Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 109 4 Solution Created 2026-10-03 Updated 2026-10-06
We prove both the general modular intersection bound for a set family and its sharper uniform version, the Frankl-Wilson theorem. Let be prime, let have size , and let be a family of subsets of . Suppose each member's cardinality modulo is outside , whereas every intersection of two distinct members has cardinality modulo in . Then the general modular intersection bound iswith for . If every member has the same size and , the uniform Frankl-Wilson theorem sharpens this toHere , so in particular . Primality matters because the argument takes place over the finite field .
For each , define the polynomialAt the indicator vector of another member , this polynomial is zero if and nonzero if . Replace every positive power of by in its expanded monomials, obtaining a multilinear polynomial of degree at most . This multilinear reduction on the Boolean cube preserves values at every zero-one vector.
The resulting polynomials are linearly independent. Indeed, if , evaluating at leaves ; the nonzero diagonal value forces for every . But multilinear polynomials of degree at most lie in the vector space with basis , whose dimension is . 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 , suppose coefficients , indexed by -sets in , satisfyThen all coefficients are zero. For this is immediate. For , fix distinct and subtract the equations for and , where is a -set avoiding . The result isApply mathematical induction with degree on the remaining coordinates; its size condition holds because . This forces for every such . Since all -sets are linked by one-element exchanges, all coefficients equal some . An original equation then gives , and the characteristic assumption gives .
For every with , now form the multilinear polynomialThis is the multilinear reduction of , so it vanishes at every indicator vector of a -set. These are linearly independent: in a nonzero linear relation choose the largest degree of a coefficient-bearing . The degree- terms have coefficients . The inclusion lemma makes these nonzero unless all degree- coefficients vanish, a contradiction. It applies because , , and .
The are also independent jointly with the . Evaluating a relation at the indicator vectors of the family first forces every coefficient of to vanish, since the vanish there; their own independence then kills the remaining coefficients. All these polynomials have degree at most , soSubtracting the lower-degree contribution proves the sharp Frankl-Wilson theorem. For the general bound already gives , so no auxiliary polynomials are needed.
For the even cross-intersection assertion, work instead in the vector space over a finite field and letThe parity assumption is exactly in . By bilinearity, all vectors in are orthogonal to all vectors in , giving . The coordinate dot product is nondegenerate over , so . There is no requirement that be zero; only this dimension formula is used. Each distinct set has a distinct indicator vector, and a dimension- binary vector space has 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 in . Applying the same orthogonal complement over the binary field argument to their spans, now in dimension , 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.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 109 4 ii Solution Created 2026-09-28 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 109 4 Solution 2026-09-28
For each , form the multilinearization on the Boolean cube ofAt the characteristic vector of a set , this polynomial vanishes for and is nonzero for . Hence the restricted functions are linearly independent. On the -slice, every square-free monomial of degree below can be raised to degree using the relation , so the degree-at-most- function space is spanned by the square-free degree- monomials. Linear independence gives the theorem.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 109 2 iv Solution Created 2026-09-24 Updated 2026-09-25
Fix a prime number . Distinct members of an intersecting -uniform family have intersection size inwhereas every member has size modulo . The Frankl-Wilson theorem with therefore givesFor fixed ,This is the asserted asymptotic weakening of the Erdős-Ko-Rado theorem.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 109 4 ii Solution Created 2026-09-24 Updated 2026-10-03
Partition the family into complementary pairs . Choose at most one member from each pair to obtain with . Distinct members of are not disjoint, and the hypothesis excludes intersection size . Since their intersection sizes lie between and , they therefore lie modulo inEvery member has size , which is outside . The Frankl-Wilson theorem givesand hence
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 109 4 i Solution Created 2026-09-24 Updated 2026-09-25
One uniform form of the Frankl-Wilson theorem is as follows. Let be prime and let have elements. If satisfiesthenThe proof assigns to each set a degree- polynomial that vanishes on the incidence vectors of all other members but not on its own. These functions are linearly independent in the space spanned by square-free monomials of degree , yielding the dimension bound.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 145 4 b Solution Created 2026-09-24 Updated 2026-10-03
The required prime-power form of the Frankl-Wilson theorem is: if , no is divisible by , and every intersection of distinct members has size divisible by , thenIt follows by associating to each its incidence vector augmented by a constant coordinate and applying the Frankl-Wilson polynomial independence lemma to the layers of multilinear intersection polynomials. The hypotheses make the diagonal evaluations nonzero modulo and every -fold off-diagonal evaluation zero; independence leaves at most polynomials. Applying the theorem gives the desired bound.
Quadratic sign-vector diameter formula 2026-10-14
Take , and sign vectors . The Frobenius inner product of two outer products is , proving the formula. The squared Euclidean distance is largest precisely when , where the inner product is minus one. The common diagonal can be deleted, leaving an isometric representation in dimension if off-diagonal coordinates are multiplied by . The Frankl-Wilson theorem then bounds each smaller-diameter class.