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

102 7. More Algebra: Polynomials and Fields
f p = ad · (X p ) d + ···+ a1 · X p + a0 = f(X p ).
For the more general exponent p k we use induction. For k = 0, there is
nothingtoprove,sincef = f(X). The case k = 1 has just been treated. Now
assume k ≥ 2, and calculate in Zp[X], by using the induction hypothesis and
the case k =1:
f pk
=
�
f pk−1� p
=(f(X pk−1
)) p = f((X p ) pk−1
)=f(X pk
),
as desired. ⊓⊔
7.2 Division with Remainder and Divisibility for
Polynomials
Let R be a ring. Even if R is a field, the ring R[X] is never a field: for every
f =(a0,a1,...)wehaveX · f =(0,a0,a1,...) �= (1, 0, 0,...) = 1, and hence
X does not have a multiplicative inverse in R[X]. We will soon see how to
use polynomials to construct fields. However, for many polynomials we may
carry out a division with remainder, just as for integers.
Proposition 7.2.1 (Polynomial Division with Remainder). Let R be
aring,andleth ∈ R[X] be a monic polynomial (or a nonzero polynomial
whose leading coefficient is a unit in R). Then for each f ∈ R[X] there are
unique polynomials q, r ∈ R[X] with f = h · q + r and deg(r) < deg(h).
Proof. “Existence”: First assume that h is monic, and write h =(a0,...,ad)
with ad = 1. We prove the existence of the quotient-remainder representation
f = h · q + r by induction on d ′ =deg(f). If d ′