Finding a Galois group over F_7

[PsychonautQQ]

Homework Statement

Find the Galois group of f(x) = x^7-x^6-2x+2 over $F_7$.

Homework Equations

The Attempt at a Solution

1 is a root of f(x) so dividing f(x) / (x-1) we get the quotient x^6-2. Now all elements of $F_7$ satisfy a^6 = 1 since it's multiplicative group is of order 6, and thus no elements will be a root of x^6-2. Thus I will add an element r such that r^6 = 2 to $F_7$ and now a*(r^6) will be roots for all a in $F_7$ and the splitting field is Q(r).

As far as Gal(Q(r):Q) goes, any automorphism in this group will fix all of Q, and thus the only thing it can move is r and it can only go to another root of it's minimal polynomial (x^6-2), so I believe there is a unique automorphism that will for each element of $F_7$, where 1(r)=r, 2(r)=2r, 3(r)=3r, etc. Thus the structure of the Galois group will be isomorphic to the multiplicative group of units in $F_7$.

Is this correct?

 

[fresh_42]

If we define $c := \sqrt[6]{2}$ as the real root of $x^6-2 \in \mathbb{C}[x]$. What are the other roots? Shouldn't $r$ to be chosen a primitive sixth root? And what is the dimension of $\mathbb{F}_7(r)$ over $\mathbb{F}_7$? Do you have six linear independent roots?

 

[PsychonautQQ]

I was thinking that all the elements in the multiplicative group of $F_7$ are sixth roots of unity, but none of them are roots. Therefore if we adjoin one element r that is a root, then r multiplied by anyone of the six elements in the multiplicative group of F_7 will also be a root. The dimension of $F_7(r) over F_7$ will be six because the minimal polynomial of r is x^6-2.

However by my logic I don't know if all the roots (1r) (2r) ... (6r) will be linearly independent, the reason I am sceptical of this is because not all of them have order 6, for example 2^3 = 1.

if r is a primitive sixth root, and c is as you defined, then are you saying that the splitting field I am looking for is Q(c,r) and that my idea of having r be the sixth real root of 2 in $F_7$ won't provide me with all the roots for some reason?

 

[fresh_42]

It is correct what you've said, i.e. $f(x)=(x-1)(x-r)(x-2r)\cdot \ldots \cdot (x-6r)$ and $|G| = |Gal(\mathbb{F}_7(r);\mathbb{F}_7)| = [\mathbb{F}_7(r) : \mathbb{F}_7]=6$ and also that we simply need $r$ to be any solution of $x^6=2$. (Forget my remark on a primitive root and $c$.)

Now this means, that $\{1,r,r^2,\ldots,r^5 \}$ is a basis of $\mathbb{F}_7(r) \supseteq \mathbb{F}_7$ and $G$ is most likely the cyclic group $\mathbb{Z}_6$. Why should $\{r,2r,\ldots,6r\}$ be a basis? The elements are all $\mathbb{F}_7$-linear dependent: $ar = (ab^{-1})br = \lambda \cdot (br)$ with $\lambda = (ab^{-1}) \in \mathbb{F}_7$.

The task is to compute all automorphisms $\varphi : \mathbb{F}_7(r) \longrightarrow \mathbb{F}_7(r)$ with $\varphi (\mathbb{F}_7) = id_{\mathbb{F}_7}$. Or at least one, because we expect one to already generate all six.

Edit: Or is $|G|=2$ and $\{1,r\}$ already a basis? I don't see it yet.

 

[PsychonautQQ]

Hm, so the elements r,2r,3r,4r,5r,1 are not F_7 independent, good call. I do not believe that 1,r would be a basis, there should be 6 basis elements because the minimal polynomial of r is degree 6 so F(r):F is 6 (i´m on a dang keyboard can´t find the brackets n stuff sry). Perhaps you are correct with 1,r,r^2,r^3,r^4,r^5 are a basis. I'll do some calculations and get back to you, but that looks correct.

 

[fresh_42]

Hm, so the elements r,2r,3r,4r,5r,1 are not F_7 independent, good call. I do not believe that 1,r would be a basis, there should be 6 basis elements because the minimal polynomial of r is degree 6 so F(r):F is 6 (i´m on a **** keyboard can´t find the brackets n stuff sry). Perhaps you are correct with 1,r,r^2,r^3,r^4,r^5 are a basis. I'll do some calculations and get back to you, but that looks correct.

That's what I think, too. But I haven't seen a fast way to rule out that e.g. $r^2 = a+br$ for some $a,b \in \mathbb{F}_7$ and I'm not so far into field theory, that I automatically can say, why it is impossible. In any case, the elements $r, \ldots ,r^5$ are all in $\mathbb{F}_7(r)$, so they must have a representation in any basis. There is probably a good argument why they are already linear independent. I simply don't have it in mind. And maybe it's easier to show that $r \mapsto r^2$ or similar is an automorphism which we actually are looking for.

 

[PsychonautQQ]

The minimal polynomial of r in $F_7$ is an irreducible polynomial of degree 6 and therefore {1,r,r^2,r^3,r^4,r^5} is a basis for F(r) over F.

Isn't this a thing that goes without saying almost? Or perhaps this is a thing that goes without saying if the character of the base field is zero.

 

[fresh_42]

Isn't this a thing that goes without saying almost? Or perhaps this is a thing that goes without saying if the character of the base field is zero.

Well, it isn't "obvious". And I tend to take nothing for granted which I can't remember the theorem of or at least the principle behind it, i.e. an outline of a proof. Here it takes some considerations on the invariants (trace, determinant, degree) of $f$, but you're right.
The characteristic isn't the issue here, and ours is $7$, not $0$.