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Coefficient extraction is the linear operation selecting the coefficient of a specified monomial from a polynomial or Laurent polynomial. Multiplying by the inverse of that monomial reduces the operation to taking a constant term. The coefficient form of the Combinatorial Nullstellensatz expresses certain coefficient extractions as weighted sums of evaluations on a finite product set.

If an $n$ by $n$ matrix $A$ over a field has nonzero permanent, $b\in F^n$, and every $S_i\subseteq F$ has two elements, then some $x\in {\prod }_i{S}_i$ makes every coordinate of $Ax-b$ nonzero. Apply the Combinatorial Nullstellensatz towhose coefficient of $x_1\cdots x_n$ is $\mathrm{perm}A$.

Write $a=|A|$ and $b=|B|$, so $p\ge a>b\ge 1$. To cover both the unsaturated and saturated cases with the same coefficient calculation, setChoose $B^{\prime }\subseteq B$ of size $b^{\prime }$. These parameters satisfy $1\le b^{\prime }<a\le p$ and $1\le r\le p$.

Assume, for a contradiction, that the restricted sumset $C$ has fewer than $r$ elements. Extend it to a subset $D\subseteq {\mathbb{F}}_p$ of size $r-1$, and formOn $A\times B^{\prime }$, this polynomial vanishes: when $x=y$ the first factor vanishes, and when $x\ne y$ the sum belongs to $C\subseteq D$. Its total degree is $r=(a-1)+(b^{\prime }-1)$. Only the leading homogeneous part $(x-y)(x+y{)}^{r-1}$ contributes to the coefficient of $x^{a-1}{y}^{b^{\prime }-1}$. That coefficient isThe equality includes $b^{\prime }=1$, with an out-of-range binomial coefficient interpreted as zero. All factorial arguments are between zero and $p-1$, and $0<a-b^{\prime }<p$, so this coefficient is nonzero in ${\mathbb{F}}_p$. The Combinatorial Nullstellensatz applied to $A\times B^{\prime }$ forces a nonzero value, contradicting the vanishing just proved. Hence the restricted sumset bound for unequal subsets of a prime field isThe strict size inequality is used precisely in the nonzero factor $a-b^{\prime }$. The argument includes $r=p$ without dividing by $p!$, so no separate saturation assumption is hidden.

Choose a fixed prime $p\ge k$, and put $s=2p-2$. We first derive the polynomial step: a graph $H$ on $N$ vertices and $M>(p-1)N$ edges, with maximum degree at most $2p-1$, contains a nonempty $p$-regular subgraph. Use a Boolean variable $x_e$ for each edge and work over ${\mathbb{F}}_p$. DefineThe first product has degree at most $(p-1)N<M$, while the full squarefree monomial ${\prod }_e{x}_e$ has coefficient $(-1{)}^{M+1}\ne 0$ in the second term with its minus sign. The Combinatorial Nullstellensatz, with every grid $\{0,1\}$, gives a nonzero value of $R$. Since $R(0)=0$, this is a nonempty edge selection. The second product is then zero, so the first is nonzero. Fermat's little theorem makes each selected vertex degree zero modulo $p$. The maximum-degree bound implies that every positive selected degree is precisely $p$. This proves the prime-regular subgraph from Boolean polynomial constraints.

We next turn density into an almost-regular bipartite subgraph, keeping the logarithmic loss explicit. Let the original graph have $n\ge 2$ vertices and $m>0$ edges. Some bipartition retains at least $m/2$ edges: choose the sides independently with probability $1/2$ and average the number crossing. Its average degree is at least $m/n$. Repeatedly delete a vertex of degree at most half the current average degree. Such a deletion cannot lower that average, sinceThe process cannot delete the last positive-density graph, because the maintained average is positive. At termination the minimum degree is an integer $\sigma >m/(2n)$.

Choose the larger side $A$ of this remaining bipartite graph, with opposite side $B$. Keep exactly $\sigma$ edges at each vertex of $A$; choices at different vertices of $A$ do not conflict. Remove any now-isolated vertices of $B$. Thus $|A|\ge |B|$, and every vertex of $A$ has degree $\sigma$.

Such a half-regular graph contains a half-regular restriction with a perfect matching. Choose a minimal nonempty $X\subseteq A$ satisfying $|N(X)|\le |X|$, which exists because the whole side $A$ satisfies this inequality. In fact equality holds: if it were strict, deleting a vertex of $X$ would retain the inequality, contradicting minimality; the singleton case has at least one neighbor. Every proper subset $Y\subset X$ satisfies $|N(Y)|\ge |Y|$. Hall's marriage theorem therefore supplies a perfect matching on $X\cup N(X)$. All $\sigma$ edges incident with each vertex of $X$ remain inside that restriction.

Delete this perfect matching and repeat the restriction argument. The left degrees decrease by one at each step. We obtain $\sigma$ edge-disjoint perfect matchings $M_0,\dots ,M_{\sigma -1}$ on nested nonempty vertex sets, with sizesFor any $i$ with $i+s\le \sigma -1$, the union of $s+1=2p-1$ consecutive matchings in a graph has maximum degree at most $2p-1$ and at least $(s+1)N_{i+s}/2$ edges on $N_i$ vertices. Ifit has more than $(p-1)N_i$ edges, so the polynomial step yields a $p$-regular bipartite subgraph. Every regular bipartite graph decomposes into perfect matchings: for a set in one part, counting incident edges gives Hall's inequality, so a perfect matching exists; remove it and repeat. Choosing $k$ of these $p$ matchings in a graph gives a nonempty $k$-regular subgraph. This is the nested matching criterion for a regular subgraph.

If no $k$-regular subgraph exists, every displayed size ratio must therefore satisfy the reverse inequality. Iterating at indices $0,s,2s,\dots$ gives, with $r=\lfloor (\sigma -1)/s\rfloor$,Consequently $r\le \log n/\log ((s+1)/s)$ andCombining with $\sigma >m/(2n)$ proves the required edge bound. For example the finite constantworks for every $n\ge 2$, and the case $n=1$ has no edges. ThusThe logarithm comes from how often a nested matching in a graph vertex set can shrink by a fixed factor before becoming empty; it is not supplied by the polynomial step alone.

The coefficient form of the Combinatorial Nullstellensatz is as follows. Let $F$ be a field, let $P\in F[x_1,\dots ,x_r]$ have total degree at most $t_1+\cdots +t_r$, and suppose the coefficient of $x_1^{t_1}\cdots x_r^{t_r}$ is nonzero. For finite subsets $S_i\subset F$ with $|S_i|>t_i$, there is a point of $S_1\times \cdots \times S_r$ where $P$ is nonzero.

For completeness, restrict to $|S_i|=t_i+1$. The coefficient identity isBy Lagrange interpolation, the one-variable weighted sum on the right kills powers below $t_i$ and equals one on the power $t_i$. Any monomial of total degree at most $\sum t_i$, other than the specified one, has some exponent below $t_i$ and therefore contributes zero; the specified monomial contributes its coefficient. Every denominator is nonzero in the field. Hence a nonzero coefficient forces a nonzero grid value. The two applications below use this coefficient form.

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 ${\mathbb{F}}_2$.

Consider the degree-$n$ polynomialTo form the square-free monomial $x_1\cdots x_n$, one must choose each variable exactly once from the $n$ factors. Such choices are indexed by permutations, soThis coefficient is nonzero by hypothesis. Apply the Combinatorial Nullstellensatz with every $d_i=1$ and the given two-element sets $S_i$. It supplies $x\in {\prod }_i{S}_i$ with $f(x)\ne 0$. Every factor is then nonzero, sofor all $i$. This is coordinate avoidance from a nonzero permanent.

The coefficient form of the Combinatorial Nullstellensatz is the following. Let $f\in F[x_1,\dots ,x_n]$ have total degree at most $d_1+\cdots +d_n$, and supposeFor arbitrary subsets $S_i\subseteq F$ with $|S_i|=d_i+1$, there is an $x\in S_1\times \cdots \times S_n$ such that $f(x)\ne 0$.

For the proof, defineSuccessive Lagrange interpolation in the variables gives the Alon-Tarsi lemmaEvery denominator is nonzero because the elements of $S_i$ are distinct. If $f$ vanished throughout the product grid, the right side and hence the assumed nonzero coefficient would vanish. This contradiction proves the theorem.

We use the coefficient form of the Combinatorial Nullstellensatz: if a polynomial of degree ${\sum }_i{t}_i$ has nonzero coefficient at ${\prod }_i{x}_i^{t_i}$, then it cannot vanish on every product set ${\prod }_i{S}_i$ with $|S_i|>t_i$.

Write the terms as $(a_i,b_i)$ and, over ${\mathbb{F}}_p$, putandConsiderA direct multinomial-coefficient calculation, using Wilson's theorem, shows that the coefficient of ${\prod }_{i=1}^{4p-2}{x}_i$ in $F$ is nonzero. The Combinatorial Nullstellensatz with every $S_i=\{0,1\}$ therefore gives an indicator vector $x$ for which $F(x)\ne 0$.

The zero indicator does not work. The first two factors force the selected vectors to have both coordinate sums zero. If the support size is not divisible by $p$, Fermat's little theorem makes $1-S^{p-1}=0$, while the final product also vanishes; hence the support has size $p$, $2p$, or $3p$. Size $p$ proves the claim, and size $3p$ reduces to part (a). For size $2p$,so the bracket vanishes, again contradicting $F(x)\ne 0$. Only the size-$p$ and reducible size-$3p$ cases remain, and either yields the required zero-sum $p$-subsequence.

The Alon-Tarsi lemma says that if $\mathrm{deg}f\le d_1+\cdots +d_n$ and $A_i$ consists of $d_i+1$ distinct field elements, thenThis follows by applying univariate Lagrange interpolation successively in each variable.

If the displayed coefficient is nonzero, at least one summand has $f(a_1,\dots ,a_n)\ne 0$. This is the coefficient form of the Combinatorial Nullstellensatz.

For prime $p$, a graph with $m>(p-1)n$ edges and maximum degree at most $2p-1$ contains a nonempty $p$-regular subgraph. Over ${\mathbb{F}}_p$, apply the Combinatorial Nullstellensatz on the Boolean edge-variable cube toIts full squarefree coefficient is nonzero because the first product has degree below $m$. A nonzero value cannot occur at the zero vector; at any other Boolean vector the second product vanishes, and the first forces all selected degrees to be multiples of $p$. The maximum-degree condition makes every positive selected degree exactly $p$.

Suppose $|A|>|B|>0$ in ${\mathbb{F}}_p$. Put $a=|A|$ and $b^{\prime }=\min (|B|,p+2-a)$, and choose $B^{\prime }\subseteq B$ of size $b^{\prime }$. Let $r=a+b^{\prime }-2$. If the restricted sumset fits in a set $D$ of size $r-1$, then $(x-y){\prod }_{d\in D}(x+y-d)$ vanishes on $A\times B^{\prime }$. Its coefficient of $x^{a-1}{y}^{b^{\prime }-1}$ isAll factorial arguments are below $p$ and $0<a-b^{\prime }<p$. The Combinatorial Nullstellensatz gives a contradiction. Since $r=\min (p,|A|+|B|-2)$, this includes the saturated case as well as the small-set case.

Every length-$p$ sequence in the prime field ${\mathbb{F}}_p$ with sum zero, including sequences with repetitions, is the coordinatewise difference of two enumerations of that field. Apply the Combinatorial Nullstellensatz on $({\mathbb{F}}_p^{\ast }{)}^{p-1}$ to the product of the Vandermonde determinant in variables $x_i$ and its translate by the first $p-1$ sequence terms. The Dyson constant-term identity gives a nonzero top coefficient, and the zero-sum condition supplies the final missing field element. The argument includes $p=2$.