Let $m$ and $n$ be positive integers. Suppose that we are given $m$ homogeneous polynomial equations (with complex coefficients) such that each of $n$ variables occurs at least once in all of the polynomials. No other variables occur in any of these polynomials – and they are not necessarily all of the same degree. If $m$ is less than $n$, will these $m$ equations always have at least one common root in the field of complex numbers that is non-trivial? By "non-trivial", I mean a root in which not all the variables are equal to zero.

  • $\begingroup$ Unless I'm misunderstanding you (possible), you don't even need your condition about each of the n variables occuring at least once. The statement is that in n dimensional complex projective space, the intersection of m < n hypersurfaces is always nonempty. That is, yes. (Moral: a degree d hypersurface is linearly equivalent to d copies of a hyperplane, and the intersection of m hyperplanes in projective n space is always nonempty. For a rigorous proof apply dimension theory to the affine cover, the intersection is nonempty because it contains the origin. I can expand on this if you want.) $\endgroup$ Dec 31 '15 at 21:54
  • $\begingroup$ I added this condition to rule out systems like 2{x(1)^2}, 3{x(1)^2} and also x(2)^2+x(3)^2+x(4)^2 where m=3 and n=4-because the first two polynomials have no non-trivial common root.. But I see now that this need not cause any trouble. You are right. I know very little about projective geometry and was just trying to puzzle out the algebra. Many thanks. $\endgroup$ Jan 1 '16 at 21:05
  • $\begingroup$ I highly recommend Shafarevich Basic Algebraic Geometry book I if you want to learn the basics of projective algebraic geometry. The proposition I mentioned will be covered in the first chapter, I think. $\endgroup$ Jan 2 '16 at 16:34

Theorem: In n dimensional complex projective space over an algebraically closed field (such as the complex numbers), the intersection of m < n hypersurfaces is always nonempty.

Proof: Suppose that hypersurfaces are given by equation $f_1, \ldots f_m$. Consider these as equations cutting out points in the affine space whose lines will make up the projective space ($A^n \setminus 0 \to P^n$). Here each $X_i = V(f_i)$ contains the origin, so the intersection of the $X_i$ is nonempty. However, by dimension theory, the minimum dimension of the components of $X = \cap X_i$ is $n - m \geq 1$. Thus, $X$ is at least one dimensional and nonempty, so it contains a point not passing through the origin. This means that the corresponding projective algebraic set is non-empty.

Lemma: (From dimension theory) (Generalizing the theorem of algebra about the finiteness of roots of a polynomial) If $X$ is an algebraic variety of dimension $k$ in $A^n$, and $Y$ is a hypersurface, then the irreducible components of $X \cap Y$ have dimension $\geq k - 1$.

Proof: Relatively hard commutative algebra. (Depending on your definition of dimension.) But morally, $Y$ is given by an equation $f$. Restricting this equation to $X$ as $f|X$, the zeros of $f|_X$ correspond to points of the intersection $X \cap Y$. Now, generally speaking, looking the zeros of a polynomial equation on a variety $X$ can't cut down the dimension by more than 1. (Krull's Principal Ideal Theorem.)

  • $\begingroup$ Wow! I never expected such a detailed response. This approach to the problem seems so much more powerful than the pure algebraic elimination theory-as presented by Van der Waerden-which I was trying to use. $\endgroup$ Jan 2 '16 at 21:46

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