Dietzfelbinger M. Primality testing in polynomial time ... - tiera.ru

126 8. Deterministic Primality Testing in Polynomial Time
Lemma 8.5.5. If I(u, f) and I(u, g), thenI(u, fg).
Proof. We apply the hypothesis, exponentiation rules in Zp[X] andZp, and
Lemma 7.2.5(b) to see that
(fg) u = f u · g u ≡ f(X u ) · g(X u )=(fg)(X u ) (mod X r − 1). ⊓⊔
Lemmas 8.5.2 – 8.5.5 taken together imply that I(u, f) holdsforf an arbitrary
product of linear terms X +a, 1≤ a ≤ ℓ, i.e., f ∈ P ,andu an arbitrary
product of n’s and p’s. This set of exponents is central for the considerations
to follow, so we give it a name as well:
U = {n i p j | i, j ≥ 0}. (8.5.15)
The overall result of this section can now be stated as follows:
Lemma 8.5.6. For f ∈ P and u ∈ U we have (in Zp[X]):
f u ≡ f(X u ) (mod X r − 1). ⊓⊔
8.5.3 A Field F and a Large Subgroup G of F ∗
By Proposition 7.6.4 there is some monic irreducible polynomial h ∈ Zp[X]
of degree d =ordr(p) that divides X r−1 + ···+ X + 1 and hence X r − 1.
We keep this polynomial h fixed from here on, and turn our attention to the
structure F = Zp[X]/(h), which is a field of size p d by Theorem 7.4.5.
Some remarks are in place. As with p and Zp[X], Algorithm 8.2.1 does
not refer to h at all; the existence of h is only used for the analysis. Thus, it
is not necessary that operations in F can be carried out efficiently. Further,
we should stress that as yet there are no restrictions we can establish on the
degree d of h. Although we assume in (δ) thatordr(n) is not too small, it
might even be the case that d =ordr(p) = 1. (Example: For r = 101, p =
607 ≡ 1(modr), n = 16389 = 27 · 607 ≡ 27 (mod r), the value ordr(n) =
100 is as large as possible, but nonetheless ordr(p) = 1.) Only later we will
see that in the situation of the theorem it is not possible that deg(h) =1.
At the center of attention from here on is the subset of F that is obtained
by taking the elements of P modulo h; more precisely, let
� �
G = { f mod h | f ∈ P } = (X +a) βa �
�
�
mod h � βa ≥ 0for1≤a≤ ℓ .
1≤a≤ℓ
(see (8.5.11)). — We first note that G actually is a subset of F ∗ .
(8.5.16)
Lemma 8.5.7. The linear polynomials X + a, 1 ≤ a ≤ ℓ, are different in
Zp[X] and in F , and they satisfy X + a mod h �= 0.