MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is an elementary fact that when the number of variables exceeds the number of linear equations then the system has a nontrivial solution.

I want to know whether there is such thing about homogeneous polynomials of degree $2$ or $3$, that is, for every natural number $n$ there exists $N \in \mathbb{N}$ s.t. for every $m>N$, the system of equations consisting $n$ homogeneous polynomials of degree $2$ in $\mathbb{C}[x_1,...,x_m]$ (or of degree $3$ in $\mathbb{R}[x_1,...,x_m]$ or $\mathbb{C}[x_1,...,x_m]$) has a non-trivial solution?

share|cite|improve this question
up vote 5 down vote accepted

1) It is not entirely true what you wrote about linear equations. If your equations are contradictory, then no matter how many additional variables you add, you will not find a solution. For instance $x+y=0$, $x+2y=1$, $2x+y=1$ has no solution regardless possible additional variables.

2) If you mean homogenous equations, that is, the constant terms are zero, then your claim is right. Well, it is obviously right since taking all variables to be zero gives you a solution. However, it is true that in this case if the number of variables is larger than the number of equations, then you have a non-trivial solution.

3) For homogenous systems of equations over $\mathbb C$ the same bound holds. Here is why: since it is homogenous, again, taking all variables to be zero gives a solution. Therefore the hypersurfaces determined by the equations have a common point and by the affine dimension theorem (which is essentially saying that one equation can cut down the dimension of the solution set by at most one) the solution set has positive dimension.

4) Over $\mathbb R$ you run into the problem of not having solutions no matter what. If your equations contain something like $x^2+y^2+...=0$ involving all the variables, then you are out of luck.

share|cite|improve this answer
For 1) yes, definitely, I meant a consistent set of equations. For the first line of my question I intended to give the background in a few words. – Ehsan M. Kermani Mar 17 '11 at 9:49
With respect to 4, I'll add that if your equations are real, homogeneous, of odd degree, and there are more variables than equations, then there is a non-trivial real solution. By adding linear equations, you can assume that there is one more variable then equation. Bezout's theorem says that generically, the number of complex solutions in projective space will be the product of the degrees, which is odd. Since the non-real solutions come in complex conjugate pairs, there must be a real solution. For non-generic equations, you can always approximate by generic systems. – Dustin Cartwright Mar 17 '11 at 18:30
Dustin Cartwright: Thanks for your complete answer. That is a careful point. – Ehsan M. Kermani Mar 19 '11 at 10:24

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.