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I'm trying to understand Galois theory and any help on this question I'm working on would be very much appreciated.

Let $E$ be the splitting field of $x^3-5$ over $\Bbb Q$. Compute $\mathrm{Gal}(E:\Bbb Q)$, and find all subfields of $E$.

Here is my attempt:

$E$ is the splitting field of $x^3-5$ which has roots $\sqrt[3]{5},w\sqrt[3]{5},w^2\sqrt[3]{5}$, where $w$ is the primitive cubed root of unity.

The extension is normal and separable $\Rightarrow$ $|\mathrm{Gal}(E:\Bbb Q)|=|E:\Bbb Q|=6$, by a minimal polynomial argument. So now we know there are six automorphisms which fix the the ground field in our Galois group.

Next we know that these automorphisms will map the roots of a polynomial to the other roots of that polynomial.

$\sqrt[3]{5},w\sqrt[3]{5},w^2\sqrt[3]{5}$ are the roots of $x^3-5$ so we have an automorphism which fixes the ground field being :

h: $\sqrt[3]{5}\rightarrow \sqrt[3]{5}$

h: $\sqrt[3]{5}\rightarrow w \sqrt[3]{5}$

h: $\sqrt[3]{5}\rightarrow w^2 \sqrt[3]{5}$

We also have that $w$ is a root of $x^3-1$ which has as its other roots 1 and $w^2$, so another automorphism in the group is

$b:w\rightarrow w$

$b:w\rightarrow w^2$

$b:w\rightarrow 1$

which gives the six elements of our Galois group.

There are only two groups of order 6: $\Bbb Z_6, S_3$, we know it's $S_3$ as we are looking at permutations of a 3 element set .

It has two non trivial subgroups $S_2, A_3$. $A_3$ I think corresponds to $\Bbb Q(\sqrt[3]{5})$ I think and the other to $\Bbb Q(w)$, because of the order of the respective groups and the degree of the extensions over $\Bbb Q$. But I'll be honest that's just an intuitive feeling I don't really understand this bit.

user26857
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excalibirr
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1 Answers1

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The minimal polynomial of ω is $x^2+x+1$, which has roots $ω$ and $ω^2$. You can't send ω to 1, as that wouldn't be a homomorphism.

Then, the galois grup G is generated by $\tau$ which sends $\sqrt[3]5$ to $\sqrt[3]5 \omega$, and $\sigma$ which sends $\omega$ to $\omega^2$. This gives rise to 6 automorphisms, and by Galois theory you know that's all since the degree of the extension is 6 too. You can now find what group G is isomorphic to.

Locally unskillful
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  • Why cant $\tau$ send $\sqrt[3]{5}$ to $w^2\sqrt[3]{5}$ ? – excalibirr Apr 14 '19 at 16:58
  • That is also valid – Locally unskillful Apr 14 '19 at 17:03
  • ah okay so its just an arbitrary choice whether we choose $\tau$ to map to either ? – excalibirr Apr 14 '19 at 17:08
  • In this case, you can choose either one yes. – Locally unskillful Apr 14 '19 at 17:10
  • I'm confused, in this case I can choose either one, but when is there a time when I can not choose either ? – excalibirr Apr 14 '19 at 17:18
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    In this case, your galois group has order 6. You know that from the beginning. Any automorphiam is determined by where you send $\sqrt[3]5$ and ω. These are sent to roots of their minimal polynomials. Then the most obvious thing you can do is send $ω$ to $ω^2$, thus automorphism of order 2. You may also send $\sqrt[3]5$ to either $\sqrt[3]5ω$ or $\sqrt[3]5 ω^2$, as either automorphism has order 3 (immaterial which one you choose). Then 2 times 3 is 6, so you have all 6 required automorphisms. – Locally unskillful Apr 14 '19 at 17:30