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Algebra problems

  1. Apr 11, 2006 #1

    AKG

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    1. Find the group

    [tex]G = Gal(\mathbb{F}_{p^{n^2}}/\mathbb{F}_{p^n})[/tex]


    There is a 1-1 correspondence between elements of this group, and intermediate fields. An intermediate field is a finite field of order m, where pn | m | p. The m that satisfy this are pk for each k in {n, n+1, ..., n²-1, n²}. So the group in question has order n²-n+1. I'm very iffy on this material, so where should I go from here? Do I find

    [tex]Gal(\mathbb{F}_{p^{k+1}}/\mathbb{F}_{p^k})[/tex]

    for n < k < n², and then G will be somehow made up of these Galois groups?

    2. Is x²=2 solvable in 7-adic numbers?

    For all k, x² = 2 (mod 7k) has two solutions, since 7k is congruent to either 1 or 7 (mod 8), depending on whether k is prime. Moreover, for each k, x² = 2 (mod 7k) has two distinct solutions, y and -y (which are incongruent since all powers of 7 are odd), and if j < k, then y (mod 7j) and -y (mod 7j) are solutions to x² = 2 (mod 7j).

    So if y1 is a solution to x² = (mod 7), then there exists y2 such that y2 = y1 (mod 7) and y2² = 2 (mod 7²). Inductively, if yi is a solution to x² = (mod 7i), then there exists yi+1 such that yi+1 = yi (mod 7i) and yi+1² = 2 (mod 7i+1). Likewise, starting with -y1 I can get a different sequence. The two sequences should give rise to 7-adic numbers y and -y which satisfy y² = 2 as 7-adic numbers, right?
     
  2. jcsd
  3. Apr 11, 2006 #2

    matt grime

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    You know what the Frobenius morphism/map is?
     
  4. Apr 12, 2006 #3

    AKG

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    I vaguely remember it being [itex]a \mapsto a^n[/itex], but I can't remember what the domain or codomain are supposed to be, but I know it has something to do with finite fields.
     
  5. Apr 12, 2006 #4

    matt grime

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    i very strongly urge you to find out about the frobenius p'th power map since that tells you what the galois action is.

    of course this is unnecessary: every finite field is multiplicatively generated by one element (the mult group is cyclic) which gives the answer, and explains your observation about intermediate fields.

    or thirdly note that (except in characteristic 2) F_q^2 is gotten by adding in a the roots of x^(q-1) +1 to F_q (and so on) so that you can expressly write out that F_q^d is the splitting field over F_q of x^{q^d-1}+1


    these are essentially all the same observation, and are just extensions of the fermat's little theorem: that poly has no roots over F_q since every element in F_q to the power q-1 is 1 (note q=/=2^r here, for q=2 use x^2+x+1)
     
    Last edited: Apr 12, 2006
  6. Apr 25, 2006 #5

    AKG

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    Okay, the Galois group I'm looking for was denoted G in my first post, but let's denote it H now. It is a subgroup of the Galois group [itex]Gal(\mathbb{F}_{p^{n^2}}/\mathbb{F}_p)[/itex], and I will denote this group G. H is the subgroup of G that fixes elements of [itex]\mathbb{F}_{p^n}[/itex]. G is cyclic of order n² generated by [itex]\sigma[/itex], the Frobenius pth power map. The group [itex]Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)[/itex] is cyclic of order n, generated again by [itex]\sigma[/itex]. So to figure out H, we want to see which powers of [itex]\sigma[/itex] as a generator of G fix [itex]\mathbb{F}_{p^n}[/itex]. Clearly, it is the nth powers, and so

    [tex]H = \{\sigma ^n,\, \sigma ^{2n},\, \dots ,\, \sigma ^{n^2} = 1\} = \langle \sigma ^n \rangle \cong \mathbb{Z}/n\mathbb{Z}[/tex]

    Correct?
     
  7. Apr 26, 2006 #6

    matt grime

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    Yes (plus extra characters to get above the minimum)
     
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