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This is an example from my lecture notes. I'm not sure what it is called so sorry for the poor title. As far as I can tell it is an application of the Fundamental Theorem of Galois Theory.

So Let K : F be a finite Galois extension with $G = \Gamma (K : F)$ (so $|G| = |K : F|)$. H a subgroup of G. And $E=Fix(H)$

Example 21

$g = x^3-2$, $F = \mathbb{Q}$, roots $t$, $wt$, $w^2t$, where $t^3 = 2$, $w$ is a primitive third root of unity. $K = \mathbb{Q}(t, w)$, $|K : \mathbb{Q}| = 6$.

$G = \{1; \sigma _1 = (1, 2, 3); \sigma ^2_1 = (1, 3, 2); \sigma _2 =(2, 3); \sigma_1 \sigma_2 = (1, 2); \sigma_1^2 \sigma_2 = (1, 3)\}.$

There are 6 different subgroups. I'll only ask about 2 since I think I can get the rest from them.

$H_1 =\{1, (1,3)\}$, so clearly $|H|=2$, now we need to work out $E$. The way I can see to do this is as follows,

Take a general element of $K, k= a+bt+cwt+dw^2t $, apply $(1,3)$, which swaps the 1st root for the 3rd and 3rd for the 1st so

$(1,3)k= a+bw^2t + cwt + dt$, so the 2nd root remains the same and so $E = F(wt)$. Now I know the answer is right but I'm not sure if the method I have done is correct.

Now another groups is:

$H_2 = \{1, (1, 2, 3), (1, 3, 2)\}$ And this is where my method falls down since I know $E= F(w)$, but I cannot work out how to do it.

Can someone please explain how do work out $E$ for $H_2$ and if my method is correct for the $H_1$, since I don't think it is.

Thanks, Bobby

P.S. I've got another question which has $x^4-x^2-1$ which I have the answers to which I can give if that makes it easier to explain.

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A $\mathbb{Q}$-basis for $K$ is $\{1,t,t^2,w,wt,wt^2\}$ so that a general element will have the form $a+bt+ct^2+dw+ewt+fwt^2$. You noted that the degree of the extension is $6$, which means that there should be $6$ basis vectors. – Jared Apr 17 '13 at 16:15
up vote 0 down vote accepted

It will probably be helpful to describe the action of the Galois group on the field elements a little more explicitly than just considering how it permutes the roots. Let's look at what the generators do to both $t$ and $w$: $$\begin{array}{rcl}\sigma_1:t&\longmapsto&wt\\w&\longmapsto&w\\[.2in]\sigma_2:t&\longmapsto&t\\w&\longmapsto&w^2=-w-1\\\end{array}$$

See if you can verify this. With this information in hand, we can now see what an element like $\sigma_1^2\sigma_2$ does to the generators:


Now, considering your first example, let's apply $\sigma_1^2\sigma_2$ to a general element in $K$: $$\begin{align}\sigma_1^2\sigma_2(a+bt+ct^2+dw+ewt+fwt^2)&=a-bwt-bt+cwt^2-dw-d+ewt+ft^2\\&=(a-d)-bt+ft^2-dw+(e-b)wt+cwt^2\end{align}$$

so that the element is fixed if and only if $b=d=0$ and $c=f$. This gives an element of the form $a+ct^2+ewt+cwt^2=a+ewt+ct^2(w+1)=a+ewt-c(wt)^2$ which is exactly all elements in the field $F(wt)$ as you've said.

See if you can follow this method for $H_2$, where the action of $\sigma_1^2$ fixes $w$ and sends $t$ to $w^2=-wt-t$.

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Ok I think I have it then. $\sigma_1^2: t \rightarrow w^2t$, $w \rightarrow w$ $\sigma_1^2(a+bt+ct^2+dw+ewt+fwt^2) =$ $a-bwt-bt+cwt^2+dw+et-fwt^2-ft^2 = a+(e-b)-ft^2+dw-bwt+(c-f)wt^2$ So $b=e=f=c=0$, and so we are left with: $a+dw = F(w)$ – Bobby Apr 17 '13 at 19:29
Looks good to me. In general, although a bit computationally heavy, this method can help you find the fixed fields of any subgroup of the Galois group. Sometimes the fixed fields are rather surprising, as they are not just obtained by adjoining certain elements of the basis. – Jared Apr 18 '13 at 1:04

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