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Source. The Frankl–Rosenberg theorem quoted on published p. 3; the cited original is A finite set intersection theorem, European Journal of Combinatorics 2 (1981), 127–129. The following finite elementary proof is supplied by the compilation. It covers exactly the positive-uniformity case needed for lemma_3_1.

Statement. Let $a>0$, $m\ge2$ be integers, and let $\mathcal F$ be a finite family of $a$-element subsets of a finite set. Suppose $|F\cap G|\equiv b\pmod m$ for distinct members and $a\not\equiv b\pmod m$. Then their incidence vectors are linearly independent over $\mathbb Q$.

Proof. Suppose a rational dependence exists. Clear denominators and divide out the greatest common divisor to obtain integers $\lambda_F$, not all zero, with greatest common divisor one and $\sum_F\lambda_F\mathbf1_F=0$. Taking inner product with the all-ones vector gives $a\sum_F\lambda_F=0$, hence $\sum_F\lambda_F=0$. Taking inner product with $\mathbf1_G$ and reducing modulo $m$ gives

$$0\equiv a\lambda_G+b\sum_{F\ne G}\lambda_F =(a-b)\lambda_G\pmod m.$$

Because $m\nmid a-b$, some prime $p$ dividing $m$ satisfies $v_p(m)>v_p(a-b)$. The divisibility $m\mid(a-b)\lambda_G$ then implies $p\mid\lambda_G$ for every $G$. This contradicts the primitive choice of the coefficients. Therefore no dependence exists.

Range. Positivity of $a$ is explicit because the all-ones argument uses it. In Lemma 3.1, $a=2n-4>0$ and $m=n-6\ge5$. The original external paper is not independently reviewed, and no statement about zero-uniform families is needed.

Bears on. #174.