# Homework Help: Calculating a pretty simple Galois Group

1. Oct 31, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Describe the structure of the Galois group for G(Q(c):Q) where c is a 14th primitive root of unity

2. Relevant equations

3. The attempt at a solution
the minimal polynomial for c over Q is f(x)=x^6-x^5+x^4-x^3+x^2-x+1. Is the galois group isomorphic to Z*_14? Or maybe that's only true for prime numbers?

Anyway, c has order 6, so c^1,.....,c^6=1 will all be roots of it's minimal polynomial f(x). The galois group will have a map b(1)=c^i for each 0>i>6. A part of me wants to say the galois group is the cyclic group of order 6, but the alternating polynomial makes me think it's a bit trickier than that. Could the galois group be some sort of alternating cyclic group? If that makes any sense...

Thanks PF!

2. Oct 31, 2016

### andrewkirk

$\mathbb Q(c)$ is the minimal field containing $\mathbb Q\cup\{c\}$. Since $c^7=-1$, and $c^0,...,c^6$ are distinct points on the unit circle in the complex plane, we can write the generic element of this field as:
$$\sum_{k=0}^6 a_kc^k$$
We are interested in automorphisms of that field, that fix (are the identity map on) $\mathbb Q$. Let $\phi$ be such an automorphism. Then we know that
$$q\in\mathbb Q\Rightarrow \phi(q)=q$$

So the structure of the Galois group will be entirely determined by the possible actions of such fixing automorphisms on $c$.

Can the group contain any $\phi$ such that $\phi(c)\in\mathbb Q$?
What else, if anything, can we say about $\phi(c)$?

3. Nov 1, 2016

### PsychonautQQ

We can see that phi will take c to another power of c? Therefore the galois group is isomorphic to the cyclic group of order 6?

4. Nov 1, 2016

### andrewkirk

I sense that you are thinking of the set of six fixing automorphisms:
$$\{\phi_k:\mathbb Q(c)\to \mathbb Q(c)\ :\ \phi(c)=c^k,\ 1\leq k\leq 6\}$$
Are you sure that is a group? Is it closed under function composition? What is $\phi_4\circ\phi_2$?

5. Nov 2, 2016

### PsychonautQQ

phi_4 * phi_2 = phi_2, and phi_4 isn't the identity so this must not be a group. I don't know what else it could be though, surely there is a unique mapping that takes the identity to each power of c?

What if given c^k and phi_i, then phi_i(c^k) = c^(k+i)? This would be a group I believe

Last edited: Nov 2, 2016
6. Nov 2, 2016

### Staff: Mentor

Right. And what again was the identity in your set?
I think @andrewkirk's method in post #2 is the better approach to start at than guessing a solution. Why didn't you try? You could start there, list all you already know about $c$, like the definition of $c$, decomposition of the minimal polynomial $f(x)$ in $\mathbb{Q}(c)[x]$ (with additional roots $c_i$) and such things.

By the way and out of personal interest: How did you get from $x^{14}-1$ to the minimal polynomial $f(x)=x^6-x^5+x^4-x^3+x^2-x+1\;$? In $x^{14}-1 = f(x) \cdot q(x)$ what is $q(x)\;$?

7. Nov 2, 2016

### andrewkirk

That's not what I get. Let's see how $\phi_4\circ\phi_2$ acts on $c$:

$\phi_4\circ\phi_2(c)=\phi_4(\phi_2(c))=\phi_4(c^2)=$ what power of $c$? (Hint: use the fact that $\phi_4$ is a field homomorphism)

Could there be some extra fixing automorphisms we need to include in our set?

8. Nov 2, 2016

### PsychonautQQ

phi_4(c^2) = phi_4(c)*phi_4(c)=c^4*c^4=c^8=c^2.
No?
Extra fixing homomorphisms? It's possible that there are homomorphisms that fix c but maybe not some power of c?

No, I don't believe phi(c) can ever be contained in Q, so therefore phi(c) can never be sent to c^0 I think.

We know that phi(c) will take c from one root of c's minimal polynomial in Q to another I believe? So phi(c) can go to any exponent of c except for any multiple of 6 because that would mean c^6n = 1 ?

Am I on the right track with anything here?

9. Nov 2, 2016

### PsychonautQQ

Hmm.. I'm trying to remember how I figured that out, it involved wolfram alpha though and I THINK (i think :-/) I used the idea that if c is a primitive 14th root of unity than -c is a 7th primitive root of unity

10. Nov 2, 2016

### Staff: Mentor

Thanks. (And it works without Wolfram, too. )

11. Nov 2, 2016

### Staff: Mentor

I'm pretty sure it would help a lot to write down the two decompositions of $f$, as element of $\mathbb{Q}[x]$ and over its decomposition field. Assuming roots of the irreducible factors will certainly be beneficial.
And if you concentrate upon the difference between your post (#1) and @andrewkirk's answer (post #2), you will find a crucial hint which might help to leave your deadlock. It's easy to overlook, but it is there!

12. Nov 2, 2016

### andrewkirk

There is a problem in this statement. Check this against the definition of $c$ that is given in the problem statement. If the mis-step doesn't quickly become apparent, try to lay out in full detail your reasoning behind this step.

13. Nov 4, 2016

### PsychonautQQ

Over Q f(x) = x^6-x^5+.....-x+1 and over Q(c) f(x) = (x-c)(x-c^2)*.....*(x-c^6) I believe.

To go back and look at andrekirk's post, I don't believe that phi(c) can ever be in the rationals, phi(c) must take c from one of the roots of f(x) to another one, and none of the roots of f(x) are in Q.

As for what else we know about phi(c), uhm, I think that for each c^i 0>i>7 there will be a phi such that phi(c) = c^i. Yes, we also know that phi(q)=q for all q in Q.

Have I said anything new/useful? haha :P thanks ya'll

14. Nov 4, 2016

### Staff: Mentor

I think the questions in this special case are:
• How does $\mathbb{Q}(c)$ be different from, say $\mathbb{Q}(d)$ with a seventh root of unity $d$, which is also in $\mathbb{Q}(c)$ but not the other way around?
• What to do with the conjugates $c \mapsto -c$ as they are both roots?

15. Nov 13, 2016

### PsychonautQQ

Right, my bad, c^8 does not equal c^2, I think it equals -c^4 though? So Phi_4 * Phi_2(c) = -c^4?

is d a primitive seventh root? If so, then it's minimal polynomial is 1+x+....+x^6 and I believe it's Galois group would be the cyclic group of order 7? Also, Q(d) would make a 7-gon on the complex plane rather than 14-gon. It seems that the minimal polynomial of Q(c) has some sign flipping going on. Oh shoot, could the galois group be D_7? This is an educated guess :D. If so, I still don't understand why completely.

16. Nov 13, 2016

### Staff: Mentor

This would be my guess, too. However, the question immediately emerges (with primitive roots $c^{14}=1\, , \,d^7=1$):

If $Aut_\mathbb{Q}(\mathbb{Q}(c)) \cong D_7$ and $Aut_\mathbb{Q}(\mathbb{Q}(d)) \cong \mathbb{Z}_7$, what is $e$ in $Aut_\mathbb{Q}(\mathbb{Q}(e)) \cong \mathbb{Z}_{14}\,$? Or how do the minimal polynomials of $c$ and $e$ over $\mathbb{Q}$ differ? And the same question will remain, if one exchanges $D_7$ and $\mathbb{Z}_{14}$. Since $\mathbb{Z}_{14}\cong\mathbb{Z}_7\times\mathbb{Z}_2$ and $D_7 \cong \mathbb{Z}_7 \rtimes \mathbb{Z}_2$, how does the semi-directness reflect in terms of minimal polynomials?

I think here lies the key to understanding. But I admit that I never really have thought about examples of this kind.

17. Nov 13, 2016

### andrewkirk

It's not clear to me where these guesses are coming from. It helps to be methodical in a problem like this. Let's review what we know about $c$. It is a primitive 14th root of unity, so that $c^{14}=1$, and $c^k\neq 1$ for $k\in\{1,2,...,13\}$. Can you use that to work out what the value of $c^7$ is?

18. Nov 14, 2016

### Staff: Mentor

Yep. Silly mistake I made. I should have calculated the degree of the extension first. (To look at the decomposition of $x^{14}-1$ or the actual formula helps not to get blinded by the fourteen points on the unit circle.)

19. Nov 16, 2016

### PsychonautQQ

On second thought, it says here in my notes that Gal[Q(c):Q] where c is a primitive nth root of unity is isomorphic to Z*_n (the multiplicative group of elements relatively prime to n). Are you sure it's D_7?

20. Nov 16, 2016

### Staff: Mentor

I'm sure it's not! Haven't you read the previous post #18? I already admitted that I have been distracted by the symmetries of the 14 roots on the unit circle without doing any calculations (and unfortunately any thoughts either). Your book is right. And if you write down the elements of $\mathbb{Z}_{14}^*$ you immediately get the automorphisms and their group table, which is one way to distinguish between the two groups with as many elements.