 Low-degree evaluation rank on a uniform layer

Let $0\le s\le k\le n$ and form the integer matrix with rows indexed by $S\subseteq [n]$, $|S|\le s$, columns indexed by $k$-subsets $T$, and entry $M_{S,T}=1_{S\subseteq T}$. Its matrix rank over any finite field ${\mathbb{F}}_p$ is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Over the rational numbers, the row of a $j$-subset $S$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of the rows of all $s$-subsets containing $S$. Thus the rows of degree exactly $s$ span, giving the bound over the rational numbers. Every larger square minor therefore has integer determinant zero, and remains zero modulo $p$. The matrix rank cannot increase under this reduction. This argument avoids dividing by an integer that might vanish in the finite field, and gives the sharp uniform-layer step in the Frankl-Wilson theorem.