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I was wondering if anyone could help me understand a statement in the proof of Theorem 4 of this paper: https://www.jstor.org/stable/pdf/24906443.pdf. I'm aiming to prove the theorem for $q=3$ and $\alpha= \beta = \gamma = 1$.

$F_3$ is a field of size $3$. $A$ is a subset of $F_3^n$ such that no three elements (with at least 2 distinct) sum to $0$ in $A$.

$S_n^d$ is defined to be the span of monomials in $x_1, ... , x_n$ with degree at most $d$, and $m_d$ is the dimension of $S_n^d$. Here $d$ is a real number between $0$ and $2n$.

$V$ is the space of polynomials in $S_n^d$ that vanish on the complement of $-A$. I say the support of a polynomial $P$ to mean the elements $x$ such that $P(x) \neq 0$.

Then if $P \in V$ has support $\Sigma$ of maximal size and $|\Sigma| < dim(V)$, then there exists another polynomial $Q \in V$ that vanishes on $\Sigma$.

However I am struggling to see why this is the case or to come up with some polynomial to demonstrate this. Any help would be appreciated!

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The condition that a polynomial vanish at a particular point is a single linear equation in the coefficients. You then have $|\Sigma|$ linear equations in $m_d$ variables (the coefficients), which automatically has a solution if $|\Sigma|<m_d$.

Alternatively, more basis freely: The map that sends a polynomial in $V$ to its list of values on $\Sigma$ is a linear map. If the dimension on $V$ is larger than $|\Sigma|$ this linear map automatically has non-trivial kernel.

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