Worked solution: The polynomial method resolves the cap set problem (Croot-Lev-Pach, Ellenberg-Gijswijt, 2016)

Step 4 of 7: Translating cap sets into the vanishing-polynomial language

In plain words

The cap-set condition is a special case of Proposition 2 in disguise. Setting $\alpha=\beta=\gamma=1$ (which indeed sums to $0$ in $\mathbb{F}_3$ ), the hypothesis of the lemma becomes precisely: no three points of $A$ sum to zero unless they are all equal -- exactly the definition of a cap set.

a+b \notin -A \text{ for all } a \ne b \in A

Detailed analysis

Ellenberg and Gijswijt (2017, Theorem 4 and Corollary 5) specialise Proposition 2 to $\alpha=\beta=\gamma=1$ , which indeed satisfies $\alpha+\beta+\gamma=0$ . A set $A$ with no non-trivial solution to $a_1+a_2+a_3=0$ in $A$ means: for $a_1,a_2 \in A$ , a solution $a_3 = -a_1-a_2$ exists in $A$ only if $a_1=a_2=a_3$ . Consequently, for any polynomial $P$ that vanishes everywhere outside $-A$ , the hypothesis $P(a+b)=0$ of Proposition 2 is automatically satisfied for $a \ne b \in A$ , because $a+b$ then lies outside $-A$ : this is exactly $a+b \notin -A \text{ for all } a \ne b \in A$ .