Prime Numbers

2.2 Polynomial arithmetic 91
Along with the polynomial gcd we shall need a polynomial inverse. In
keeping with the notion of integer inverse, we shall generate a solution to
s(x)f(x)+t(x)g(x) =d(x),
for given f,g, whered(x) =gcd(f(x),g(x)).
Algorithm 2.2.2 (Extended gcd for polynomials). Let F be a field. For
given polynomials f(x),g(x) in F [x], not both zero, with either deg f(x) ≥
deg g(x) or g(x) =0, this algorithm returns (s(x),t(x),d(x)) in F [x] such that
d =gcd(f,g) and sg+th = d. (For ease of notation we shall drop the x argument
in what follows.)
1. [Initialize]
(s, t, d, u, v, w) =(1, 0,f,0, 1,g);
2. [Extended Euclid loop]
while(w = 0) {
q =(d − (d mod w))/w; // q is the quotient of d ÷ w.
(s, t, d, u, v, w) =(u, v, w, s − qu, t − qv, d − qw);
}
3. [Make monic]
Set c as the leading coefficient of d;
(s, t, d) =(c −1 s, c −1 t, c −1 d);
return (s, t, d);
If d(x) = 1 and neither of f(x),g(x) is0,thens(x) istheinverseoff(x)
(mod g(x)) and t(x) istheinverseofg(x) (modf(x)). It is clear that if naive
polynomial remaindering is used, as described above, then the complexity of
the algorithm is O(D 2 ) field operations, where D is the larger of the degrees
of the input polynomials; see [Menezes et al. 1997].
2.2.2 Finite fields
Examples of infinite fields are the rational numbers Q, the real numbers
R, and the complex numbers C. In this book, however, we are primarily
concerned with finite fields. A common example: If p is prime, the field
Fp = Zp
consists of all residues 0, 1,...,p− 1 with arithmetic proceeding under the
usual modular rules.
Given a field F and a polynomial f(x) inF [x] of positive degree, we
may consider the quotient ring F [x]/(f(x)).TheelementsofF [x]/(f(x)) are
subsets of F [x] of the form {g(x) +f(x)h(x) : h(x) ∈ F [x]}; we denote
this subset by g(x) +(f(x)). It is a coset of the ideal (f(x)) with coset
representative g(x). (Actually, any polynomialinacosetcanstandinasa
representative for the coset, so that g(x) +(f(x)) = G(x) +(f(x)) if and
only if G(x) ∈ g(x)+(f(x)) if and only if G(x) − g(x) =f(x)h(x) forsome