{1/8}$ and $\zeta_8$? $\endgroup$ – Mariano Suárez-Álvarez Jun 22 '13 at 5:21
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Let $G=\mathbb{Z}_8\rtimes \mathbb{Z}_8^\times$. Elements of $G$ are pairs $(a,b)$ with $a\in\mathbb{Z}_8$ and $b\in\mathbb{Z}_8^\times$. The group operation is given by $(a,b)(a',b')=(a+ba',bb')$. Since $(a,b)=(a,1)(0,b)$, it is generated by elements of the form $(a,1)$ and $(0,b)$. Note that the inclusion maps $i_1:\mathbb{Z}_8\rightarrow G$ and $i_2:\mathbb{Z}_8^\times\rightarrow G$ given by $i_1(a)=(a,1)$ and $i_2(b)=(0,b)$ are homomorphisms. Since $\mathbb{Z}_8^\times$ is generated by 3 and 7, we can use the above facts to conclude that $G$ is generated by the elements $\alpha=(1,1)$, $\beta=(0,3)$ and $\gamma=(0,7)$.

Since $i_1$ is a homomorphism, $\alpha^8=(0,1)$. Since $i_2$ is a homomorphism, $\beta^2=\gamma^2=1$ and $\beta\gamma=\gamma\beta$. Furthermore, $\beta\alpha=(0,3)(1,1)=(3,3)$ and $\alpha^3\beta=(1,1)^3(0,3)=(3,1)(0,3)=(3,3)$ so $\beta\alpha=\alpha^3\beta$. Similarly, $\alpha^7\gamma=\gamma\alpha$. Using the fact that $G$ has order 32, one may show these give all the relations. By comparing with your presentation for $\mathrm{Gal}(F/\mathbb{Q})$, we see that $G\cong \mathrm{Gal}(F/\mathbb{Q})$.

Alternately, we could have made this identification directly by using the fact that $G$ acts on $F$ by $(a,b) \zeta_8=\zeta_8^b$ and $(a,b)5^{1/8}=\zeta_8^a 5^{1/8}$.

We can give a slightly different description as follows. Let $H$ be the subgroup of $GL_2(\mathbb{Z}_8)$ consisting of matrices of the form $ \left[ \begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$ with $a\in \mathbb{Z}_8$ and $b\in\mathbb{Z}_8^\times$. Define $\phi:G\rightarrow H$ by $\phi(a,b)= \left[\begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$. One may check that $\phi$ is an isomorphism, so the Galois group can be identified with the group $H$ of invertible matrices over $\mathbb{Z}_8$ of the form $ \left[ \begin{array}{ c c } b & a \\ 0 & 1 \end{array} \right]$.

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  • $\begingroup$ Very very beautiful! Thank you so much! By the way, I wonder what is the motivation driving you to consider such construction of $G$ in the first place? $\endgroup$ – hxhxhx88 Jun 22 '13 at 11:51

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