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Math Help - quadratic again.. may be tricky

  1. #1
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    quadratic again.. may be tricky

    hi, i think this one is a bit trickier.

    for an odd prime number "p", let F_p be the field with "p" elements. i.e. the integers {0,....p-1} with addition and multiplication defined modulo "p".
    So how many quadratic forms are there on the vector space F^n_p and why?

    any clues here please?
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  2. #2
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    Quote Originally Posted by joanne_q View Post
    So how many quadratic forms are there on the vector space F^n_p and why?
    What do you mean by "how many quadradic forms..."? What do you mean by a "quadradic form"?
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  3. #3
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    thats the problem i have myself.
    i dont quite understand the question so im trying to see if any of you guys have an idea of what i am meant to be doing here
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  4. #4
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    here is the question as on the paper:

    for an odd prime number "p", let F_p be the field with "p" elements. i.e. the integers {0,....p-1} with addition and multiplication defined modulo "p", how many quadratic forms are there on the vector space F^n_p and why?
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  5. #5
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    a quadratic form by the way would be the quadratic equation corresponding to the symmetric matrix :

    so for example, a 2x2 matrix would produce the follow quadratic form:
    ax + 2hxy + by = 1.

    i hope that makes it a bit more clearer..

    any ideas on how to solve the original question then?
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  6. #6
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    Searching the internet reveals that a quadradic form is a symettric quadradic polynomial, for example, x^2+y^2+4xy. Thus we you are talking about a vector space \mathbb{F}_p^n it seems to me are you trying to count all quadric forms of x_1,...,x_n variables. So it has the form: a_1x_1^2+...+a_nx_n^2+b_1x_1x_2+.... Now just do a counting argument. Count all such coefficients. The a_i for each 1\leq i \leq n can be one of 0,1,...,p-1, thus in total we have p^n combinations. Now the b_i are all combinations of x_i,x_j with i\not = j there are C_{n,2}=n(n-1)/2 such possibilities. For each coefficient we have p possibilities that we have p^{n(n-1)/2}. Altogether we have p^{n(n-1)/2+n)}=p^{n(n+1)/2}. But in counting this we have included that zero polynomial which is not a quadradic form so we just subtract 1 from that.
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