# Fixed fields for Galois groups

• Apr 6th 2006, 06:43 AM
DMT
Fixed fields for Galois groups
I need some help with a Galois problem.

I need to find the splitting field for
$\displaystyle x^7 - 1$ over Q, then find its Galois group, then the subgroups of the Galois group, and then the fixed fields for the subgroups.

Now I thought I understood this, but I got into trouble at the end, which started to make me realize I must have been doing something wrong earlier as well.

The roots of this are obviously (besides 1)
$\displaystyle w, w^2, w^3, w^4, w^5, w^6, w=e^{2\pi i/7}$
and $\displaystyle w+w^2+w^3+w^4+w^5+w^6=0, w^7=1$

Which gives me $\displaystyle Q(e^{2\pi i/7})$ as the field extension, which a Galois group of order 6 isomorphic with Z6 and Z7*, all the elements of which are generated by setting w to one of the other roots. This gives me two subgroups, one with 2 elements and one with 3 elements.

I was completey unable to figure out how to find the fixed fields for these subgroups in any meaningful way. And then I started to think my Galois group wasn't even correct. Most of the permutations seemed to change Q, which means they aren't correct, though I thought this was the standard technique with the cyclotomics (but I'm really new to this stuff).

Now, now I'm thinking that there must be something I'm missing. I see a few opitions here. One is that $\displaystyle cos(2\pi/7)$ can be expressed in terms of a cube root, which would mean I only need to add that cube root and i to Q to get the rigth field, which gives me the same isomorphy anyway.

But this isn't some "obvious" cosine to me, so a question is how would i recognize that this was a root like that if it were true without a table (given an arbitrary cyclotomic)?

The other idea I had was that there might be some obvious relations between various of the roots, like being conjugates of each other or something like that, which would seem to have to be the case now that I think about it ... I hadn't thought about that before, but obviously if one complex number is a root, the conjugate must also be a root, which gives me 3 roots and their conjugates. I guess the question is will that make everything cancel the way it should wihtout me needing to figure out if the cosine is a cube root?

Any comments, help would be appreciated.

Thanks.
• Apr 7th 2006, 11:01 PM
rgep
I'll write z for the primitive 7-th root of unity. The Galois group is $\displaystyle G = ({\mathbf Z}/7)^*$ acting by $\displaystyle \sigma_a : z \mapsto z^a$ and is abstractly isomorphic to a cyclic group of order 6.

The subgroups of G are of order 2 and 3, generated by -1 and 2 say.

For the subgroup of order 3, consider $\displaystyle A =z + z^2 + z^4$. This is invariant under the action $\displaystyle z \mapsto z^2$ and so is in the fixed field. It has a conjugate $\displaystyle A' = z^3 + z^6 + z^5$ which is obtained by applying $\displaystyle \sigma_3$. We observe that $\displaystyle A+A' = -1$ and on multiplying out, $\displaystyle AA' = 2$. We conclude that the subfield is $\displaystyle {\mathbf Q}(A) = {\mathbf Q}(\sqrt{-7})$.

For the subgroup of order 2, consider $\displaystyle B = z + z^{-1}$. This has conjugates $\displaystyle B' = z^3 + z^{-3}$ and $\displaystyle B'' = z^2 + z^{-2}$. We have $\displaystyle B+B'+B'' = -1$, $\displaystyle BB' + B'B'' + B''B = -2$ and $\displaystyle BB'B''=1$. This identifies the subfield $\displaystyle {\mathbf Q}(B)$ as a cubic extension of the rationals. Of course you can save yourself time by seeing that $\displaystyle B = 2\cos (2\pi/7)$.

Since 2 and 3 are coprime, the subgroups of orders 2 and 3 intersect in the trivial subgroup, so the corresponding fields generate the whole extension: that is, $\displaystyle \mathbf Q(z) = \mathbf Q(A,B)$.
• Apr 8th 2006, 12:15 AM
DMT
Thanks, I actually managed to get most of that, but the last part is still giving me some trouble. I did see the cosine relation, but I don't see how that ituitively says that the group was a cubic extension of the rationals? Is there some princicple for some cosines that I shoudl know to have seen this?

As for your relation comparing the conjuagates ... how does looking at that tell you that it is a cubic extension of the rationals? And how would you know to combine all the various B's in that way to look?

In other words, I'm looking for some general principles here to help with finding the fixed intermediate fields. Thanks.
• Apr 8th 2006, 08:44 AM
rgep
If H is a subgroup of the Galois group G, then $\displaystyle y = x^H = \sum_{h\in H} x^h$ is invariant under H; if now $\displaystyle G:H$ is a set of coset repsentatives of H in G, then the $\displaystyle y^c$ for $\displaystyle c \in G:H$ are a set of conjugates of y and the elementary symmetric functions give the coefficients of an equation for y.

In this case you know that $\displaystyle B = \cos 2\pi/7$ will be cubic, and the expression as a cosine makes it easier to compute directly what the relation is by considering $\displaystyle B' = \cos 3.2\pi/7$ in terms of B.