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A question which, I suppose, is as easy as abc for an expert. For a given finite field $F$ of odd characteristic (if needed, the characteristic is $3$ and the size of $F$ is large), do there exist two quadratic forms on $F^5$ without non-trivial common zeroes? More generally, what should be the relation between $n$ and $k$ in order for $k$ quadratic forms on $F^n$ without non-trivial common zeroes to exist?

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You shouldn't say "as easy as abc" to a number theorist ;-) –  Laurent Moret-Bailly Jan 15 '11 at 9:00
    
@Laurent, do you mean that the number-theory tag should be removed as well? :-) –  Wadim Zudilin Jan 15 '11 at 9:30
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up vote 7 down vote accepted

Chevalley-Warning: If you have a system of forms of degrees $d_1,...,d_k$ in $n$ variables, they will have a common non-trivial zero if $n > \sum d_i$. For $d_i=2$, the condition is $n > 2k$.

There is a full proof in the wikipedia page: http://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem

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... and if $n=2k$, then they can have no common zeroes: fix a quadratic non-residue $d$ and consider the forms $x_1^2-dx_2^2,\ x_3^2-dx_4^2,\ ...,\ x_{n-1}^2-dx_n^2$. Thanks! –  Seva Jan 15 '11 at 6:57
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