1. ## quadratic again.. may be tricky

hi, i think this one is a bit trickier.

for an odd prime number "p", let $\displaystyle F_p$ 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 $\displaystyle F^n_p$ and why?

2. Originally Posted by joanne_q
So how many quadratic forms are there on the vector space $\displaystyle F^n_p$ and why?

3. 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

4. here is the question as on the paper:

for an odd prime number "p", let $\displaystyle F_p$ 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 $\displaystyle F^n_p$ and why?

5. 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:
ax² + 2hxy + by² = 1.

i hope that makes it a bit more clearer..

any ideas on how to solve the original question then?

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