Contents - Student subdomain for University of Bath

216 APPENDIX A. ALGEBRAIC BACKGROUND
so, despite the apparent asymmetry of the construction, c ≡ b (mod N) as
well.
In fact, we need not restrict ourselves to X being an integer: X, a and b
may as well be polynomials (but M and N are still integers).
Algorithm 40 (Chinese Remainder (Polynomial form))
Input: Polynomials a = ∑ n
i=0 a ix i , b = ∑ n
i=0 b ix i , and integers M and N
(with gcd(M, N) = 1).
Output: A polynomial = ∑ n
i=0 c ix i satisfying Theorem 35.
Compute λ, µ such that λM + µN = 1;
#The process is analogous to Lemma 1 (page 45)
for i := 0 to n do
c i :=a i + λM(b i − a i );
A.4 Chinese Remainder Theorem for Polynomials
The theory of the previous section has an obvious generalisation if we replace
Z by K[y] for a field K, and an equivalent application to sections 4.2–4.3.
Theorem 36 (Chinese Remainder Theorem for polynomials) Two congruences
X ≡ a (mod M) (A.6)
and
X ≡ b (mod N), (A.7)
where M and N are relatively prime polynomials in K[y], are precisely equivalent
to one congruence
X ≡ c (mod MN). (A.8)
By this we mean that, given any a, b, M and N, we can find such a c that
satisfaction of (A.6) and (A.7) is precisely equivalent to satisfying (A.8). The
converse direction, finding (A.6) and (A.7) given (A.8), is trivial: one takes a
to be c (mod M) and b to be d (mod N).
Algorithm 41 (Chinese Remainder for Polynomials)
Input: a, b, M and N ∈ K[y] (with gcd(M, N) = 1).
Output: c satisfying Theorem 36.
Compute λ, µ such that λM + µN = 1;
#As in Lemma 1 (page 45). Note µ isn’t needed in practice.
c:=a + λM(b − a);