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Proving isomorphism

  1. Apr 8, 2005 #1
    how do i prove that Aut n(K) is isomorphic to the symmetric group Sq^n.

    K is a finite field of q elements. Aut n(K) is the group of polynomial automorphisms over K.

    n i just the number of variables/indeterminants.

    so i guess i have to somehow show that Aut n(K) are the permutations of the elements of K. so i guess i have to do something with the coeficients of a polynomial...
     
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  3. Apr 8, 2005 #2

    Hurkyl

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    Any function from a finite field to itself is (equivalent to) a polynomial...
     
  4. Apr 9, 2005 #3
    i'm sorry? how does this help me?


    i rewrite the question, becuase it was stated a bit sloppy:

    i'm supposed to prove that Aut n(K) is isomorphic to the symmetric group Sq^n.

    K is a finite field of q elements. Aut n(K) is the group of polynomial mappings over K.

    n i just the number of variables/indeterminants.

    Aut n(K) is the so called cremona group. the set of polynomial mappings over K.

    here is a definition of a polynomial map:
    http://mathworld.wolfram.com/PolynomialMap.html

    every mapping from K^n -> k^n is polynomial(i have already proven this).


    i guess i should have mentioned that i work with the automorphisms as "mappings". this
    is because i can subsitute all X^q by X.
     
  5. Apr 9, 2005 #4

    Hurkyl

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    It means you can just say "automorphisms" instead of "polynomial automorphisms".


    The more I look at the problem statement, the more it confuses me, and I think I've identified the problem:

    Automorphisms of what? I don't recognize the terminology n(K).

    If it wasn't for the fact you haven't seemed to be able to connect it to what you're trying to prove, I would have thought you're trying to describe (in a roundabout way) a permutation of the elements of k^n.
     
    Last edited: Apr 9, 2005
  6. Apr 9, 2005 #5
    just forget about the automorphisms. use the term map instead.

    n i just the number of variables/indeterminants.

    aut n(K) the group of polynomial mappings over K.

    here is a definition of a polynomial map:
    http://mathworld.wolfram.com/PolynomialMap.html

    any mapping from K^n -> k^n is polynomial(i have already proven this).
     
  7. Apr 9, 2005 #6

    Hurkyl

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    I repeat -- you are doing a very poor job describing what you mean by "Aut n(k)". If I don't know what that means, I can't help you. More importantly, if you don't know what that means, you can't solve the problem.

    In this latest post, you told me that "Aut n(K)" means the group of all functions from k^n --> k^n. That is clearly not isomorphic to a symmetric group.
     
  8. Apr 9, 2005 #7
    i don't get it. what is the problem? aut n(k) are the polynomial mappings over k. i've defined what a polynomial mapping is(the link):
    http://mathworld.wolfram.com/PolynomialMap.html

    i'm supposed to show that this group is isomorphic to Sn^q. n is the number of variables/indeterminants. q is the number of elements in k.


    what is it you don't get? any of the terms? because aut n(K) IS isomorphic to the symmetric group Sq^n. i just have to prove it...
     
  9. Apr 9, 2005 #8

    Hurkyl

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    The group of polynomial mappings from K^n --> K^n is not isomorphic to a symmetric group.

    As we've noted, the terms "polynomial map from K^n --> K^n" and "function from K^n --> K^n" are synonymous, because K is finite.

    And, we know that the number of functions from a set P to a set Q is [itex]|Q|^{|P|}[/itex]. In this case, that would be [itex]q^{nq^n}[/itex].

    However, the symmetric group on [itex]q^n[/itex] elements consists of [itex](q^n)![/itex] permutations.

    The groups have different sizes, so they're clearly not equal.


    An example of a polynomial map from K^n --> K^n is the map that sends every point of K^n to the zero vector. Is that really something that's supposed to be in your Aut n(K)? Aut n(K), then, wouldn't even be a group!
     
    Last edited: Apr 9, 2005
  10. Apr 9, 2005 #9
    it has to be isomorphic, otherwise my professor is just f*cking with me. he gave us this exercise last week and i'm still struggling with it.

    we were supposed to show that aut n(K) is isomorphic to Sq^n and then that
    |aut n(k)| = (q^n)!.

    i guess i have to ask him in class on monday. this is really confusing...

    i think it has something to do with the fact that one can subsitute all apperances of X^q with X.

    what you are saying makes sense, because i have already proven that
    end n(K) consists of q^nq^q elements. where end n(K) is the set of polynomial mappings k^n -> k^n.
     
  11. Apr 9, 2005 #10

    Hurkyl

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    I think I'm convinced that by n(k), you mean the set of all n-tuples with elements in k... whether you know you mean that is a different question...


    Anyways, do you know the difference between an endomorphism and an automorphism?
     
  12. Apr 10, 2005 #11
    auto - iso to itself
    endo - homo to itself.
    the group end n(K) is obviously bigger than aut n(K).

    aut n(K) is the cremona group. the group of polynomial mappings over a finite field K. K has q elements. in this group one can subsitute X^Q by X. the link i provided shows what a polynomial map is.
     
  13. Apr 10, 2005 #12

    Hurkyl

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    What is n(K)? Is it the set of all n-tuples with elements in K?


    Now, as you've stated, an automorphism of n(K) is an isomorphism. And as I've stated, not just any old map K^n -> K^n is isomorphic... aut n(K) cannot consist of all maps K^n -> K^n.
     
    Last edited: Apr 10, 2005
  14. Apr 10, 2005 #13

    aut n(K) is the group of invertible polynomial mappings from K^n -> K^n.
     
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