The Alon-Tarsi lemma says that if

de g f \leq d_{1} + \dots + d_{n}

and

A_{i}

consists of

d_{i} + 1

distinct field elements, then

[x_{1}^{d_{1}} \dots x_{n}^{d_{n}}] f = \sum_{a_{i} \in A_{i}} \frac{f ( a _{1} , \dots , a _{n} )}{\prod _{i} \prod _{b \in A_{i} ∖ {a_{i}}} ( a _{i} - b )} .

This 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}) \neq = 0

Solved by gpt-5.6-sol high.

b

Solution

h_{j} (x) = 0

, normalized so that

h_{j} (0) = 1

, and put

P (x) = \prod_{j = 1}^{m} h_{j} (x) .

Then

P (0) = 1

and

P

vanishes at every other point of

F_{q}^{n}

. Reduce

P

x_{i}^{q} - x_{i}

in every variable. This preserves its function on

F_{q}^{n}

, does not increase total degree, and gives the unique representative with each variable degree at most

q - 1

The unique reduced polynomial for the delta function at zero is

\prod_{i = 1}^{n} (1 - x_{i}^{q - 1}),

whose total degree is

(q - 1) n

. Hence

m = de g P \geq (q - 1) n .

Uniqueness follows equally from the Alon-Tarsi lemma on the grids

A_{i} = F_{q}

Solved by gpt-5.6-sol high.

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