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

Ellenberg and Gijswijt (2017, Proposition 2, generalising Lemma 1 of Croot-Lev-Pach 2016) fix $\mathbb{F}_q$ elements $\alpha,\beta,\gamma$ with $\alpha+\beta+\gamma=0$. If $P \in S_n^{\le d}$ satisfies $P(\alpha a+\beta b)=0$ for every $a \ne b \in A$, then $|\{a \in A : P(-\gamma a) \ne 0\}| \le 2 m_{d/2}$. The proof expands $P(\alpha x+\beta y)$: every monomial of total degree at most $d$ has one of its two halves ($x$-part or $y$-part) of degree at most $d/2$, so $P(\alpha x+\beta y) = \sum_{m \in M_n^{d/2}} m(x)F_m(y) + \sum_{m \in M_n^{d/2}} m(y)G_m(x)$ for some polynomials $F_m, G_m$ indexed by monomials $m$ of degree at most $d/2$. Evaluating at $x=a, y=b \in A$ writes the matrix $B$ with entries $B_{ab} := P(\alpha a+\beta b)$ as a sum of $2m_{d/2}$ rank-one matrices, so $\operatorname{rank}(B) \le 2m_{d/2}$.