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The vector space of formal polynomials over a field $k$ in $n$ variables with total degree of a polynomial at most $d$ has the monomial basis $x_1^{a_1}\cdots x_n^{a_n}$ for nonnegative exponents with ${\sum }_i{a}_i\le d$. Introducing a slack exponent and using stars and bars counts $\left(\genfrac{}{}{0ex}{}{n+d}{n}\right)$ basis elements. These are formal polynomials; over a finite field, distinct formal polynomials need not define distinct polynomial functions unless suitable degree restrictions hold.

For a finite-field Kakeya set $A\subseteq {\mathbb{F}}_q^n$, a smaller cardinality would give a nonzero polynomial of total degree of a polynomial at most $q-1$ vanishing on $A$, by the dimension of a bounded-total-degree polynomial space. Its restriction to each complete affine line in a vector space has $q$ roots of a polynomial and degree less than $q$, hence vanishes identically. The top homogeneous polynomial part then vanishes in every direction. A polynomial of degree less than $q$ in each variable cannot vanish everywhere on ${\mathbb{F}}_q^n$ unless it is zero, giving a contradiction. For the prime field, this is the polynomial nonvanishing below the field size.

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.

Work in the prime field ${\mathbb{F}}_p$ and set $r=p-1$. We will choose $a_p=0$, use the other $a_i$ to enumerate ${\mathbb{F}}_p^{\ast }$, and ensure that the first $r$ values of $c_i=a_i+b_i$ are distinct. The zero-sum condition will then force the last value to be the missing field element.

Consider the polynomialIt has total degree of a polynomial $r(r-1)$. Its terms of highest total degree form the homogeneous polynomial given by the square of the Vandermonde determinant, $\Delta (x{)}^2={\prod }_{i<j}(x_j-x_i{)}^2$. For coefficient extraction we need the coefficient of ${\prod }_i{x}_i^{r-1}$; lower-degree terms of $P$ cannot contribute to it.

Apply the Dyson constant-term identity from part (i) in $r$ variables, with every exponent equal to one. Pairing its factors for $i,j$ givesThe constant term is $r!$. Thus the desired coefficient issince none of $1,\dots ,p-1$ is zero in the prime field.

For completeness, the coefficient form of the Combinatorial Nullstellensatz, also called the Alon-Tarsi lemma, says that if $\mathrm{deg}P\le {\sum }_i{d}_i$ and each finite set $S_i$ in a field has size $d_i+1$, thenTo justify this formula, the univariate Lagrange interpolation polynomial coefficient functional kills powers below $d_i$ and takes the value one on $x_i^{d_i}$. Apply the product of these functionals to each monomial of $P$. Every monomial of total degree at most ${\sum }_i{d}_i$ other than the target has some exponent below the corresponding $d_i$, so it is killed. The target survives with coefficient one. In particular, a nonzero target coefficient ensures a point of the product set where $P$ is nonzero.

Use $S_i={\mathbb{F}}_p^{\ast }$ and $d_i=r-1$ for every $i$. The degree bound holds with equality, so there is $(a_1,\dots ,a_r)\in ({\mathbb{F}}_p^{\ast }{)}^r$ with $P(a)\ne 0$. Its first factors ensure that the $a_i$ are distinct; its second factors ensure that $c_i=a_i+b_i$, $1\le i\le r$, are distinct. With $a_p=0$, the $a_i$ enumerate the whole prime field.

Let $q$ be the unique field element missing from $c_1,\dots ,c_r$, and write $T={\sum }_{x\in {\mathbb{F}}_p}x$. Since ${\sum }_i{b}_i=0$,The same sum is $T-q$, hence $q=b_p$. Taking $c_p=b_p=a_p+b_p$ completes the enumeration. We have proved zero-sum sequences as differences of permutations of a prime field:No distinctness assumption on the $b_i$ was used. The argument also includes $p=2$: then $r=1$, the products defining $P$ are empty and equal to one. Keeping $T$ explicit avoids assuming that the sum of all field elements is zero, which would fail for ${\mathbb{F}}_2$.

We prove the polynomial nonvanishing below the field size by mathematical induction on $n$. The case $n=1$ is the Lagrange root bound over a field: a nonzero polynomial of degree below $p$ cannot have all $p$ elements of ${\mathbb{F}}_p$ as roots.

For the induction step, suppose that $P$ vanishes on all of ${\mathbb{F}}_p^n$ and writeThe upper limit is valid because the total degree of a polynomial is less than $p$. Fixing the first $n-1$ variables produces a univariate polynomial of degree below $p$ which vanishes at every $x_n\in {\mathbb{F}}_p$. It is therefore the zero polynomial, so every $P_j$ vanishes on all of ${\mathbb{F}}_p^{n-1}$. Each $P_j$ also has total degree below $p$, and the induction hypothesis gives $P_j=0$ for every $j$. Thus $P=0$.

Taking the contrapositive, every nonzero such $P$ has some $a\in {\mathbb{F}}_p^n$ for which

Let $P\in {\mathbb{F}}_p[x_1,\dots ,x_n]$ have total degree of a polynomial less than $p$. If the associated polynomial function vanishes on all of ${\mathbb{F}}_p^n$, then $P$ is the zero polynomial. Induct on $n$: writeFor each fixed $(x_1,\dots ,x_{n-1})$, the resulting univariate polynomial of degree less than $p$ has all $p$ field elements as roots, so every coefficient $P_j$ vanishes everywhere. The induction hypothesis makes every $P_j$ the zero polynomial.