1.4. Equidistribution in finite fields 61

group. Note that a polynomial of degree zero is the same thing as a constant

function, thus Poly≤0(V → R/Z) ≡ R/Z.

An important special case of polynomials are the classical polynomials,

which take values in F (which we identify with the

pth

roots of unity in

R/Z in the obvious manner); the space of such polynomials of degree at

most d will be denoted Poly≤d(V → F); this is clearly a vector space over

F. The classical polynomials have a familiar description, once we use a basis

to identify V with

Fn:

Exercise 1.4.1. Let P :

Fn

→ F be a function, and d ≥ 0 an integer. Show

that P is a (classical) polynomial of degree at most d if and only if one has

a representation of the form

P (x1,...,xn) :=

i1,...,in≥0:i1+···+in≤d

ci1,...,in x11

i

. . .

xnni

for some coeﬃcients ci1,...,in ∈ F. Furthermore, show that we can restrict

the exponents i1,...,in to lie in the range {0,...,p − 1}, and that once one

does so, the representation is unique. (Hint: First establish the d = 1 case,

which can be done, for instance, by a dimension counting argument, and

then induct on dimension.)

Exercise 1.4.2. Show that the cardinality of Poly≤d(V → F) is at most

p(d+dim(V )),

d

with equality if and only if d p.

Now we study more general polynomials. A basic fact here is that mul-

tiplying a polynomial by the characteristic p lowers the degree:

Lemma 1.4.1. If P ∈Poly≤d(V → R/Z), then pP ∈Poly≤max(d−p+1,0)(V →

R/Z).

Proof. Without loss of generality, we may take d ≥ p−1; an easy induction

on d then shows it suﬃces to verify the base case d = p − 1. Our task is now

to show that pP is constant, or equivalently that pΔeP = 0 for all e ∈ V .

Fix e. The operator 1 + Δe represents a shift by e. Since pe = 0, we

conclude that

(1+Δe)pP

= P . On the other hand, as P has degree at most

p − 1, ΔeP

p

= 0, and so

((1 +

Δe)p

− 1 − Δe)P

p

= 0.

Using the binomial formula, we can factorise the left-hand side as

(1 +

p − 1

2

Δe + · · · +

Δe−2)(pΔeP p

) = 0.

The first factor can be inverted by Neumann series since Δe acts nilpotently

on polynomials. We conclude that pΔeP = 0 as required.