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Use the balanced cut-vector distance formula with $p=13$. There are $\left(\genfrac{}{}{0ex}{}{51}{25}\right)=247959266474052$ distinct cuts. A subset of strictly smaller diameter contains no pair with side intersection thirteen, so deleting their shared coordinate one and applying the Frankl-Wilson theorem bounds its size by $\left(\genfrac{}{}{0ex}{}{51}{12}\right)=158753389900$. At least 1562 smaller-diameter parts are needed, exceeding $1325+1$. The cut vectors occupy an affine hyperplane in ${\mathbb{R}}^{1326}$, which identifies isometrically with ${\mathbb{R}}^{1325}$ and gives the stated explicit counterexample.

For a prime $p$, take rank-$2p$ subsets of $[4p]$ with no distinct pair meeting in $p$ points. Split by membership of coordinate one. In the containing part, remove that coordinate; in the avoiding part, first take complements and then remove it. Both transformed families have rank $2p-1$ on $4p-1$ points. Their distinct intersections avoid residue $p-1$ modulo $p$. The uniform Frankl-Wilson theorem bounds each by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$, yielding the displayed stronger form of the requested bound.

If $n\ge 2d+1$ and the field characteristic is zero or greater than $d+1$, the matrix recording inclusion of $d$-sets in $(d+1)$-sets has full row rank. Subtracting equations on sets differing in one coordinate reduces the kernel problem to degree $d-1$ on $n-2$ coordinates. Mathematical induction then makes all coefficients equal, and the factor $d+1$ forces them to vanish. This gives the auxiliary-polynomial independence used in the uniform Frankl-Wilson theorem.

Kahn and Kalai represented uniform subsets by quadratic sign vectors so that diameter pairs correspond to one forbidden intersection size. The Frankl-Wilson theorem bounds each smaller-diameter part, forcing exponentially many parts while the ambient dimension grows only quadratically.

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.

For a prime number $p$, the vertices are $k$-subsets of $[N]$. An independent set avoids the only off-diagonal intersection congruent to $k$ modulo $p$, so the Frankl-Wilson theorem bounds its size by $\left(\genfrac{}{}{0ex}{}{N}{p-1}\right)$. A clique has constant intersection $p-1$ and has size at most $N$, by the same theorem modulo a prime larger than $k$. This explicit graph gives an asymmetric Ramsey number lower bound and an exponential chromatic number lower bound.

Let $p$ be a prime number, $0\le a<p$, $1\le s<p$, and $n\ge a+s$. A multilinear polynomial over ${\mathbb{F}}_p$ of degree less than $s$ that vanishes on every Boolean vertex whose weight is not congruent to $a$ modulo $p$ vanishes everywhere. Alternating sums over intervals of $s$ free coordinates eliminate the possible nonzero weight levels one at a time. This proves independence of the auxiliary functions used for the uniform Frankl-Wilson theorem bound.

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.

Use the uniform Frankl-Wilson theorem: if $p$ is a prime number, $L\subseteq {\mathbb{F}}_p$ has $s$ distinct residues, $s\le \min \{r,n-r\}$, and an $r$-uniform set family $\mathcal{F}$ has distinct-member intersection sizes in $L$ modulo $p$ while $r\mathrm{mod}p\notin L$, thenHere is a polynomial method in combinatorics proof. For each $A\in \mathcal{F}$, form the intersection polynomialover ${\mathbb{F}}_p$, and use multilinear reduction on the Boolean cube so its degree is at most $s$. At the characteristic vector of $B\in \mathcal{F}$, this evaluates to zero for $A\ne B$ and is nonzero for $A=B$. Consequently these evaluation functions are linearly independent.

The low-degree evaluation rank on a uniform layer bounds their space by $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. To justify that step without division by a possibly zero residue, form the integer incidence matrix with rows $T\subseteq [n]$, $|T|\le s$, columns $r$-sets $B$, and entries $1_{T\subseteq B}$. Over the rational numbers, a row with $|T|=j$ equals ${\left(\genfrac{}{}{0ex}{}{r-j}{s-j}\right)}^{-1}$ times the sum of the rows of all $s$-sets containing $T$. Thus its rational matrix rank is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger minor is an integer determinant equal to zero and stays zero modulo $p$, so the same upper bound holds over ${\mathbb{F}}_p$. This proves the theorem.

For the forbidden midpoint problem, split $\mathcal{A}$ according to whether its members contain coordinate 1. Leave the containing half unchanged; complement every member of the avoiding half. Both resulting families consist of $2p$-sets containing 1, and complementation preserves their internal intersection sizes becauseWithin either transformed family, distinct intersections lie between 1 and $2p-1$ and are not $p$. Their residues are therefore in $L=\{1,\dots ,p-1\}$, whereas the set-size residue is $2p\equiv 0$. Apply the Frankl-Wilson theorem to each half and add:The factor 2 comes from splitting into two families. The unsplit family can contain disjoint sets, whose intersection residue zero equals the set-size residue and prevents this direct application. Removing the shared coordinate after splitting also gives the stronger complement splitting for forbidden midpoint intersections bound $2\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the construction, fix a $(p+1)$-set $B$ and let $R=[4p]\setminus B$, of size $3p-1$. The fixed-core construction avoiding midpoint intersections isTwo members of $\mathcal{F}$ intersect in at least $p+1$ points; the same is true for two complements. An intersection across the two halves has size $|X\setminus Y|\le p-1$. No intersection is $p$. The halves are disjoint because one contains all of $B$ and the other none, so

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 Frankl-Wilson theorem in its uniform form says: let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have size $s\le \min (k,n-k)$, and let $\mathcal{F}\subseteq \left(\genfrac{}{}{0ex}{}{[n]}{k}\right)$ have $k\mathrm{mod}p\notin L$, while $|A\cap B|\mathrm{mod}p\in L$ for every distinct pair of members. Then

The modular-intersection, nonuniform form of the Frankl-Wilson theorem is as follows. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$ satisfy $|A|\mathrm{mod}p\notin L$ for each $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for distinct $A,B\in \mathcal{F}$. ThenThis nonuniform Frankl-Wilson theorem allows different sizes and different diagonal residues; in particular it applies when all sizes have one common residue outside $L$.

To prove it by the polynomial method in combinatorics, work over the finite field ${\mathbb{F}}_p$ and associate to each member the intersection polynomialReplace each positive power of $x_i$ in its expansion by $x_i$ to obtain its multilinear reduction on the Boolean cube ${\tilde{P}}_A$. This preserves evaluation on every characteristic vector, whose coordinates are zero or one, and leaves degree at most $s$. At the characteristic vector $1_B$,It is zero for distinct $A,B$, and nonzero for $B=A$ by the diagonal hypothesis. If ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at each $1_B$ forces $c_B=0$. Thus the multilinear polynomials are linearly independent. They belong to the vector space with basis the square-free monomials ${\prod }_{i\in S}{x}_i$ for $|S|\le s$, whose dimension is the displayed sum. This proves the nonuniform form, including $s=0$, where distinct members are impossible and the empty product is one.

For the sharp uniform form, we need the low-degree evaluation rank on a uniform layer. Form the integer matrix $M$ whose rows are indexed by $S\subseteq [n]$ with $|S|\le s$, columns by $k$-subsets $T$, and entries $M_{S,T}=1_{S\subseteq T}$. These are evaluations of the square-free monomials. Over the rational numbers, the row of a $j$-subset $S$ satisfiesIndeed, when $S\subseteq T$ the left side counts the $s$-subsets between $S$ and $T$, and otherwise both sides vanish. Since $s\le k$, the coefficient is a positive integer. Thus the degree-$s$ rows span all the rows over the rational numbers, giving ${\mathrm{rank}}_{\mathbb{Q}}M\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every square submatrix of larger size has integer determinant zero, so its determinant remains zero modulo $p$. HenceThis reduction of matrix rank is the reason one must not try to divide by $\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)$ directly in ${\mathbb{F}}_p$.

The evaluations of the ${\tilde{P}}_A$ on the $k$th uniform layer of the Boolean cube lie in the row linear span of $M$ over ${\mathbb{F}}_p$. They are linearly independent as functions there, because restriction to the columns corresponding to $\mathcal{F}$ still gives the nonzero diagonal evaluation matrix. Their number is consequently at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. This completes the proof of the uniform Frankl-Wilson theorem as well.

For the odd-cardinality problem the sharper Oddtown theorem follows directly from linear algebra. Let $v_A=1_A\in {\mathbb{F}}_2^n$ be the characteristic vector. The parity conditions sayFor any linear combination ${\sum }_A{c}_A{v}_A=0$, take its dot product with $v_B$. All off-diagonal terms vanish, leaving $c_B=0$. Hence these vectors are linearly independent in an $n$-dimensional vector space, givingThe singleton sets show that the bound can be attained. Notice why the constant term in the general nonuniform Frankl-Wilson theorem is unnecessary here: these dot products already give a diagonal evaluation matrix using only the $n$ coordinate functions.

For even $n$, partition $[n]$ into the pairs $\{1,2\},\{3,4\},\dots ,\{n-1,n\}$ and take every union of a selection of these pairs. This gives exactly $2^{n/2}$ distinct sets, including the empty set. Each has even size; every pairwise intersection is again a union of entire pairs and therefore has even size. This attains the proposed bound.

For any set family satisfying the even-cardinality conditions, let $W$ be the linear span of its characteristic vectors over ${\mathbb{F}}_2$. Now the self-dot products vanish as well as the off-diagonal ones. By bilinearity, $u\cdot v=0$ for all $u,v\in W$, so $W\subseteq W^{\perp }$: it is a self-orthogonal binary subspace.

For completeness the standard bilinear form on ${\mathbb{F}}_2^n$ is nondegenerate, because a vector orthogonal to every coordinate vector has every coordinate zero. Thus the map $v\mapsto (w\mapsto v\cdot w)$ from ${\mathbb{F}}_2^n$ onto the dual space $W^{\ast }$ is surjective: every linear functional on $W$ extends to the whole vector space by extending a basis, and every such functional on ${\mathbb{F}}_2^n$ is a dot product with a vector. Its kernel is $W^{\perp }$. The rank-nullity theorem consequently gives $\dim W^{\perp }=n-\dim W$. The inclusion $W\subseteq W^{\perp }$ implies $\dim W\le n/2$. Finally, distinct members have distinct characteristic vectors, all in $W$, and a $k$-dimensional vector space over ${\mathbb{F}}_2$ has exactly $2^k$ elements. ThereforeThis proves the Eventown theorem and also explains the suggested dimension contradiction if the set family had more than $2^{n/2}$ members.

Work over the finite field ${\mathbb{F}}_p$. For each set family member form the intersection polynomialAt the characteristic vector of a set $B$, its value is ${\prod }_i(|A\cap B|-{\ell }_i)$. The given restrictions make $f_A({\chi }_B)=0$ for $A\ne B$ and $f_A({\chi }_A)\ne 0$. Hence these functions are linearly independent: evaluate a purported relation at each set family member to eliminate its coefficient.

Use multilinear reduction on the Boolean cube, replacing every positive power of $x_i$ by $x_i$. This preserves all evaluations on $\{0,1{\}}^n$ and cannot increase degree. The multilinear polynomials of degree at most $s$ have basis $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$. Their dimension is ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$, and thereforewhere $\left(\genfrac{}{}{0ex}{}{n}{j}\right)=0$ for $j>n$. The proof also covers a list whose distinct integers have repeated residues: no assumption of distinct residues is needed for this first bound.

To obtain the sharper conclusion, suppose the common size residue $r$ avoids $0,\dots ,s-1$. If $s>n$, no set on this ground set can have the required residue, so the set family is empty. Otherwise, add the modular-size auxiliary polynomialsagain understood as functions after multilinear reduction on the Boolean cube. They have degree at most $s$, and all vanish on the set family's characteristic vectors of sets.

The $q_S$ are linearly independent. Indeed, evaluate ${\sum }_S{c}_S{q}_S=0$ on ${\chi }_T$ with $|T|\le s-1$:The first factor is nonzero in the finite field. Starting with $T=\varnothing$ and proceeding in increasing cardinality shows $c_T=0$ for every such $T$.

Moreover, the combined list consisting of the $f_A$ and the $q_S$ is linearly independent. Evaluation at each ${\chi }_A$ first kills the coefficient of $f_A$; the preceding argument then kills all remaining coefficients. Comparing this list with the dimension of the degree-at-most-$s$ space yieldsThis is a nonuniform common-residue version of the Frankl-Wilson theorem; exact equality of all member sizes was never used.

The exclusion of the small residues is necessary. A concrete example with $s<n$ is $n=4$, $p=5$, $s=3$, the residue list $0,1,3$, and the set family of all two-element subsets. Every member has size residue $r=2$, outside that list, while distinct members intersect in zero or one element. ButHere the omitted condition fails at $j=2$. Thus the example satisfies all the other hypotheses, rather than relying on a vacuous $s>n$ case.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.

The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

We prove both the general modular intersection bound for a set family and its sharper uniform version, the Frankl-Wilson theorem. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a family of subsets of $[n]$. Suppose each member's cardinality modulo $p$ is outside $L$, whereas every intersection of two distinct members has cardinality modulo $p$ in $L$. Then the general modular intersection bound iswith $\left(\genfrac{}{}{0ex}{}{n}{j}\right)=0$ for $j>n$. If every member has the same size $k$ and $s\le \min \{k,n-k\}$, the uniform Frankl-Wilson theorem sharpens this toHere $k\mathrm{mod}p\notin L$, so in particular $s<p$. Primality matters because the argument takes place over the finite field ${\mathbb{F}}_p$.

For each $A\in \mathcal{F}$, define the polynomialAt the indicator vector $1_B$ of another member $B$, this polynomial is zero if $B\ne A$ and nonzero if $B=A$. Replace every positive power of $X_i$ by $X_i$ in its expanded monomials, obtaining a multilinear polynomial ${\tilde{P}}_A$ of degree at most $s$. This multilinear reduction on the Boolean cube preserves values at every zero-one vector.

The resulting polynomials are linearly independent. Indeed, if ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at $1_B$ leaves $c_B{\tilde{P}}_B(1_B)=0$; the nonzero diagonal value forces $c_B=0$ for every $B$. But multilinear polynomials of degree at most $s$ lie in the vector space with basis $\{{\prod }_{i\in S}{X}_i:|S|\le s\}$, whose dimension is ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The bound on the dimension of a vector space gives the general modular intersection bound for a set family. This is the polynomial method in combinatorics: the intersection conditions produce an evaluation matrix with nonzero diagonal and zero off-diagonal entries.

To prove the sharper uniform bound, we first establish injectivity of inclusion between adjacent set layers. Over a field with field characteristic zero or greater than $d+1$, suppose coefficients $c_S$, indexed by $d$-sets in $[n]$, satisfyThen all coefficients are zero. For $d=0$ this is immediate. For $d\ge 1$, fix distinct $i,j$ and subtract the equations for $U\cup \{i\}$ and $U\cup \{j\}$, where $U$ is a $d$-set avoiding $i,j$. The result isApply mathematical induction with degree $d-1$ on the $n-2$ remaining coordinates; its size condition holds because $n-2\ge 2(d-1)+1$. This forces $c_{S\cup \{i\}}=c_{S\cup \{j\}}$ for every such $S$. Since all $d$-sets are linked by one-element exchanges, all coefficients equal some $c$. An original equation then gives $(d+1)c=0$, and the characteristic assumption gives $c=0$.

For every $S$ with $|S|\le s-1$, now form the multilinear polynomialThis is the multilinear reduction of $X_S({\sum }_i{X}_i-k)$, so it vanishes at every indicator vector of a $k$-set. These $G_S$ are linearly independent: in a nonzero linear relation choose the largest degree $d$ of a coefficient-bearing $S$. The degree-$(d+1)$ terms have coefficients ${\sum }_{S\subset T,|S|=d}{c}_S$. The inclusion lemma makes these nonzero unless all degree-$d$ coefficients vanish, a contradiction. It applies because $d\le s-1$, $p>s$, and $n\ge 2s$.

The $G_S$ are also independent jointly with the ${\tilde{P}}_A$. Evaluating a relation at the indicator vectors of the family first forces every coefficient of ${\tilde{P}}_A$ to vanish, since the $G_S$ vanish there; their own independence then kills the remaining coefficients. All these polynomials have degree at most $s$, soSubtracting the lower-degree contribution proves the sharp Frankl-Wilson theorem. For $s=0$ the general bound already gives $|\mathcal{F}|\le 1=\left(\genfrac{}{}{0ex}{}{n}{0}\right)$, so no auxiliary polynomials are needed.

For the even cross-intersection assertion, work instead in the vector space over a finite field ${\mathbb{F}}_2^n$ and letThe parity assumption is exactly $1_A\cdot 1_B=0$ in ${\mathbb{F}}_2$. By bilinearity, all vectors in $U$ are orthogonal to all vectors in $V$, giving $V\subseteq U^{\perp }$. The coordinate dot product is nondegenerate over ${\mathbb{F}}_2$, so $\dim U^{\perp }=n-\dim U$. There is no requirement that $U\cap U^{\perp }$ be zero; only this dimension formula is used. Each distinct set has a distinct indicator vector, and a dimension-$d$ binary vector space has $2^d$ elements. Therefore the even cross-intersection bound isThis is the even cross-intersection bound, including empty families, for which the product is zero.

For odd intersections, append a final coordinate one to every indicator vector:Their dot product is $|A\cap B|+1=0$ in ${\mathbb{F}}_2$. Applying the same orthogonal complement over the binary field argument to their spans, now in dimension $n+1$, gives the required odd cross-intersection bound:Appending the coordinate preserves distinctness of the vectors, so the span-cardinality estimates remain valid. This proves the odd cross-intersection bound directly by reducing the odd condition to an even one in a larger vector space.