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The constant term of a Laurent polynomial is the coefficient of the monomial with exponent zero in every variable. Taking the constant term in just one variable leaves a Laurent polynomial in the other variables; successive extractions recover the full constant term. This operation is linear, but need not preserve products. The Dyson constant-term identity is a multivariate example of coefficient extraction.

When all exponents are positive, the Dyson constant-term identity satisfies $D(a)={\sum }_{i}D(a-e_i)$. This follows from the rational identity ${\sum }_i{\prod }_{j\ne i}(1-X_i/X_j{)}^{-1}=1$, obtained from Lagrange interpolation polynomial at zero. When $a_i=0$, taking the constant term in $X_i$ deletes that coordinate, giving the boundary recurrence.

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 ${\mathbb{F}}_2$.

Let $D_n(a)$ denote the constant term, retaining the dimension as a subscript when it changes. We use the Good recurrence for the Dyson constant term. For algebraically independent variables, Lagrange interpolation polynomial applied to the constant polynomial one givesAt $t=0$, this becomes the rational identityWhen every $a_i>0$, multiplying by the given Laurent polynomial cancels one factor in row $i$ of the product and givesThis is a polynomial identity after cancellation; no choice of an expansion of a rational function is involved.

The boundary case is essential. If $a_i=0$, the factors in row $i$ are absent. The only factors involving $X_i$ are then $(1-X_j/X_i{)}^{a_j}$ for $j\ne i$, and all their powers of $X_i$ are nonpositive. To obtain power zero in $X_i$, one must select the constant term one from each of them. Taking the constant term in $X_i$ therefore deletes that variable and exponent:For $n=1$, the empty product is one, so $D_1(a_1)=1$. The all-zero exponent vector also gives one.

Now put $M_n(a)=m!/{\prod }_i{a}_i!$, where $m={\sum }_i{a}_i$. This multinomial coefficient obeys the same deletion rule for a zero exponent. For positive exponents,Induction on $n$, and within each dimension on $m$, consequently determines $D_n$ uniquely and identifies it with $M_n$. This proves the Dyson constant-term identity:

Let $D(a)$ denote the stated constant term and suppose every $a_i>0$. The rational identitycan be verified after clearing denominators, or by Lagrange interpolation. Multiplying it by the Dyson product and taking constant terms gives the recursionThe multinomial coefficientobeys the same recursion by the multinomial form of Pascal's identity.

If $a_k=0$, taking the constant term in $X_k$ forces the zero term from every factor involving $X_k$ and reduces the expression to the $(n-1)$-variable Dyson product with $a_k$ omitted. The same boundary reduction holds for $M(a)$. Finally $D(0,\dots ,0)=M(0,\dots ,0)=1$. Induction on $n$ and on $a_1+\cdots +a_n$ therefore proves the Dyson constant-term identity

Identify the additive group $G$ with the finite field ${\mathbb{F}}_q$, where $q=p^{\gamma }$. The sets $A$ and $B$ each have size $k<p$, and the field has odd characteristic. The Snevily matching theorem for an elementary abelian group states precisely that for two $k$-element subsets of the additive group of such a field, there is a bijection $\pi :A\to B$ for which the sums $a+\pi (a)$ are pairwise distinct.

For context, its polynomial proof antisymmetrizes the Vandermonde polynomialover all orderings of $B$. The coefficient furnished by the Dyson constant-term identity is a nonzero multiple of $k!$; it cannot vanish because $k<p$. Therefore at least one ordering $(b_1,\dots ,b_k)$ makes the Vandermonde product nonzero, which says exactly that

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 $p=2$.