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First of all, I've read some questions that have similar title, but I didn't find any answers that correspond to what I have in mind.

So, I am an undergraduate student starting my investigations in Galois Theory, and I have sort of computational questions:

Suppose we are looking for the intermediate fields of the Galois extension $\mathbb{Q}(\zeta_7)/\mathbb{Q}$, where $\zeta_7$ is a non-real root of $x^{7}=1$.

I know that the Fundamental Theorem of Galois Theory gives us the correspondence between these intermediate fields and the subgroups of the group $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q})$.

So, this group is isomorphic to $(\mathbb{Z}/7\mathbb{Z})^{\times}\cong \mathbb{Z}/6\mathbb{Z}$, which has 4 subgroups, namely: the identity, generated by 2, generated by 3 and the whole group.

Ok, from this we get that there are 2 proper subfields in the extension with degrees 2 and 3, corresponding to these proper subgroups.

Now, some thoughts occurred to me:

We know that $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong \mathbb{Z}/6\mathbb{Z}$, but we do not immediately know which automorphisms of the Galois group correspond to the elements of $\mathbb{Z}/6\mathbb{Z}$. In other words, we do not know immediately which of our known automorphisms are, say, in the proper subgroup of index 2 (information we would use to find the intermediate fields, by checking which elements of $\mathbb{Q}(\zeta_7)$ get fixed by them). Of course I could check the order, for example, of the automorphism that sends $\zeta_7\mapsto (\zeta_7)^{2}$ and find the corresponding element of $\mathbb{Z}/6\mathbb{Z}$, but, from the examples I've been doing, I know that this takes a while to do (maybe not so much in this example, but in some bigger extension it certainly does).

Isn't there any way of looking for these automorphisms in a more geometrical way? In such a way that it would be quite obvious which elements of $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q}$ correspond to which elements of $\mathbb{Z}/6\mathbb{Z}$?

By geometrical I mean, for example, looking at the unit circle and imagining such an automorphism moving things around. I've tried to think that way but I didn't get much out of it. Is this possible? If so, I believe this would help us too with the job of finding the corresponding fixed field, (which too involves long calculations) right?

Thanks in advance to all.

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    $\begingroup$ It is actually easier to see the Galois group as $\mathbb Z/\mathbb 7\mathbb Z^{\times}$. Then finding an isomorphism to $(\mathbb Z/6\mathbb Z,+)$ amounts to finding a generator for $\mathbb Z/\mathbb 7\mathbb Z^{\times}$. Finding generators is hard - little is know about how to find a generator for $\mathbb Z/p\mathbb Z^{\times}$ other than slight improvements on brute force. $\endgroup$ – Thomas Andrews Sep 30 '15 at 14:12
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    $\begingroup$ In this case, (the restriction of) complex conjugation is an automorphism of order $2$, and there is only one element of $\Bbb Z / 6 \Bbb Z$ of that order. $\endgroup$ – Travis Sep 30 '15 at 14:13
  • $\begingroup$ So, there is no "easy way" to see things geometrically? This example was an illustration, but i ask more generally, if there is, at least for cyclotomic extensions, this kind of view .. I say this because I've heard some guy in my departament talking about this, but I hadn't got the chance to talk to him yet and I'm very curious about this $\endgroup$ – Shoutre Oct 1 '15 at 1:47
  • $\begingroup$ This might be a little too advanced, depending on your background, but my answer to this question perhaps may shed some light: math.stackexchange.com/questions/1392519/… $\endgroup$ – Alex Youcis Oct 1 '15 at 18:56
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    $\begingroup$ @Shoutre I mean, in a non-canonical way you can always take a faithful representation of the Galois group (i.e. an injective morphism into some $\mathrm{GL}_n(\mathbb{C})$) which allows one to think about the Galois group as being a group of linear transformations. That is not canonical though. $\endgroup$ – Alex Youcis Oct 4 '15 at 12:01

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