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
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
Exercise 1.4.1. Let P :
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) :=
ci1,...,in x11
. . .
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 )),
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
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
= P . On the other hand, as P has degree at most
p 1, ΔeP
= 0, and so
((1 +
1 Δe)P
= 0.
Using the binomial formula, we can factorise the left-hand side as
(1 +
p 1
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.
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