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

7.5 Roots of Polynomials 111
X, has the inverse 02, which corresponds to 2X. Indeed, X · 2X =2X 2 =
2 · (−1) = 2 · 2 ≡ 1, calculated in Z3[X]/(X 2 + 1). Similarly, in Z5[X]/(X 2 +
4X + 1), the polynomial X has the inverse (4X +1),sinceX · (4X +1)=
4X 2 + X ≡ 4(X 2 +4X +1)+1≡ 1.
It should be mentioned that (in contrast to the more difficult task of
factoring polynomials into irreducible factors) a version of the Extended Euclidean
Algorithm 3.2.4 known from the integers yields an efficient procedure
to calculate inverses in the field F [X]/(h) −{0}. Since this algorithm is not
relevant for our task, we do not describe it here.
The question remains if there are sufficiently many irreducible polynomials
to make Theorem 7.4.5 a useful approach to obtaining finite fields. It is
a well-known fact, to be proved by methods not described in this book, that
for every field F and for every d ≥ 0 there is at least one irreducible (monic)
polynomial of degree d over F , which then leads to the construction of a field
that consists of d-tuples of elements of F .IfF is finite, the cardinality of
this field is |F | d . Starting with the fields Zp for p aprimenumber,weobtain
fields of cardinality p d for every prime number p and every exponent d. Inthe
other direction, it is not hard to show by basic methods from linear algebra
that if there is a finite field of cardinality q then q is the power of a prime
number p.
The field F [X]/(h) has the interesting property that it contains a root of
the polynomial h. This fact will be very important later.
Proposition 7.4.6. Let F and h be as in the previous theorem, and let F ′ =
F [X]/(h) be the corresponding field. Then the element ζ = X mod h ∈ F ′ is
arootofh, i.e., in F ′ we have h(ζ) =0.
(Note that if deg(h) ≥ 2thenζ = X ∈ F ′ − F .Ifdeg(h) =1,thenh = X + a
for some a ∈ F and ζ = −a.)
Proof. We use Proposition 7.3.3 for calculating modulo h in F [X], and Example
7.1.14(b) to obtain
h(ζ) =h(X mod h) modh = h(X) modh = h mod h =0. ⊓⊔
7.5 Roots of Polynomials
From calculus it is well known that if we consider nonzero polynomials over
R, then linear functions x ↦→ ax+b have at most one root, quadratic functions
x ↦→ ax 2 + bx + c have at most two, cubic polynomials have at most three,
and so on. We note here that this is a property that holds in all fields. The
basis for this observation is simply division with remainder, which shows that
if a is a root of f then f contains X − a as a factor.
Theorem 7.5.1. Let F be a field, and let f ∈ F [X] with f �= 0, i.e., d =
deg(f) ≥ 0. Then