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Finding a Galois group over F_7

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the Galois group of f(x) = x^7-x^6-2x+2 over ##F_7##.

    2. Relevant equations


    3. 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?
     
  2. jcsd
  3. Dec 4, 2016 #2

    fresh_42

    Staff: Mentor

    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?
     
  4. Dec 5, 2016 #3
    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 any one 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?
     
  5. Dec 5, 2016 #4

    fresh_42

    Staff: Mentor

    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.
     
  6. Dec 5, 2016 #5
    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.
     
  7. Dec 5, 2016 #6

    fresh_42

    Staff: Mentor

    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.
     
  8. Dec 6, 2016 #7
    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.
     
  9. Dec 6, 2016 #8

    fresh_42

    Staff: Mentor

    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##.
     
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