Prime Numbers

144 Chapter 3 RECOGNIZING PRIMES AND COMPOSITES
R = Zn[x]/(x 2 − ax + b). To somewhat demystify this concept, we explicitly
list a complete set of coset representatives:
{i + jx : i, j are integers with 0 ≤ i, j ≤ n − 1}.
We add coset representatives as vectors (mod n), and we multiply them via
x 2 = ax − b. Thus,wehave
where
(i1 + j1x)+(i2 + j2x) =i3 + j3x
(i1 + j1x)(i2 + j2x) =i4 + j4x,
i3 = i1 + i2 (mod n), j3 = j1 + j2 (mod n),
i4 = i1i2 − bj1j2 (mod n), j4 = i1j2 + i2j1 + aj1j2 (mod n).
We now prove Theorem 3.6.3. Suppose p is an odd prime with
∆
p = −1.
Then ∆ is not a square in Zp, so that the polynomial x 2 − ax + b, which
has discriminant ∆, is irreducible over Zp. Thus,R = Zp[x]/(x 2 − ax + b) is
isomorphic to the finite field F p 2 with p 2 elements. The subfield Zp (= Fp) is
recognized as those coset representatives i + jx with j =0.
In F p 2 the function σ that takes an element to its p-th power (known
as the Frobenius automorphism) has the following pleasant properties, which
are easily derived from the binomial theorem and Fermat’s little theorem (see
(3.2)): σ(u + v) =σ(u)+σ(v), σ(uv) =σ(u)σ(v), and σ(u) =u if and only if
u is in the subfield Zp.
We have created the field F p 2 so as to provide roots for x 2 − ax + b, which
were lacking in Zp. Which coset representatives i + jx are the roots? They
are x itself, and a − x (= a +(p − 1)x). Since x and a − x are not in Zp and
σ must permute the roots of f(x) =x 2 − ax + b, wehave
in the case
∆
p = −1 :
x p ≡ a − x (mod (f(x),p)),
(a − x) p ≡ x (mod (f(x),p)).
(3.10)
Then xp+1 − (a − x) p+1 ≡ x(a − x) − (a − x)x ≡ 0(mod(f(x),p)), so that
(3.8) implies Up+1 ≡ 0(modp).
The proof of (3.9) in the case where p is a prime with
∆
p = 1 is easier.
In this case we have that x 2 − ax + b has two roots in Zp, so that the ring
R = Zp[x]/(x 2 −ax+b) is not a finite field. Rather, it is isomorphic to Zp×Zp,
and every element to the p-th power is itself. Thus,
in the case
∆
p =1 :
x p ≡ x (mod (f(x),p)),
(a − x) p ≡ a − x (mod (f(x),p)).
(3.11)
Note, too, that our assumption that gcd(p, b) = 1 implies that x and a − x are
invertible in R, sincex(a − x) ≡ b (mod f(x)). Hence x p−1 =(a − x) p−1 =1
in R. Thus, (3.8) implies Up−1 ≡ 0(modp). This concludes the proof of
Theorem 3.6.3.