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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2$\begingroup$ You shouldn't say "as easy as abc" to a number theorist ;-) $\endgroup$– Laurent Moret-BaillyCommented Jan 15, 2011 at 9:00
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$\begingroup$ @Laurent, do you mean that the number-theory tag should be removed as well? :-) $\endgroup$– Wadim ZudilinCommented Jan 15, 2011 at 9:30
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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
answered Jan 14, 2011 at 21:34
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1$\begingroup$ ... 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! $\endgroup$– SevaCommented Jan 15, 2011 at 6:57