Finite-field Kakeya polynomial bound Created 2026-10-06 Updated 2026-10-14
For a finite-field Kakeya set , a smaller cardinality would give a nonzero polynomial of total degree of a polynomial at most vanishing on , by the dimension of a bounded-total-degree polynomial space. Its restriction to each complete affine line in a vector space has roots of a polynomial and degree less than , hence vanishes identically. The top homogeneous polynomial part then vanishes in every direction. A polynomial of degree less than in each variable cannot vanish everywhere on unless it is zero, giving a contradiction. For the prime field, this is the polynomial nonvanishing below the field size.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 12 5 Solution Created 2026-10-03 Updated 2026-10-06
Apply the polynomial method in combinatorics to the finite-field Kakeya set . We will actually obtain the stronger estimateLet be the vector space over the prime field of multivariate polynomials in variables of total degree of a polynomial at most . By the dimension of a bounded-total-degree polynomial space, its monomial basis consists of with and . Adding a slack exponent gives nonnegative exponents summing to ; the stars and bars count shows thatIf , the evaluation linear map has a nonzero element in its kernel of a linear map. Thus there is a nonzero polynomial of total degree of a polynomial vanishing at every point of . Its degree cannot be zero, since contains a line and is nonempty. Write for its nonzero top homogeneous polynomial part.
For every nonzero , the directional hypothesis supplies an affine line in a vector spaceDirections are one-dimensional vector subspaces; rescaling a representative does not change this line. The univariate polynomial has degree at most and vanishes for all values of . The root bound for a polynomial makes it the zero polynomial. Its coefficient of is exactly , so for every nonzero . Because , too.
To finish, we prove the relevant polynomial nonvanishing below the field size. A polynomial over of degree at most in each variable cannot vanish on all of unless it is zero. Induct on . The one-variable case is the root bound for a polynomial. For more variables, writeFixing the first coordinates gives a univariate polynomial with roots of a polynomial, so all its coefficients are zero. Hence every vanishes on , and the induction hypothesis makes every zero. Applying this to , whose total degree of a polynomial is less than , contradicts its choice as nonzero.
Therefore . Finally,which proves the requested lower bound. The crucial observation in the finite-field Kakeya polynomial bound is that complete lines force the highest homogeneous polynomial part to vanish in every direction.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 147 1 iii Solution 2026-10-03
We prove the polynomial nonvanishing below the field size by mathematical induction on . The case is the Lagrange root bound over a field: a nonzero polynomial of degree below cannot have all elements of as roots.
For the induction step, suppose that vanishes on all of and writeThe upper limit is valid because the total degree of a polynomial is less than . Fixing the first variables produces a univariate polynomial of degree below which vanishes at every . It is therefore the zero polynomial, so every vanishes on all of . Each also has total degree below , and the induction hypothesis gives for every . Thus .