# The Combinatorial Nullstellensatz

I’ve been taking a TCC course this term on Additive Combinatorics, delivered via video link from Bristol by Julia Wolf. At some point once the dust of this term has settled, I might write some things about the ideas of the course I’ve found most interesting, in particular the tools of discrete Fourier analysis to get a hold on some useful combinatorial properties of subsets of $\mathbb{Z}/n\mathbb{Z}$ for example.

For this post, I want to talk instead about a topic that was merely mentioned in passing, the Combinatorial Nullstellensatz. The majority of this post is based on Alon’s original paper, which can be found here, and Chapter 9 of Tao and Vu’s book Additive Combinatorics. My aim is to motivate the theorem, give a proof, introduce one useful application from additive combinatorics, and solve Q6 from IMO 2007 as a direct corollary.

What does Nullstellensatz mean? Roughly speaking, it seems to mean ‘a theorem specifying the zeros’. We will be specifying the zeros of a polynomial. We are comfortable with how the zeros of a complex-valued polynomial of one variable behave. The number of zeros is given precisely by the degree of the polynomial (allowing appropriately for multiplicity). It is generally less clear how we might treat the zeros of a polynomial of many variables. The zero set is likely to be some surface, perhaps of dimension one less than the number of variables. In particular, it no longer really makes sense to talk about whether this set is finite or not. The Combinatorial Nullstellensatz gives us some control over the structure of this set of zeros.

The idea behind the generalisation is to view the Fundamental Theorem of Algebra as a statement not about existence of roots, but rather about (combinatorial) existence of non-roots. That is, given a polynomial P(x) of degree n, for any choice of (n+1) complex numbers, at least one of them is not a root of P. This may look like a very weak statement in this context, where we only expect finitely many roots anyway, but in a multivariate setting it is much more intuitively powerful.

Recall that the degree of a monomial is given by the sum of the exponents of the variables present. So the degree of $4x^2 y^3 z$ is 6. The degree of a polynomial is then given by the largest degree of a monomial in that polynomial. A polynomial $P(x_1,\ldots,x_n)$ over a field F with degree d might have lots of monomial terms of degree d. Suppose one of these monomials is $x_1^{d_1}\ldots x_n^{d_n}$, where $\sum d_i=d$. Then one version of the Combinatorial Nullstellensatz asserts that whenever you take subsets of the base field $S_i\subset F$ with $|S_i|\ge d_i+1$, then there is a point with $x_i\in S_i$ such that $P(x_1,\ldots,x_n)=0$.

In other words, you can’t have a box (ie product of sets) of dimension $d_1+1 \times d_2+1 \times\ldots\times d_n+1$ on which the polynomial is zero.

Unsurprisingly, the proof proceeds by induction on the number of variables. Alon’s result proceeds via a more general theorem giving information about the possibility of writing multinomial polynomials as linear combinations of polynomials in one variable.

We would like to start this induction by fixing the x_n co-ordinate, then viewing P as a polynomial in $x_1,\ldots,x_{n-1}$ only. One problem with this approach is that the largest degree monomials in P are not necessarily still the largest degree monomials in P with x_n fixed. So we need to apply a division algorithm argument.

I’m going to miss some steps so as to keep this of suitable blog post length. The key idea is to apply the division algorithm to P with respect to the simplest polynomial that is zero on all of $S_n$, which we define as:

$g(x_n)=\prod_{s_n\in S_n}(x_n-s_n).$

We can decompose as

$P(x_1,\ldots,x_n)=q_n(x_1,\ldots,x_n)g(x_n)+\sum_{j=0}^{|S_n|-1}r_{n,j}(x_1,\ldots,x_{n-1})x_n^j.$

So now we ask where the term $x_1^{d_1}\ldots x_n^{d_n}$ is coming from, bearing in mind that $d_n<|S_n|$. The lower order terms in g cannot contribute to this, as  they cannot be of maximal degree. Also, the first term in $q_n(\mathbf{x})g(x_n)$ cannot contribute as the exponent of $x_n$ is too large. So the term in question must be coming from $r_{n,d_n}(x_1,\ldots,x_{n-1})x_n^{d_n}$. So now we can apply the induction hypothesis to the polynomial $r_{n,d_n}$ to find $x_1\in S_1,\ldots, x_{n-1}\in S_{n-1}$ such that $r_{n,d_n}(x_1,\ldots,x_{n-1}$ is non-zero. With these values, we can view the remainder as a polynomial in $x_n$ of degree $|S_n|>d_n$, and so there is an $x_n\in S_n$ such that

$\sum_{j=0}^{|S_n|}r_{n,j}(x_1,\ldots,x_{n-1})x_n^j)\neq 0.$

This concludes the proof by induction.

I want to discuss two relatively simple applications. The first is the Cauchy-Davenport Theorem, which one might view as the first non-trivial theorem in additive combinatorics, giving a bound on the size of a sumset.

Theorem (Cauchy-Davenport): Given A, B non-empty subsets of $Z_p$ for p a prime, then

$|A+B|\geq \min\{p,|A|+|B|-1\}.$

( $A+B:=\{c: c=a+b,a\in A,b\in B\}$ )

Note that the result isn’t especially surprising. Providing some sort of ordering to the elements of A and B might be a sensible way to proceed. Certainly if they were sets in $\mathbb{Z}$, this would give a proof immediately.

Proof: Only the case |A|+|B| <= p is interesting. Following Alon’s argument, suppose that |A+B| <= |A|+|B|-2, and let C=A+B. Set $f(x,y)=\prod_{c\in C}(x+y-c)$, so f(a,b)=0 for all $a\in A,b\in B$.

Then the coefficient of $x^{|A|-1}y^{|B|-1}$ in f is $\binom{|A|+|B|-2}{|A|-1}$ as we have to choose which of the terms in the product supply an x and which supply a y. This is non-zero (in $Z_p$ recall) since the upper integer is less than p. The Combinatorial Nullstellensatz then gives a contradiction.

My second example is from the IMO in Vietnam which I attended. I spent a lot of time thinking about this problem, but made no progress.

IMO 2007 Question 6: Let n be a positive integer. Consider

$S=\{(x,y,z) | x,y,z\in \{0,1,\ldots,n\}, x+y+z>0\}$

as a set of $(n+1)^3-1$ points in 3D space. Determine the smallest number of planes, the union of which contains S but does not include (0,0,0).

Answer: 3n. Consider the planes x+y+z = k for k varying between 1 and 3n. The aim is to prove that you cannot do it with fewer.

To prove this, suppose we can do with fewer planes, say k. We write the equation of a plane as

$ax+by+cz-d=0.$

Note that the d’s are non-zero as (0,0,0) must not be a solution. Then take the product of all these degree one polynomials together and subtract a multiple of

$\prod_{i=1}^n (x-i)(y-i)(z-i),$

with the multiple chosen so the resulting polynomial has a root at (0,0,0). (This constant must be non-zero to cancel the non-zero product of the d’s.) This resulting polynomial is degree 3n by construction, and $x^ny^nz^n$ has a non-zero coefficient, but it is zero on the box $[0,n]^3$, which contradicts Combinatorial Nullstellensatz.