Prime Numbers

94 Chapter 2 NUMBER-THEORETICAL TOOLS
we saw above, we can “create” the finite field Fpk by coming up with an
irreducible polynomial f(x) inFp[x] ofdegreek. Wethussayalittleabout
how one might do this.
Every element a in Fpk has the property that apk = a, thatis,ais a root
of xpk − x. In fact this polynomial splits into linear factors over Fpk with no
repeated factors. We can use this idea to see that xpk − x is the product of
all monic irreducible polynomials in Fp[x] of degrees dividing k. Fromthiswe
get a formula for the number Nk(p) of monic irreducible polynomials in Fp[x]
of exact degree k: One begins with the identity
dNd(p) =p k ,
d|k
on which we can use Möbius inversion to get
Nk(p) = 1
k
p d µ(k/d). (2.5)
d|k
Here, µ is the Möbius function discussed in Section 1.4.1. It is easy to see that
the last sum is dominated by the term d = k, sothatNk(p) is approximately
p k /k. That is, about 1 out of every k monic polynomials of degree k in Fp[x]
is irreducible. Thus a random search for one of these should be successful in
O(k) trials. But how can we recognize an irreducible polynomial? An answer
is afforded by the following result.
Theorem 2.2.8. Suppose that f(x) is a polynomial in Fp[x] of positive
degree k. The following statements are equivalent:
(1) f(x) is irreducible;
(2) gcd(f(x),xpj − x) =1for each j =1, 2,...,⌊k/2⌋;
(3) xpk ≡ x (mod f(x)) and gcd(f(x),xpk/q − x) =1for each prime q|k.
This theorem, whose proof is left as Exercise 2.15, is then what is behind the
following two irreducibility tests.
Algorithm 2.2.9 (Irreducibility test 1). Given prime p and a polynomial
f(x) ∈ Fp[x] of degree k ≥ 2, this algorithm determines whether f(x) is
irreducible over Fp.
1. [Initialize]
g(x) =x;
2. [Testing loop]
for(1 ≤ i ≤⌊k/2⌋) {
g(x) =g(x) p mod f(x); // Powering by Algorithm 2.1.5.
d(x) =gcd(f(x),g(x) − x); // Polynomial gcd by Algorithm 2.2.1.
if(d(x) = 1) return NO;
}
return YES; // f is irreducible.