Determine the subgroup $\text{Gal}(E/\mathbb Q[i\sqrt 3])$ where $E$ the splitting field of $(t^3-2)(t^3-5)$ and the intermediate fields. We clearly have that $E=\mathbb Q(\sqrt[3]2,\sqrt[3]5,e^{\frac{2i\pi}{3}})$, but how can I simplify this ? For the moment, I just want to get the subgroups, but I have the impression that $\text{Gal}(E/\mathbb Q[i\sqrt 3])$ is a group with $18$ elements, and so it looks very long to compute the group and the subgroups, but there is probably a trick.
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What you want is the intermediate fields of $L/\mathbb{Q}[i\sqrt{3}]$? Are you sure is $i$ and not the third root of unity $\xi$? – Aaron Maroja Dec 28 '14 at 13:35
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Yes, you're right, I corrected it. – idm Dec 28 '14 at 13:40
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Well, you could start finding $Aut_\mathbb{Q[\xi\sqrt{3}]}L$ as usual, determine what is the group that fixes $\xi\sqrt{3}$. You sure there are 18 elements? – Aaron Maroja Dec 28 '14 at 13:42
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1A related question. You may benefit from studying that also. Admittedly the focus is different there. – Jyrki Lahtonen Dec 28 '14 at 15:59
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In fact my problem is that's it's over $\mathbb Q(i\sqrt 3)$ and not over $\mathbb Q$. But don't we have that $e^{\frac{2i\pi}{3}}=\frac{1}{2}+i\frac{\sqrt 3}{2}\in\mathbb Q(i\sqrt 3)$ ? and so that $[E:\mathbb Q(i\sqrt 3)]=6$ ? – idm Dec 31 '14 at 10:29
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Why would it be 6? To my understanding $g=X^3-2$ is irreducible and you now have to check if $X^3-5$ is also irreducible after you are allowed to use the roots of g. – kummerer94 Dec 31 '14 at 11:57
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Let's write $a=\root3\of2$, $b=\root3\of5$, $c=e^{2\pi i/3}$, and let $F={\bf Q}(c)$. Can you prove that the Galois group of $F(a)/F$ is cyclic of order 3? and the same for the group of $F(b)/F$? and then that the group of $F(a,b)/F$ is the product of these two groups?
Gerry Myerson
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