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The Frankl-Wilson theorem is a striking example of the polynomial method in combinatorics: an intersection restriction creates a family of linearly independent low-degree polynomials, whose dimension bounds the set family. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}$ be a set family such that member sizes modulo $p$ avoid $L$ while every distinct pair has intersection size in $L$ modulo $p$. Its general modular form givesIf all members have the same size $k$ and $s\le k$, the uniform refinement gives $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

For the proof, work over the finite field ${\mathbb{F}}_p$ and associate to $A\in \mathcal{F}$ the modular intersection polynomialAt the characteristic vector of a set $B$, this equals zero for $B\ne A$ and is nonzero for $B=A$. A linear relation among the evaluation functions therefore has every coefficient zero. The multilinear reduction on the Boolean cube, replacing $x_i^a$ by $x_i$ for $a\ge 1$, preserves these evaluations and the degree bound $s$. There are ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ possible square-free monomials, proving the general form.

For uniform members, the low-degree evaluation rank on a uniform layer supplies the sharper dimension bound. To verify this without division modulo $p$, form the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns indexed by $k$-subsets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, a row with $|S|=j$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of all size-$s$ rows containing $S$. Its matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is consequently zero and remains zero modulo $p$, giving the same bound over ${\mathbb{F}}_p$. The nonzero diagonal evaluation matrix of the $P_A$ then forces $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$.

The theorem, proved by Frankl and Wilson in 1981, extends the theme of the Ray-Chaudhuri–Wilson theorem from ordinary intersection sizes to residues. Modular restrictions can be weaker than exact intersection restrictions yet still force unexpectedly small families. Here are three consequences from one explicit modular intersection graph.

Take $N=4p-1$, $k=2p-1$, and let the vertices be all $k$-subsets of $[N]$, with an edge exactly when the intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so applying the Frankl-Wilson theorem with $L={\mathbb{F}}_p\setminus \{p-1\}$ givesA clique has one exact intersection size. Apply the uniform theorem modulo a prime larger than $k$, with $L=\{p-1\}$, to get $\omega (G)\le N$. Since there are $v=\left(\genfrac{}{}{0ex}{}{N}{k}\right)$ vertices, this gives the explicit asymmetric Ramsey number bound $R(N+1,B+1)>v$. It illustrates how algebraic constructions produce graphs with simultaneously restricted cliques and independent sets.

For a geometric application, map each vertex to $x_A=1_A/\sqrt{2p}$ in ${\mathbb{R}}^N$. Then $|x_A-x_B{|}^2=(k-|A\cap B|)/p$, so edges have unit Euclidean distance. Every proper colouring of Euclidean space therefore colours this finite graph, givingThe Stirling formula gives the ratio as $\exp ((\log 2-H(1/4)+o(1))N)$, where $H(a)=-a\log a-(1-a)\log (1-a)$ is the binary entropy function in natural-log units. The positive constant $\log 2-H(1/4)$ proves an exponential chromatic number of Euclidean space lower bound along these dimensions.

Finally, a quadratic embedding converts this same forbidden intersection into a forbidden diameter pair. Set $u_A=21_A-1$ and $Q_A=u_A{u}_A^T$. The quadratic sign-vector diameter formula follows from the Frobenius inner product:For distinct members the smallest possible absolute inner product is one, attained exactly when the intersection is $p-1$. Thus maximum-distance pairs are exactly edges of $G$. The matrices have common diagonal; using their off-diagonal coordinates, multiplied by $\sqrt{2}$, realizes them isometrically in $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ dimensions. They are distinct because equal outer products require $u_A=\pm u_B$, and complements have size $2p$, not $k$.

Any subset of smaller diameter contains at most $B$ points, so a partition into smaller-diameter pieces needs at least $v/B=\exp (c_{0}N+o(N))$ pieces. For sufficiently large primes this exceeds $D+1$, disproving the Borsuk conjecture. This is the mechanism behind the Kahn-Kalai counterexample to the Borsuk conjecture: the number of required pieces grows exponentially in the original set-system parameter while dimension grows quadratically. Their 1993 paper establishes the resulting high-dimensional counterexample.

The uniform modular Frankl-Wilson theorem states the following. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a rank-$k$ family on $[N]$, with $0\le s\le \min \{k,N-k\}$. Assume $k\mathrm{mod}p\notin L$ and $|A\cap B|\mathrm{mod}p\in L$ for every distinct $A,B\in \mathcal{F}$. ThenThe restrictions concern distinct pairs; the diagonal size residue is explicitly excluded from $L$.

For each member $A$, over ${\mathbb{F}}_p$ form the intersection polynomialAt the characteristic vector $1_B$, its value vanishes when $B\ne A$ and is nonzero when $B=A$. Thus these functions are linearly independent on the whole rank-$k$ layer: evaluate any linear relation at each family member in turn. Replacing every positive power of $z_i$ by $z_i$ preserves all Boolean evaluations, so the functions are restrictions of multilinear polynomials of degree at most $s$.

We need the sharp low-degree evaluation rank on a uniform layer, not the larger whole-cube dimension ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Let $M$ be the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns by rank-$k$ sets $B$, and entries $1_{S\subseteq B}$. Over the rationals, if $|S|=j\le s$, thenThis coefficient counts the possible $T$ inside each containing column set and is positive because $s\le k$. Consequently the degree-$s$ rows span all rows over $\mathbb{Q}$, giving rank at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. Every larger square minor is therefore the integer zero. Reducing those minors modulo $p$ shows the rank over ${\mathbb{F}}_p$ is also at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. This transfer avoids dividing by a binomial coefficient that might vanish modulo $p$. The independent functions $f_A$ lie in this row space, so their number is at most its rank. This proves the theorem in full.

For the requested forbidden-intersection application, split $\mathcal{A}$ into sets containing coordinate one and sets avoiding it. In the first part, delete one. Distinct original sets have intersection size between one and $2p-1$, excluding $p$. Their reduced intersections therefore lie between zero and $2p-2$, excluding $p-1$, so their residues belong to $L=\{0,\dots ,p-2\}$. The reduced set size is $2p-1$, whose residue is $p-1\notin L$. Frankl-Wilson on the ground set of size $4p-1$ bounds this part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the avoiding part, first complement every set. The complements contain one, retain size $2p$, and satisfyDelete one and apply the same argument. The complement splitting for forbidden midpoint intersections yields the stronger boundThe split is essential: a family may contain complementary pairs with intersection zero, preventing direct use of $L=\{1,\dots ,p-1\}$ on the original layer.

The Borsuk conjecture asserted that a bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. Here is the Kahn-Kalai counterexample to the Borsuk conjecture, implemented by balanced cut vectors. For each $X\in [4p{]}^{(2p)}$, define $v_X\in {\mathbb{R}}^{\left(\genfrac{}{}{0ex}{}{4p}{2}\right)}$ whose coordinate indexed by an unordered pair $\{i,j\}$ is one when exactly one endpoint lies in $X$, and zero otherwise. Complementary sets give the same cut. Choose the representative with $1\in X$, obtaining exactlydistinct points: a cut determines its two parts, and fixing the side containing one removes its only ambiguity. Every cut has $(2p{)}^2=4p^2$ occupied coordinates, so all points lie in the affine hyperplane with coordinate sum $4p^2$. Its dimension isWe can identify this hyperplane isometrically with ${\mathbb{R}}^d$; no claim of full affine span is needed.

For two side sets with $t=|X\cap Y|$, their symmetric difference has $4p-2t$ points. The two cuts disagree exactly on pairs with one endpoint in that symmetric difference, giving the balanced cut-vector distance formulaThe diameter is $2p$, attained exactly when $t=p$; such pairs exist even among representatives containing one. Thus a smaller-diameter part contains no pair with side intersection $p$. Since all representative sets already contain one, deleting that common coordinate and applying the preceding Frankl-Wilson argument bounds each part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. There is no extra factor two at this step. This is the cut construction described in the original Kahn-Kalai paper; the distance and part-size calculations above give the required justification directly.

An explicit choice is $p=13$. The point set lies in dimensionThe exact counts areTheir ratio requires at least $\lceil M/B\rceil =1562$ smaller-diameter parts, exceeding $d+1=1326$. In particular,Rescaling the point set to diameter one changes none of these partition counts. Therefore the Borsuk counterexample in dimension 1325 is fully explicit. This is a concrete dimension, not a claim that it is the smallest possible counterexample dimension.

The Borsuk conjecture asserted that every bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. We construct a Kahn-Kalai counterexample to the Borsuk conjecture.

For every $2p$-subset $A$ of $[4p]$, let $v_A\in \{-1,1{\}}^{4p}$ be $1$ on $A$ and $-1$ outside it, and defineBecause $x_A=x_{A^c}$, retain one representative of each complementary pair. The resulting set $X$ has $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{4p}{2p}\right)$ points. For $s=|A\cap B|$,andAll $x_A$ have the same norm, so their distance is largest exactly when this inner product is smallest, namely when $s=p$.

Every smaller-diameter part of $X$ therefore corresponds to a family with no pair having intersection $p$. Part ii bounds such a part by $2\left(\genfrac{}{}{0ex}{}{4p}{p-1}\right)$. Any smaller-diameter partition consequently needs at leastparts. By Stirling formula, this ratio grows like $(27/16{)}^p$ up to a polynomial factor, whereas $d+1=O(p^2)$. For every sufficiently large prime $p$, the required number of parts exceeds $d+1$, disproving the conjecture.