Prime Numbers

4.5 The primality test of Agrawal, Kayal, and Saxena (AKS test) 211
necessarily. However, the following result gives a criterion that ensures that
fr,d remains irreducible when reduced modulo p.
Lemma 4.5.5 (Kummer). If r is a prime, d is a positive divisor of r − 1,
and p is a prime with the order of p (r−1)/d modulo r equal to d, then fr,d(x)
remains irreducible when considered in Fp[x].
A proof of this result using that ηr,d and its conjugates form an integral basis
of the ring of integers in Q(ηr,d) may be found in [Adleman and Lenstra 1986].
We present a different proof using Gauss sums.
Proof. Consider the splitting field K of (xr − 1)(xd − 1) over Fp, which
may be viewed as the homomorphic image of Z[ζr,ζd], where ζr = e2πi/r and
ζd = e2πi/d .Letζbe the image of ζr in K and let ω be the image of ζd.
Further, let η =
j∈S ζj be the image of ηr,d. Assuming that the order of
p (r−1)/d modulo r is d, we are to show that η has degree d over Fp (since we
have fr,d(η) =0andfr,d has degree d, sothatifη has degree d, thenfr,d
must be irreducible over Fp). We apply the Frobenius p-th-power map to η;
if this is done i times, we have ηpi. We are to show that the least positive k
with ηpk = η is k = d. Foreachkwe have
η pk
=
ζ pk
j
= ζ j ,
so that η pd
j∈S
j∈p k S
= η, sincep d ∈ S. Thus, the least positive k with η pk
= η is a
divisor of d. Our goal is to show that k = d, so we may assume that d>1.
Let χ be a Dirichlet character modulo r of order d; specifically, let
χ(a d ) = 1 for any nonzero residue a modulo r and let χ(p) = ζd. (Since
the order of p (r−1)/d modulo r is assumed to be d, we have completely defined
χ.) We consider the Gauss sum
r−1
τ(χ) = χ(j)ζ j r.
j=1
Note that the proof of Lemma 4.4.1 gives that |τ(χ)| 2 = r, see Exercise 4.21,
so that the image of τ(χ) inK is not zero. We reorganize the Gauss sum,
getting that
τ(χ) =
d−1
i=0 j∈piS
χ(j)ζ j d−1
r =
i=0
i
χ(p)
j∈p i S
Thus, τ(χ) is a “twisted” sum over the complex roots of fr,d(x). We take this
equation over to K, noting that
j∈piS ζj = ηpi.Butηpi1 = ηpi2 whenever
i1 ≡ i2 (mod k), so the image of τ(χ) inK is
d−1
i
ω
i=0
j∈p i S
ζ j d−1
= ω i η pi
=
i=0
k−1
m=0
η pm
d/k−1
l=0
ω m+kl =
k−1
m=0
ζ j r.
η pm
ω m
d/k−1
l=0
ω kl .