• Nov 28th 2007, 10:58 AM
joanne_q
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?

• Nov 28th 2007, 12:35 PM
ThePerfectHacker
Quote:

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

• Nov 28th 2007, 01:03 PM
joanne_q
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 :)
• Nov 28th 2007, 01:04 PM
joanne_q
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?
• Nov 28th 2007, 01:06 PM
joanne_q
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?
• Nov 28th 2007, 07:28 PM
ThePerfectHacker
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.