Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

1 Answer 1

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:

share|cite|improve this answer
... 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.