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A subset of ${\mathbb{F}}_q^n$ is a finite-field Kakeya set when it contains an affine line in a vector space in every one-dimensional direction. The line position may depend on the direction. The polynomial method in combinatorics yields the finite-field Kakeya polynomial bound on its cardinality.

A finite-field Kakeya set, also called a finite-field Besicovitch set, is a subset containing a complete affine line in a vector space $a+{\mathbb{F}}_{p}v$ in every nonzero direction $v$, with scalar multiples counted as the same direction. The line's translate may depend on the direction.

For odd $p$, take the union of the lines $y=mx-m^2/4$, $m\in {\mathbb{F}}_p$. Completing the square gives $x^2-y=(m/2-x{)}^2$. Thus that union is exactly $K_0=\{(x,y):x^2-y\text{ is a square, including zero}\}$. For each fixed $x$ there are $(p+1)/2$ such $y$, so $|K_0|=p(p+1)/2$. It contains every nonvertical slope. Add the vertical line $x=0$, whose overlap with $K_0$ has $(p+1)/2$ points. The tangent-line finite-field Kakeya construction consequently hasThe prime two is irrelevant to this asymptotic statement and can be handled by taking its whole plane.

For the high-dimensional lower bound, we give a direct polynomial method in combinatorics proof, which yields a stronger estimate. Suppose a finite-field Kakeya set $K\subseteq {\mathbb{F}}_p^n$ has fewer than $\left(\genfrac{}{}{0ex}{}{p+n-1}{n}\right)$ points. That binomial coefficient is the dimension of a bounded-total-degree polynomial space of polynomials of total degree at most $p-1$. Evaluation at the points of $K$ imposes fewer linear conditions than this dimension. By the rank-nullity theorem, there is a nonzero polynomial $P$ of total degree $d\le p-1$ vanishing on $K$.

For every nonzero $v$ choose the line $a_v+tv\subseteq K$. The polynomial $P(a_v+tv)$ in $t$ has degree at most $d<p$ and vanishes at all $p$ field elements, so it vanishes identically. Its coefficient of $t^d$ is the leading homogeneous polynomial part $P_d(v)$. Thus $P_d(v)=0$ for every nonzero $v$. The case $d=0$ is already impossible for a nonzero constant vanishing on a nonempty set; for $d>0$, the absence of a constant term in this homogeneous polynomial also gives $P_d(0)=0$.

A polynomial of degree less than $p$ in each variable cannot vanish at every point of ${\mathbb{F}}_p^n$ unless it is zero. To see this, induct on $n$: fix the first $n-1$ variables, apply the one-variable root bound in the last variable, then apply the induction hypothesis to each resulting coefficient polynomial. This forces $P_d=0$, contradicting the choice of the leading part. We have proved the finite-field Kakeya polynomial boundAt $n=15$ this implies the required bound with an explicit absolute constant:The argument is valid in every prime characteristic and does not invoke the lower bound as an unproved theorem.

Apply the polynomial method in combinatorics to the finite-field Kakeya set $A$. We will actually obtain the stronger estimateLet $\mathcal{P}$ be the vector space over the prime field ${\mathbb{F}}_p$ of multivariate polynomials in $n$ variables of total degree of a polynomial at most $p-1$. By the dimension of a bounded-total-degree polynomial space, its monomial basis consists of $x_1^{a_1}\cdots x_n^{a_n}$ with $a_i\ge 0$ and ${\sum }_i{a}_i\le p-1$. Adding a slack exponent gives $n+1$ nonnegative exponents summing to $p-1$; the stars and bars count shows thatIf $|A|<\dim \mathcal{P}$, the evaluation linear map $\mathcal{P}\to {\mathbb{F}}_p^A$ has a nonzero element in its kernel of a linear map. Thus there is a nonzero polynomial $P$ of total degree of a polynomial $d\le p-1$ vanishing at every point of $A$. Its degree cannot be zero, since $A$ contains a line and is nonempty. Write $P_d$ for its nonzero top homogeneous polynomial part.

For every nonzero $v\in {\mathbb{F}}_p^n$, 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 $P(a_v+tv)$ has degree at most $d<p$ and vanishes for all $p$ values of $t$. The root bound for a polynomial makes it the zero polynomial. Its coefficient of $t^d$ is exactly $P_d(v)$, so $P_d(v)=0$ for every nonzero $v$. Because $d>0$, $P_d(0)=0$ too.

To finish, we prove the relevant polynomial nonvanishing below the field size. A polynomial over ${\mathbb{F}}_p$ of degree at most $p-1$ in each variable cannot vanish on all of ${\mathbb{F}}_p^n$ unless it is zero. Induct on $n$. The one-variable case is the root bound for a polynomial. For more variables, writeFixing the first $n-1$ coordinates gives a univariate polynomial with $p$ roots of a polynomial, so all its coefficients are zero. Hence every $Q_j$ vanishes on ${\mathbb{F}}_p^{n-1}$, and the induction hypothesis makes every $Q_j$ zero. Applying this to $P_d$, whose total degree of a polynomial is less than $p$, contradicts its choice as nonzero.

Therefore $|A|\ge \dim \mathcal{P}$. 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.